Science, Irving K. Barber Faculty of (Okanagan)
Computer Science, Mathematics, Physics and Statistics, Department of (Okanagan)
DSpace
Tsopméné, Paul
The most important way to learn calculus is through problem-solving. While going through the solution to a problem, students are often faced with several issues. They may not see the connection between the concept taught in class and the solution. Others may not understand the solution because a step is missing or there are insufficient explanations. Or because they have weak algebra skills. The main goal of this exercise book is to address these issues to help students learn the material more efficiently and get better results. The book contains a wide variety of problems in integral calculus, linear algebra, and linear regression, with applications to differential equations, probability, and life sciences. Every problem has a very detailed solution, and the book is self-contained, as the summary for every concept is provided. [OER Description: This problem bank on integral calculus was developed to support undergradute lower division students. This open educational resource (OER) includes problem sets, detailed solutions, concept summaries, and alternative problem solutions.]
Integral Calculuswith Applications to Life SciencesExercise Book Practice Problems Concept Summaries Algebra Review Very Detailed Solution To Each Problemby Paul Tsopme´ne´Last Update: March 7, 2023Integral Calculus with Applications to Life Sciences Exercise Book by PaulTsopme´ne´ is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International (CC BY-NC-SA 4.0).iiAbout the Author Title: Assistant Professor of Teaching Since: September 2020 Institution: University of British Columbia Okanagan Department: CMPS (Computer Science, Mathematics, Physicsand Statistics) Email: paul.tsopmene@ubc.caSource File. To get the source file, please get in touch with theauthor at the following email address: paul.tsopmene@ubc.caiiiFor InstructorsDear instructors,Thank you for your interest in this open exercise book. I hope you find it avaluable addition to your existing teaching material.I would love to hear your feedback and adoption decision for your course.Your feedback will be highly appreciated and will be used for future im-provement of the book.Please take a few minutes to fill in the survey, which can be accessed in thefollowing way:Survey LinkThank you for your time and support on this open exercise book develop-ment.Best Regards,Paul Tsopme´ne´This page and survey is adapted from Introduction to Engineering Thermodynamics byDr. Claire Yan which is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.ivContentsPreface ixI Exercises 11 Integration 31.1 Antiderivatives and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . 31.2 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . 101.5 The Area Between Two Curves . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Total Change and Average Values . . . . . . . . . . . . . . . . . . . . . . . 152 Further Techniques of Integration and Numerical Integration 192.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3 Practicing Integration and Integration by Tables . . . . . . . . . . . . . . . 212.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Differential Equations 253.1 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Equilibria and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.4 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 Linear Algebra 354.1 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 354.2 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.3 Eigenvalues, Eigenvectors, and Matrix Models . . . . . . . . . . . . . . . . 41vvi CONTENTS4.4 Eigenvalues, Eigenvectors, and Matrix Models (continued) . . . . . . . . . 445 Infinite Series 475.1 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.2 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 Continuous Probability Distributions and Linear Regression 516.1 Probability Density Functions and Cumulative Distribution Functions . . . 516.2 Expected Value and Variance of Continuous Random Variables . . . . . . . 536.3 Uniform, Exponential, and Normal Distributions . . . . . . . . . . . . . . . 556.4 Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57II Solutions to Exercises 591 Integration 611.1 Antiderivatives and Indefinite Integrals . . . . . . . . . . . . . . . . . . . . 611.2 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 871.3 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1091.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . 1261.5 The Area Between Two Curves . . . . . . . . . . . . . . . . . . . . . . . . 1561.6 Total Change and Average Values . . . . . . . . . . . . . . . . . . . . . . . 1762 Techniques of Integration and Computational Methods 1872.1 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1872.2 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2082.3 Practicing Integration and Integration by Tables . . . . . . . . . . . . . . . 2262.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2432.5 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2613 Differential Equations 2673.1 Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . 2673.2 Equilibria and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2933.3 Integrating Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3073.4 Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3244 Linear Algebra 3374.1 Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 3374.2 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3534.3 Eigenvalues, Eigenvectors, and Matrix Models . . . . . . . . . . . . . . . . 365vii4.4 Eigenvalues, Eigenvectors, and Matrix Models (continued) . . . . . . . . . 3765 Infinite Series 3975.1 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3975.2 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4036 Continuous Probability Distributions and Linear Regression 4156.1 Probability Density Functions and Cumulative Distribution Functions . . . 4156.2 Expected Value and Variance of Continuous Random Variables . . . . . . . 4336.3 Uniform, Exponential, and Normal Distributions . . . . . . . . . . . . . . . 4536.4 Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468viiiPrefaceThe author wrote this exercise book when teaching MATH 103 (Integral Calculus withApplications to Life Sciences) at the University of British Columbia, Okanagan campus.The book is primarily for students taking a first-year integral calculus course (with lifesciences applications). Students taking a general Calculus II course can also use it to getdown basic content in integral calculus.The most important way to learn calculus is through problem-solving. While goingthrough the solution to a problem, students are often faced with several issues. They maynot see the connection between the concept taught in class and the solution. Others maynot understand the solution because a step is missing or there are not enough explanations.Or because they have weak algebra skills. I think that nearly everybody agrees that thelatter issue is the main reason many students struggle with calculus.My main goal in this book is to address these issues to help students learn the materialmore efficiently and get better results. To that end, I have included the following features. Concept Summaries Boxes: They allow students to review the material and makethe book self-contained. Each concept involved in a problem is summarized in a boxright before the solution. The relevant information needed to solve the problem isgiven. And the solution clearly shows how that information is used. Very Detailed Solutions: They provide an “easy-to-understand” approach. Eachproblem has a complete step-by-step solution with careful explanations for bettercomprehension. Alternate solutions (to some problems) are also given. “Help Texts”: They are located to the right of the solution sequence and helpstudents recognize which rule, property, or procedure is being applied. Algebra Review: They help students understand the solution, especially those withweaker algebra skills. Basic algebra rules and procedures (exponent rules, addingfractions, factoring, etc.) are listed next to the steps within solutions. Also, not onlyixx Prefacethe calculus steps are shown in the solution, but also the algebra steps. The idea isthat students with a poor algebra background can just jump in and learn as they go.The book is divided into two parts. The first part includes a wide variety of problemsas well as applications in life sciences. The idea is not to have as many problems as ina traditional textbook but a fair amount of problems that cover essential/basic skills inintegral calculus (and more). The second part includes the solutions. Every problem has acomplete step-by-step solution. Hyperlinks are added to ease the navigation of the book(they are mainly used to go back and forth between the two parts).To the Student: If you try to solve a problem and have no idea what to do, you canlook at the solution and try to understand it (if there is a box, you should first try toread and understand its content). Try to solve the subsequent few problems yourself beforechecking the solution. It’s not a big deal if you make a mistake; you will learn better fromthat mistake. If your final answer does not match the one in the solution, try to see if itis equivalent (since there are often several different ways of expressing the answer). Forexample, if the answer given in the book is√xxand you obtain 1√x, you are right becauserationalizing 1√xwill give you√xx.PartIExercises1Chapter1Integration1.1 Antiderivatives and Indefinite Integrals1. [Solution on page 61]Determine whether F (x) is an antiderivative of f(x).(a) f(x) = 9x8, F (x) = x9(b) f(x) = −5x−4, F (x) = x−5(c) f(x) = 2x√1+x2, F (x) =√1 + x2(d) f(x) = xex2, F (x) = ex2(e) f(x) = 1 + lnx, F (x) = x lnx2. [Solution on page 62]Determine whether each equality is correct.(a)∫3x2 dx = x3 + C(b)∫e−x dx = e−x + C(c)∫(1 + 2x)e2x dx = xe2x + C(d)∫ −x4 − 3x2 + 2x(x2 + 1)2dx =x3 + 1x2 + 1+ C3. [Solution on page 64]Evaluate the following indefinite integrals.(a)∫5 dx (b)∫pi2 dx34 Chapter 1. Integration(c)∫x8 dx(d)∫x14 dx(e)∫1x3dx(f)∫3√x dx(g)∫16√xdx(h)∫x3√xdx(i)∫4√x5 dx(j)∫ 3√x2x4dx4. [Solution on page 67]Find each indefinite integral.(a)∫4x dx(b)∫ √3x dx(c)∫12x5dx(d)∫(x2 + x3) dx(e)∫ (1x2− 8)dx(f)∫(−3x2 + 5x− 1) dx(g)∫(x2 +√x− pi4) dx(h)∫4x− 3√xdx(i)∫2−√2t+ 3√t2√tdt(j)∫2x(3− x−3) dx(k)∫(t+ 4)(2t+ 1) dt(l)∫(x+ 3)(4x− 5) dx5. [Solution on page 71]Evaluate each indefinite integral.(a)∫8ex dx(b)∫e−x dx(c)∫e7x dx(d)∫(1 + x− 6e−9x) dx(e)∫25e−8xdx(f)∫5x dx(g)∫3−7x dx(h)∫ √9 · 75x dx6. [Solution on page 73]Find each integral.1.1. Antiderivatives and Indefinite Integrals 5(a)∫2xdx(b)∫59xdx(c)∫x2(1− 4x−3) dx(d)∫x5 + 3x2 − 7x+ 1x2dx7. [Solution on page 75]Find the following integral. ∫(4x2 − 1)(3x+ 5)x3dx8. [Solution on page 75]Find the following.(a)∫(3 cosx+ 5 sinx) dx(b)∫(7 cos t− 8 sin t) dt(c)∫ (12− sec2 x) dx(d)∫ (e−2x − 9 csc2 x) dx(e)∫(2 secx tanx− 5 cscx cotx) dx(f)∫secx(secx− tanx) dx9. [Solution on page 79]Find the following.(a)∫ (2x+ cos2 x+ sin2 x)dx(b)∫11− sin2 x dx(c)∫ (4 + 4 tan2 t)dt(d)∫sin(2x)cosxdx10. [Solution on page 80]Find each integral(a)∫31 + x2dx(b)∫10x28x2 + 8dx(c)∫83√1− x2 dx(d)∫ √1− x21− x2 dx11. [Solution on page 84]State all the rules of integration you learned in this section.6 Chapter 1. Integration12. [Solution on page 86]Suppose a population of bacteria increases in size at a rate ofN ′(t) = 9e1.5t bacteria per hourAlso suppose that the initial population is 50 bacteria.(a) Find the population size N(t) at time t.(b) Find the population size five hours later.13. [Solution on page 86]The size of a bacteria colony increases at a rate ofP ′(t) =6500√tbacteria per day,where t ≥ 1. Suppose that the population is 29000 bacteria when t = 1.(a) Find the population size, P (t), at time t.(b) Find the number of days needed for the population to get to 211000.1.2 The Substitution Rule1. [Solution on page 87]Find each integral.(a)∫(x+ 1)8 dx(b)∫2x(x2 + 3)15 dx(c)∫(2x− 1)9 dx(d)∫x2(x3 − 7)6 dx(e)∫x√1− x2 dx(f)∫z2√z3 + 4 dz2. [Solution on page 92]Find each integral.(a)∫ −3x4(x5 + 1)7dx(b)∫ −4x+ 43√(x2 − 2x+ 1)2 dx(c)∫2xx2 − 16 dx(d)∫t5t6 + 3dt1.2. The Substitution Rule 7(e)∫18x2 − 6x2x3 − x2 dx(f)∫e1−x dx(g)∫x3ex4dx(h)∫x− 1e−x2+2xdx(i)∫ep1 + epdp(j)∫e2x − e−2xe2x + e−2xdx(k)∫ln(3x)xdx(l)∫(ln z)2zdz(m)∫1t ln tdt(n)∫5√x√xdx(o)∫x√x+ 7 dx(p)∫x√1 + 2xdx3. [Solution on page 98]Find each integral.(a)∫cos(4x) dx(b)∫sin(9x) dx(c)∫sec2(−x2)dx(d)∫−7x sin(5x2 + 1) dx(e)∫(x2 + 1) csc2(x3 + 3x) dx(f)∫e−2x sec(e−2x) tan(e−2x) dx4. [Solution on page 101]Find each integral.(a)∫cosx√sinx dx(b)∫sinx e− cosx dx(c)∫sin2 x cosx dx(d)∫sec2 x(1 + tan x)3dx(e)∫cot2 x csc2 x dx(f)∫tan(3x) dx(g)∫cot(θ4)dθ(h)∫cos2 x dx(i)∫cos3 x tanx dx(j)∫cosxsinx+ cscxdx5. [Solution on page 105]Find each integral.8 Chapter 1. Integration(a)∫11 + 9x2dx(b)∫1√16− 25x2 dx(c)∫e3x1 + e6xdx(d)∫tanxcosx+ secxdx6. [Solution on page 107]Suppose a population grows at a rate ofP ′(t) = 400te−t24 ,where P (t) is the population after t months. Suppose further that the initial popu-lation is 1500.(a) Find a formula for the population size after t months.(b) What is the size of the population after 5 months?7. [Solution on page 108]Let N(t) be the number of people with a disease at time t (where t is in days) in acity. Suppose N(t) is increasing at a rate ofN ′(t) =200tt2 + 5Suppose further that the number of sick people at time t = 0 is 180. Find the numberof sick people after 39 days.1.3 The Definite Integral1. [Solution on page 109]Find an approximation of the area between the function f(x) = x2 + 3, the x-axis,x = 0, and x = 1 by taking n = 4 subdivisions with left endpoints.2. [Solution on page 113]Find an approximation of the area between the function f(x) =1x, the x-axis, x = 1,and x = 3 by taking n = 4 subdivisions with right endpoints.3. [Solution on page 114]Approximate the area of the region under the graph of f(x) = ex + 1, above thex-axis, and between x = −2 and x = 3 by taking n = 5 rectangles with midpoints.1.3. The Definite Integral 94. [Solution on page 115]Find the Riemann sum forf(x) = x2 − 1, 0 ≤ x ≤ 2,with four intervals. Use the right endpoints.5. [Solution on page 116]In each case, approximate the definite integral by using four rectangles and the mid-points.(a)∫ 51√x− 1 dx(b)∫ 113(1− x2) dx6. [Solution on page 119]Express the following definite integrals as limits of Riemann sums.(a)∫ 135x4 dx (b)∫ −1−3x2e−x dx7. [Solution on page 120]Express each limit as a definite integral on the given interval.(a) limn→∞n∑i=1ln(xi)xi∆x, [1, 3](b) limn→∞n∑i=1(2xi − 3xi cosxi)∆x, [0, pi]8. [Solution on page 121]Express each limit as a definite integral.(a) limn→∞n∑i=1(in)3· 1n(b) limn→∞n∑i=1(2inln(1 +2in))· 2n9. [Solution on page 122]Find the exact value of each integral using formulas from geometry.10 Chapter 1. Integration(a)∫ 302x dx(b)∫ 40(4− x) dx(c)∫ 0−2√4− x2 dx10. [Solution on page 124]Let f be the function whose graph is shown below. Find∫ 51f(x)dx.xy0 1 51211. [Solution on page 125]Let f be the function whose graph is shown below. Find∫ 4−3 f(x)dx.xyy = f(x)−3 −1 0 4451.4 The Fundamental Theorem of Calculus1. [Solution on page 126]Find dydx.1.4. The Fundamental Theorem of Calculus 11(a) y =∫ x0t3 dt(b) y =∫ x15 sin t dt(c) y =∫ x−211 + t4dt(d) y =∫ x20(1 + e3t)dt(e) y =∫ √x2x+17 sec t dt(f) y =∫ x+x41−x2t5 dt2. [Solution on page 129]Evaluate each integral.(a)∫ 813 dx(b)∫ 1−35 dx(c)∫ 10x3 dx(d)∫ 941√xdx(e)∫ 413x dx(f)∫ 21(−x2 + 3x− 1) dx(g)∫ 21(1x2− 4x3)dx(h)∫ 3−2(x2 − 3) dx(i)∫ 414 + 6x√xdx(j)∫ 812 + t3√t2dt(k)∫ 20(2x− 3)(4x2 + 1) dx(l)∫ 40(4− x)√x dx(m)∫ 10e−2x dx(n)∫ ln 205e3x dx(o)∫ 21(x2 − 3x) dx(p)∫ e1x2 − xx2dx3. [Solution on page 139]Evaluate each integral.(a)∫ pi60cosx dx (b)∫ pi403 sec2 x dx4. [Solution on page 139]Let f be a function such that∫ 53f(x) dx = −10 and∫ 3−2f(x) dx = 4.Find the following:12 Chapter 1. Integration(a)∫ 22f(x) dx(b)∫ 5−2f(x) dx(c)∫ 35f(x) dx(d)∫ −2512f(x) dx5. [Solution on page 141]Let f be a function such that∫ 81f(x) dx = 15 and∫ 86f(x)2dx = 3.(a) Find∫ 617f(x) dx.(b) Find∫ 16(4− f(x)5)dx6. [Solution on page 142]Show that1 ≤∫ 10√1 + x2 dx ≤√27. [Solution on page 143]Evaluate each integral.(a)∫ 202x(x2 − 1)3 dx(b)∫ 10(4t− 1)50 dt(c)∫ 103√26x+ 1 dx(d)∫ 10xex2dx(e)∫ 12131x2e1x dx(f)∫ 21x2(x3 + 1)2dx(g)∫ 801√1 + 2xdx(h)∫ 20t2√1 + t3dt(i)∫ 21x+ 1√x2 + 2x− 1 dx(j)∫ −1−2xx2 + 1dx(k)∫ ln 40ex5− ex dx(l)∫ 40x√1 + 2xdx8. [Solution on page 152]Find each integral.(a)∫ pi60sin(2x− pi6)dx (b)∫ pi40√1 + tan xcos2 xdx1.5. The Area Between Two Curves 13(c)∫ pi20sinx cos(cosx) dx(d)∫ √320sin−1 x√1− x2 dx(e)∫ 1201 + x√1− x2 dx(f)∫ 10x31 + x8dx1.5 The Area Between Two Curves1. [Solution on page 156]Let f be the function whose graph is shown below. Find∫ 4−2f(x) dx using formulasfrom geometry.xy0 1 4−24−22. [Solution on page 159]Let f be the function whose graph is shown below. Find∫ 5−4f(x) dx using formulasfrom geometry.14 Chapter 1. Integrationxy0 1 2 5−2−44−23. [Solution on page 159]Find the area of the region that lies under the given curve y = f(x) over the indicatedinterval a ≤ x ≤ b.(a) y = x2ex3, 0 ≤ x ≤ 1(b) y =45− 2x, −2 ≤ x ≤ 1(c) y = sec2 x− cosx, 0 ≤ x ≤ pi4(d) y = cot(2x),pi12≤ x ≤ pi44. [Solution on page 163]Find the area of the region bounded above by y = 2x + 5 and below by y = x3 onthe interval [0, 2].5. [Solution on page 164]Find the area of the region bounded by the curves y = x2 and y = x + 2. Include asketch of the relevant region as part of your solution.6. [Solution on page 165]In each case, find the area of the region bounded by the given curves.(a) y = x2 − 8 and y = −x2 + 10(b) y = 4x− x2 and y = x2(c) y =x32and y = x21.6. Total Change and Average Values 157. [Solution on page 169]Find the area between the curve y = 2x2 − 7x+ 3 and the x-axis.8. [Solution on page 170]In each case, find the area of the region enclosed by the given curves.(a) x = 0, x = 4, y = x, y = x2 − 2x(b) x = −2, x = 1, y = ex, y = e−x(c) x = −12, x = 2, y = 1x+1, y = 13(d) y = x4 − ln(x+ 3), y = x3 − ln(x+ 3)9. [Solution on page 174]Find the area between the given curves.(a) y = e−x, y = −x, x = −2, x = 1(b) y = cosx, y = x, x = −pi, x = −pi410. [Solution on page 176]Consider the following leaf bounded above by f(x) = x−x2 and below by the functiong(x) = x2 − x.xy0 1Find the area of this leaf.1.6 Total Change and Average Values1. [Solution on page 176]In each case, find the total change in F (x) as x goes from a to b.(a) F ′(x) = x3, a = 1, b = 3(b) F ′(x) = 4xe3x2−5, a = 0, b = 216 Chapter 1. Integration2. [Solution on page 178]Let N ′(t) be the rate of growth of a population of bacteria at time t. Explain what∫ 94N ′(t) dt represents.3. [Solution on page 178]Let dhdtbe the rate of change of the height of an animal at time t.∫ 72dhdtdtrepresent?4. [Solution on page 178]A population grows at rateN ′(t) =15(3t+ 1)2Find the total number of individuals added to the population between times t = 3and t = 5. [Round your answer to the nearest integer.]5. [Solution on page 179]Let dydtbe the rate of rainfall in mm/h.(a) Explain what∫ 240dydtdt means.(b) If dydt= 4(1− cos ( pi12t)), calculate the cumulative rainfall over the interval[0, 24].6. [Solution on page 180]A particle moves along a straight line at a velocityv(t) = t2 − 7t+ 10 m/s(a) Find the net displacement (or the total change in the position) of the particlebetween times t = 3 and t = 8. [Recall that if s(t) is the position at time t ofan object traveling along a straight line, the velocity at time t is the derivativeof s(t); that is, v(t) = s′(t).](b) Find the total distance traveled by the particle between times t = 3s and t = 8s.7. [Solution on page 182]Find the average value of each function on the given interval.1.6. Total Change and Average Values 17(a) f(x) =1− x√x[1, 9](b) f(x) = x√x2 + 16 [0, 3](c) f(t) = csc2(t4) [2pi3, 2pi]8. [Solution on page 185]Suppose the snowfall in a city is modeled byf(t) = 2.3 + 2.3 cos(2pit),where t = 0 corresponds to January 1 and t = 1 corresponds to December 31. Andf(t) is in cm/month. Find the average snowfall in one year.9. [Solution on page 185]Suppose the temperature in a city at time t (in hours) is approximated by the functionT (t) = 70 + sin( pi12t)Find the average temperature in one day (0 ≤ t ≤ 24).18 Chapter 1. IntegrationChapter2Further Techniques of Integrationand Numerical Integration2.1 Integration by Parts1. [Solution on page 187]Evaluate the following integrals.(a)∫2xex dx(b)∫xe4x dx(c)∫3xe−5x dx(d)∫−2xe−3x+1 dx(e)∫(−7x+ 9)e−2x+3 dx(f)∫x ln(2x) dx(g)∫x2 lnx dx(h)∫lnx dx(i)∫8x ln(7x) dx(j)∫lnxx3dx(k)∫3x+ 2exdx(l)∫3x sinx dx(m)∫x cos(2x) dx(n)∫(4x−2) sin(−5x+7) dx2. [Solution on page 197]Evaluate the following integrals.(a)∫x2ex dx(b)∫3x2e−3x dx(c)∫x2 cosx dx3. [Solution on page 199]Evaluate the following integrals.1920 Chapter 2. Further Techniques of Integration and Numerical Integration(a)∫ 1−1(x+ 1)e2x dx(b)∫ 21(−5x2 + 4x+ 1)e−2x dx(c)∫ 41√x lnx dx(d)∫ 120tan−1(2x) dx(e)∫sin−1 x dx4. [Solution on page 204]Find the following integrals.(a)∫ex cosx dx (b)∫e2x sin(−3x) dx5. [Solution on page 206]Find the area of the region under the curve y = ln (√x) above the x-axis from x = 4to x = 9.6. [Solution on page 207]Suppose the rate of change of of a population is given byN ′(t) = (t+ 2) ln(t+ 2)where t is in days. Find the total population after 11 days. [Round your answer tothe nearest integer.]2.2 Partial Fractions1. [Solution on page 211]Find the partial fraction decomposition of each function.(a) f(x) =2x+ 6x2 − 3x(b) f(x) =−2x+ 5x(x− 1)22. [Solution on page 214]Evaluate each integral.(a)∫2x+ 6x2 − 3x dx (b)∫6xx2 − x− 2 dx2.3. Practicing Integration and Integration by Tables 21(c)∫x2 + 2x+ 2x3 − 3x2 + 2x dx(d)∫ 1012x2 + 5x− 12 dx(e)∫ −2x+ 5x(x− 1)2 dx(f)∫2x2 − 3x− 14x3 + 4x2 + xdx3. [Solution on page 221]Find the integral∫2x3 − 3x2 + 3x+ 2x2 − x− 2 dx4. [Optional] [Solution on page 222]Find the following integrals.(a)∫5x3 + 5xdx(b)∫2x3 + x2 + 4x+ 3(x2 + 3)2dx2.3 Practicing Integration and Integration by Tables1. [Solution on page 226]The following integrals are mixed. Use the appropriate integration technique(s) tofind each integral.(a)∫x2ex3dx(b)∫x2e3x dx(c)∫ √lnxxdx(d)∫x2√x+ 1 dx(e)∫(−3x+ 2) ln(4x) dx(f)∫ e1lnx√xdx(g)∫ 10x33x4 + 7dx(h)∫tan (√x)√xdx2. [Solution on page 231]The following integrals are mixed. Use the appropriate integration technique(s) tofind each integral.(a)∫1x2 − 4 dx(b)∫xx2 − 4 dx(c)∫1x2 + 4dx(d)∫x4 sin2(x5) dx(e)∫e√x dx(f)∫x7 cos(x4 + 1)dx22 Chapter 2. Further Techniques of Integration and Numerical Integration(g)∫sinxcos2 x+ cosxdx (h)∫13√x+ 10 + xdx3. [Solution on page 240]Use the table of integrals to find each integral.(a)∫ √x2 + 9 dx(b)∫10√2x2 + 50dx(c)∫ −4x√64− x2 dx(d)∫5√3x2 − 4 dx2.4 Improper Integrals1. [Solution on page 243]Determine whether each improper integral is convergent or divergent. Find the valueof each that converges.(a)∫ ∞11x4dx(b)∫ ∞45√xdx(c)∫ ∞15xdx(d)∫ ∞212x− 3 dx(e)∫ ∞31x(lnx)4dx(f)∫ ∞0cos(2x) dx2. [Solution on page 249]Determine whether each improper integral is convergent. Find the value of each thatconverges.(a)∫ 0−∞3x dx(b)∫ 0−∞3xe2x dx(c)∫ 1−∞xx2 + 7dx(d)∫ 0−∞−2x+ 53x2 − 13x+ 4 dx(e)∫ ∞−∞x3e−x4dx(f)∫ ∞−∞7e−8x dx(g)∫ ∞−∞51 + 16x2dx(h)∫ ∞−∞sin(pix) dx3. [Solution on page 259]If the rate of change of a quantity y is 1500e−0.04t, find the total change in y over thetime interval [0,∞).2.5. Numerical Integration 234. [Solution on page 259]The rate of change in the concentration of a reactant can modeled byf ′(t) = 3t2e−0.5t,where t is time in hours. Find the total change in the concentration over all the time;that is, over the time interval [0,∞).2.5 Numerical Integration1. [Solution on page 261]Use the trapezoidal rule with n = 4 to approximate the integral∫ 10√3x2 + 5 dx.2. [Solution on page 262]Use Simpson’s rule with n = 4 to approximate the integral∫ 10√3x2 + 5 dx.3. [Solution on page 264]Approximate the area under the curve y =√9− x2 and above the x-axis by usingn = 6 with the following methods:(a) The trapezoidal rule.(b) Simpson’s rule.(c) Compare the results with the area found by the formula for the area of circle.Which of the two approximation was more accurate?4. [Solution on page 266]A certain plant grows at ratef(t) =4t+ 4+ e−t2feet/monthUsing n = 8 with Simpson’s rule, find an approximation of the plant’s total growthas t changes from t = 1 to t = 5.Note: It is impossible to find the exact value of∫ 51e−t2dt. That is, it is not possibleto find an antiderivative of e−t2in terms of the functions we are familiar with. Thereare many other integrals that are impossible to evaluate in terms of the functions we24 Chapter 2. Further Techniques of Integration and Numerical Integrationknow. Examples of such integrals include the following:∫ √1 + x3 dx∫exxdx∫sin(x2) dx∫1lnxdx∫sinxxdx∫cos(ex) dxChapter3Differential Equations3.1 Separable Differential Equations1. [Solution on page 267]Solve each differential equation.(a)dydx= x(b)dydx= −3x+ 15x2 + 1(c)dydx=2x+ x14 − sec2 x(d)dydx− 7x2 = x3√x(e) e−3x − 2dydx=x√x2 + 1(f)1tdydt− sin(4t) = 02. [Solution on page 270]Solve each differential equation.(a)dydx=x4y2(b)dydx=x5 + 3y(c)dydx=x2 + 1y√x(d) y4dydx= x3 − x2(e)dydx= 5x4y(f)dydx= 7x2y − 5xy(g) (xy3 − xy2)dydx= x3y(h) exdydx= xy2e6x(i)dydt= 4t√1− y2(j)dydt= ty2 + t3. [Solution on page 276]Solve each initial value problem.2526 Chapter 3. Differential Equations(a)dydx= 6x5 − 18x2; y(0) = 2(b) x− x2 dydx=√x; y(1) = 7(c)dydx=5exey; y(0) = 0(d) x4dydx− yx 52 = 0; y(1) = e−2(e)dydx=3x+ 5y − 2 ; y(4) = 12(f)dydt= y2 − 1; y(0) = 24. [Solution on page 281]Let V (t) be the volume of a cell at time t. Suppose that the rate of change of V (t)is given bydVdt= 3 + cos (2pi(t+ 1))Suppose further that V (0) = 8. Find V (t).5. [Exponential Population Growth] [Solution on page 282]Suppose that the rate of change of a population is three times the size of the popu-lation. (This amounts to saying that the per capita growth rate is 3.) If the initialpopulation is 10 and if time is in years, find the size of the population after 4 years.6. [Solution on page 284][Restricted Growth: The von Bertalanffy Growth Equation] The length ofa fish, L(t) (in cm), at age t (in years), can be modeled by the following differentialequation due to Ludwig von Bertalanffy (1901-1972):dLdt= k(L∞ − L)for some positive constants L∞ and k. Suppose that L(0) = 3 cm.(a) Find L(t) in terms of k and L∞.(b) What is the asymptotic length of the fish? In other words, what is the length ofthe fish in the long run?7. [Restricted Growth] [Solution on page 285]Let y = y(t) be the size of a certain population at time t. Let M be the maximumpopulation size. Suppose that the rate of growth of that population is proportionalto how close the population is to that maximum, that is, to the difference betweenM and y.(a) Write a differential equation to express the rate of population growth.3.1. Separable Differential Equations 27(b) Find the general solution to the equation in part (a).(c) Suppose that M = 5000 and the initial population is y0 = 1500. Find a functiondescribing the population at time t if the growth rate constant is 25%.(d) With the same assumptions as in part (c), what will the population be in 10years?8. [Tumor Growth: Gompertz Differential Equation] [Solution on page 285]The size of a tumor, Y (t), at time t, can be modeled by the Gompertz differentialequationdYdt= a (ln b− lnY )Ywhere a and b are both positive constants. Suppose that the initial tumor size is 2.(a) Find Y (t) in terns of a and b.(b) Find the asymptotic tumor size; that is, find limt→∞Y (t).9. [Allometry] [Solution on page 288]The sizes, X and Y , of two parts of a living organism’s body change according to thefollowing differential equation:1YdYdt= a1XdXdtwhere a is a constant. (This means that the per unit growth rate of Y is a timeslarger than that of X.) Express Y as a function of X.10. [Circadian Cycle] [Solution on page 288]Let N(t) be the size of a population of cells at time t (in hours). Suppose that the percapita growth rate of that population is 3(1 + sin(pit12)). This means that N changesaccording to the differential equation1NdNdt= 3(1 + sin(pit12))Suppose the initial population is 7. Find the number of cells after 24 hours.11. [Logistic Equation for Population Growth] [Solution on page 289]Let y(t) be the size of a population at time t. Suppose that y changes according tothe differential equationdydt= 0.28(1− y195)ySuppose further that the initial population is y(0) = y0.28 Chapter 3. Differential Equations(a) Find y(t) in terms of y0. If y0 6= 0, does the size of the population in the longrun depend on y0? Why or why not?(b) Sketch the graph of y(t) in each case:i. y0 = 40 ii. y0 = 125 iii. y0 = 30012. [Solution on page 292]Consider the logistic differential equationdNdt= rN(1− NK)with N(0) = N0where r > 0, K > 0, and N0 ≥ 0 are all constants. One can prove that the solutionof this initial value problem isN(t) =K1 +(KN0− 1)e−rtif N0 6= 0N(t) = 0 if N0 = 0N(t) = K if N0 = K(a) If the carrying capacity is 180, N(0) = 15, N(2) = 100, find the limiting growthrate for small N (that is, find the constant r).(b) Estimate r if N(0) = 20, N(4) = 38, N(175) ≈ 45, and N(300) ≈ 45.3.2 Equilibria and Stability1. [Solution on page 293]Find all equilibria of each differential equation.(a)dydt= 7y(b)dydt= y2 − 9(c)dydt= y3 − 2y2 − 8y(d)dNdt=N − 5N + 2, N ≥ 0(e)dNdt= 2(N23 − 4)e−3N(f)dxdt= x ln( x12), x > 02. [Solution on page 295]Find all equilibria and determine whether they are stable or unstable.3.2. Equilibria and Stability 29(a)dydt= y − 3(b)dydt= y2 − 7y + 10(c)dydt= 4y − y3(d)dxdt=32 + x− 1, x > 03. [Solution on page 301]Let g be the function whose graph is shown below. Find all equilibria and determinewhether they are stable or unstable.−2 2 yg(y)4. [Solution on page 302]Shown is the graph of a function g. Find all equilibria and determine whether theyare stable or unstable.2 4 yg(y)5. [Solution on page 303]Find the equilibrium points of each differential equation and determine whether theyare stable, unstable, or semistable.30 Chapter 3. Differential Equations(a)dydt= −y2 + 4y − 4(b)dydt= (y2 + 2y + 1)(y − 3)(c)dydt= g(y), where g is the function whose graph is shown below.−2 4 yg(y)6. [Logistic Equation for Population Growth] [Solution on page 305]Let y(t) be the size of a population at time t. Suppose that y changes according tothe differential equationdydt= 0.28(1− y195)ySuppose further that the initial population is y(0) = y0.(a) If y0 = 0, then y(t) = 0 for all t. Explain why without solving the differentialequation.(b) If y0 = K = 195, then y(t) = 195 for all t. Explain why without solving thedifferential equation.(c) If y0 > 0, find the asymptotic behavior of y(t) (that is, find limt→∞N(t)) withoutsolving the differential equation.7. [The Allee Effect] [Solution on page 306]Consider a population whose growth is affected by the Allee effect. Then the size Nof that population changes according to the differential equationdNdt= rN(N − a)(1− NK)where r, K are positive numbers, and 0 < a < K.3.3. Integrating Factors 31(a) Find all equilibria of this differential equation and determine their stability.Construct the vector field plot (or phase plot).(b) Let N0 be the initial population. Without solving the differential equation, findthe asymptotic behavior of N(t) (that is, find limt→∞N(t)) in each case.i. 0 < N0 < a ii. a < N0 < K iii. N0 > K3.3 Integrating Factors1. [Solution on page 307]Find the general solution of each differential equation.(a)dydx+ 7y = 19(b)dydx− 2y = 5(c)dydx− 3y = x(d) 2dydx+ 4xy = 16x(e)dydx+8xy = x3(f) xdydx+ 4y = x5 − x(g) 5dydx+ 10xy + x = 0(h) 3xy + 4x3 = xdydx(i) 2tdydt− y = 2t ln t, t > 02. [Solution on page 316]Find the particular solution of each initial value problem.(a)dydx+ 5y = 12ex, y(0) = 11(b) (x2 + x)dydx+ 2xy = x+ 1, y(1) = 0(c)dydt+ (cos t)y − 3 cos t = 0, y (pi) = −13. [Solution on page 319]Let y(t) be the amount of insulin present in the bloodstream of a patient at time t.Suppose insulin is delivered intravenously into the bloodstream at a rate of 8 unitsper hour, and removed from the bloodstream at a rate proportional to y.(a) Write a differential equation that models the amount of insulin present in thebloodstream.32 Chapter 3. Differential Equations(b) Solve this equation for y.(c) Find the amount of insulin in the long run (that is, find the limit of y(t) as tapproaches infinity). Answer this question in two different ways: (1) By takingthe limit of the function obtained in part (b) as t→∞. (2) By using the phaseplot of the differential equation.4. [Solution on page 322]A patient receives a drug at a rate of A(t) units/h. At the same time, the drug isabsorbed by the patient’s body at a rate proportional to the current concentration,y(t), of the drug in the bloodstream (suppose the constant of proportionality is k).(a) Write a differential equation that models this process.(b) If A(t) = sin t and k = 1, solve the differential equation found in part (a).(c) Assuming y(0) = 4, use the solution from part (b) to find the concentration ofthe drug in the bloodstream at time t = 2.3.4 Euler’s Method1. [Solution on page 324]Consider the differential equationdydx= 6x+ 7y, y(0) = 5Use Euler’s method with step size h = 0.1 to approximate the value of y(0.3).2. [Solution on page 327]Use Euler’s method with step size h to estimate the value of y(a) for each differentialequation.(a) dydx= 4xy, y(0) = 2, a = 0.3, h = 0.1(b) dydx= x2 +√y, y(0) = 1, a = 0.5, h = 0.1(c) dydx= xexy, y(1) = 5, a = 1.4, h = 0.1(d) 2 dydx− 3y = 18 ln x, y(3) = 4, a = 3.6, h = 0.23. [Solution on page 329]Consider the differential equationdydx+ 3x2y = 4x, y(0) = 5.83.4. Euler’s Method 33Use Euler’s method with step size 0.1 to approximate the solution y(x) for x in theinterval [0, 1].4. [Solution on page 330]Consider the differential equationdydx= x2y2, y(2) = 1(a) Use Euler’s method with step size 0.02 to approximate the value of y(2.1).(b) Solve the differential equation to find the exact solution y(x). Next, find y(2.1).(c) Compare the results from parts (a) and (b).5. [Solution on page 332]Consider the differential equationdydx= 2x, y(0) = 1(a) Use Euler’s method with step size 0.2 to approximate the solution y(x) for x inthe interval [0, 1].(b) Solve the differential equation to find the exact solution y = f(x). Next, sketchthe graph of y = f(x), 0 ≤ x ≤ 1, and plot the Euler approximations found inpart (a) on the same axes. Join two successive approximations by a line segment.(c) Find the error made in part (a) in using Euler’s method to approximate the truevalue of f(0.6). (Recall that the error is the difference between the exact valueand the approximate value.)6. [Solution on page 333]Do Exercise 5 with the following differential equation.−x+ dydx= −y, y(0) = 17. [Solution on page 335]Let y = y(t) be the size of a population at time t. Suppose y changes according tothe differential equationdydt= 0.09y − 0.5y 12If the initial population is 70, approximate the population size at time t = 4 usingEuler’s method with step size h = 1.34 Chapter 3. Differential EquationsChapter4Linear Algebra4.1 Systems of Linear Equations1. [Solution on page 337]Determine whether each equation is linear. If an equation is linear, find the numberof unknowns, the coefficients, and the constant term.(a) x+ y = 0(b) −2x+ 4y − 5z = 12(c)√x− y = 19(d) −4x− 20y + cos(z) = −1(e) 5x+ y√2 = −47(f) −x+ 9xy − z = −3(g) 1x+y− z = 242. [Solution on page 338]For each equation, three points are given. Determine which one is a solution to theequation.(a) 2x+ 3y = 14 Points: (3, 3), (1, 4), and (5, 1).(b) −4x− 5y + 21z = 7 Points: (−2, 3, 0), (−2,−3, 0), and (2,−3, 0).3. [Solution on page 339]Determine whether each system is a system of linear equations. If a system is not alinear system, explain why.(a)x − 3y = 04x + 8y = 9(b)5x − 5y + 12z = ln 7x + y = −5z(c)x + 3y = 2z4x + y − z = 0−2x + ez = −2(d)34x − √5y = −441x + ln(y) = 03536 Chapter 4. Linear Algebra4. [Solution on page 340]For each linear system, determine whether the given point is a solution to the system.(a)x − 2y = 33x + y = 2(1,−1)(b)−3x + 4y = 185x − 6y = −27 (−2, 3)5. [Solution on page 341]For each system, sketch the graph of both equations on the same axes and determinewhether the system has a unique solution, or no solution, or infinitely many solutions.(a)x + y = 1−x + y = −1(b)−x + 2y = 32x − 4y = −5(c)2x + 3y = 5−3x + y = 9(d)x − y = 2−2x + 2y = −46. [Solution on page 343]Find the augmented matrix of each system.(a)3x + 4y = 56x − 7y = 8 (b)−5x + 7y = −166x = 117. [Solution on page 344]Write the linear system corresponding to each augmented matrix.(a)[1 −1 3−2 4 −8](b)[1 −2 40 −3 −9]8. [Solution on page 344]Use the Gauss-Jordan method to solve each system.(a)x + 2y = 52x − y = −10 (b)x + y = 03x − 4y = 74.2. Matrix Operations 37(c)−2x − 10y = −203x + 4y = −3 (d)2x − 3y = 75x + 7y = 39. [Solution on page 348]Use the Gauss-Jordan method to solve each system.(a)x − 3y = −4−2x + 6y = 7(b)−6x + 9y = 158x − 12y = 9(c)−5x + 10y = −15x − 2y = 3(d)7x + 14y = 35−4x − 8y = −20(e)4x + 24y = 0−3x − 18y = 010. [Solution on page 350]Consider the following system−x + 5y = a7x − 30y = bUse the Gauss-Jordan method to solve the system for the following values of a and b.(a) a = 1, b = −2 (b) a = −4, b = 18 (c) a = 12, b = −311. [Solution on page 351]Consider the following system.x − 3y = b−2x + ay = 0Find conditions on a and b such that the system has no solution, one solution, orinfinitely many solutions.4.2 Matrix Operations1. [Solution on page 353]38 Chapter 4. Linear AlgebraLetA =[1 7−3 8], B =[−2 65 −14], C = 9 −7−3 011 −22 , and D = 17 −9−4 10 19 .Compute each of the following expressions if they are defined. If an expression isundefined, explain why.(a) A+B(b) C +D(c) A+ C(d) C +D +B(e) 2C(f) 3C − 2D2. [Solution on page 354]Consider the matrices A and B from Exercise 1. Compute −5A + 2B − 3I, whereI =[1 00 1]is the 2× 2 identity matrix.3. [Solution on page 355]Find a 2× 2 matrix X such thatX +[5 −31 6]= −6[−2 81 0]4. [Solution on page 355]LetA =[1 23 4], B =[−3 54 −7], C = 2 05 −813 −9 , D = −2 1 34 8 910 −4 5 ,E =[−2 6 93 −7 7], and F = 0 9 4−2 −2 −51 −10 −1Compute each of the following expressions if they are defined. If an expression isundefined, explain why.4.2. Matrix Operations 39(a) AB(b) BA(c) BC(d) CB(e) CD(f) DC(g) CE(h) EC(i) DF(j) A2(k) B3(l) C25. [Solution on page 358]Let A and B be matrices such that A is of size 5× 7 and the number of columns ofAB is 11. What is the size of B?6. [Solution on page 358]Let A and B be matrices such that the product BA is of size 17× 32.(a) How many rows does B have?(b) How many columns does A have?7. [Solution on page 359]In ordinary algebra with real numbers, the zero product property says that if ab = 0,then a = 0 or b = 0. Does this property hold for matrices?8. [Solution on page 359]A vector is a matrix that contains one column. A vector is in the plane R2 if it is ofsize 2 × 1. If A is a 2 × 2 matrix and u is a vector in R2, then the product Au is a2× 1 matrix. So, the product of a matrix and a vector is still a vector. LetA =[1 23 4], u =[−45], and v =[3−1](a) Compute the vectors u+ v,−2v, and 6u+ 7v.(b) Compute Au and Av.(c) Compute A(3u− 4v) and A(−4u+ 7v).(d) The graph of the vector u is shown in the following figure.40 Chapter 4. Linear Algebra−45uxyGraph the vectors v,Au, and Av on the same axes.9. [Solution on page 361]Find the inverse of each matrix. If the inverse does not exit, explain why.(a) A =[1 22 6](b) A =[−6 712 −14] (c) A =[−2 3−4 8](d) A =[15 −18−10 12]10. [Solution on page 362]Let A =[k −5−4 k + 1]. Find the values of k that make A invertible.11. [Solution on page 362]Write each system as a matrix equation.(a)7x − 6y = 13−8x + 9y = 5 (b)x + y − z = 02x + z = 13y − 4z = 712. [Solution on page 363]Solve each linear system by finding and using the inverse of the coefficient matrix.(a)−3x − 6y = 10−x + 3y = 10 (b)4x − 5y = 117x + 6y = −34.3. Eigenvalues, Eigenvectors, and Matrix Models 414.3 Eigenvalues, Eigenvectors, and Matrix Models1. [Solution on page 365]In each case, determine if u and v are eigenvectors of A. If so, find the correspondingeigenvalues.(a) A =[3 −3−2 2], u =[−32], and v =[1−1].(b) A = 1 0 −11 −3 04 −13 1 , u = 302, and v = 113.2. [Solution on page 366]Let A be a 2 × 2 matrix. Let u, v, Au, and Av be vectors as shown in the followingfigure.−4 66uvAuAvxyDetermine whether u and v are eigenvectors of A. If a vector is an eigenvector, findthe corresponding eigenvalue.3. [Solution on page 367]Let A be a 2 × 2 matrix. Let u, v, w,Au,Av, and Aw be vectors as shown in thefollowing figure.42 Chapter 4. Linear Algebra8−8uvwAuAvAwxyDetermine whether u, v, and w are eigenvectors of A. If a vector is an eigenvector,find the corresponding eigenvalue.4. [Solution on page 367]Determine if the statement is True or False, and justify your answer.(a) If A is a square matrix, a nonzero vector x is an eigenvector of A if Ax is amultiple of x.(b) An eigenvector can be the zero vector.(c) The vector u =[−23]is an eigenvector of A =[7 4−3 −1].(d) The vector u =[−12]is an eigenvector of A =[3 −4−4 5].5. [Solution on page 368]Find the eigenvalues of each matrix.(a) A =[5 −3−10 6](b) A =[8 −116 −9](c) A =[13 4−16 −3](d) A =[22 7−1 14](e) A =[0.7 11.21 0.7](f) A =[−2.3 30.2 −1.2]4.3. Eigenvalues, Eigenvectors, and Matrix Models 436. [Solution on page 371]Find the eigenvalues of the following matrix.A =[−1672112−14]7. [Solution on page 372]Let A be a square matrix. Show that if u is an eigenvector of A corresponding to theeigenvalue λ, then any nonzero multiple of u is an eigenvector corresponding to thesame eigenvalue λ.8. [Solution on page 372]Suppose the Leslie matrix of a population isL = 4 6 1.40.3 0 00 0.5 0(a) Interpret this matrix.(b) Suppose the initial population vector is X(0) = 800240550. Determine the popu-lation distribution in year 1 and year 2.9. [Solution on page 375]Consider an age-structured population divided into two age groups: age 0 and age 1.Suppose we have the following data.- The percentage of individuals age 0 at time t that survive to time t+ 1 is 15%.- An individual age 0 produces, on average, 1.8 offspring yearly.- An individual age 1 produces, on average, 0.7 offspring yearly.- The population size of individuals age 0 at time t = 0 is 2000, and that ofindividuals age 1 at the same time is 800.Find the Leslie matrix and the population size of each age group at time t = 3.44 Chapter 4. Linear Algebra4.4 Eigenvalues, Eigenvectors, and Matrix Models (con-tinued)1. [Solution on page 376]LetA =[3 −3−2 2]An eigenvalue of A is λ = 5. Find the corresponding eigenvectors.2. [Solution on page 378]Let A =[8 5−12 −9]. An eigenvalue of A is λ = −4. Find the correspondingeigenvectors.3. [Solution on page 378]Find the eigenvalues and eigenvectors of each matrix.(a) A =[6 4−6 −5](b) A =[−12 −323 8](c) A =[3 2−2 7](d) A =[1.2 5.60.6 −0.4]4. [Solution on page 385]Let A be a 2 × 2 matrix. An eigenvector of A is u =[33]and the associatedeigenvalue is λ = 2.(a) Compute A10u.(b) For an integer n ≥ 1, find Anu in terms of n.5. [Solution on page 386]Let A be a 2× 2 matrix. Assume that the eigenvalues of A are λ1 = −3 and λ2 = 4,and the corresponding eigenvectors are u1 =[23]and u2 =[15]. Let v =[−65].(a) Find two numbers a and b such that au1 + bu2 = v.(b) If n ≥ 1 is an integer, find Anv in terms of n.4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 456. [Solution on page 387]LetL =[1.6 0.60.6 0]be the Leslie matrix of a population divided into two age classes. Suppose that theinitial population vector is X(0) =[1500800].(a) Find the eigenvalues and eigenvectors of L.(b) Find the population vector after 1 year.(c) Find an approximate expression for the population distribution after n years(for large n).7. [Solution on page 391]Consider an age-structured population divided into two age groups: age 0 and age 1.Suppose we have the following data.- The percentage of individuals age 0 at time t that survive to time t+ 1 is 40%.- An individual age 0 produces, on average, 1.1 offspring yearly.- An individual age 1 produces, on average, 2 offspring yearly.- The population size of individuals age 0 at time t = 0 is 2100, and that ofindividuals age 1 at the same time is 1260.(a) Find the Leslie matrix L and the number of individuals in each age class at timet = 3. [Round your answer to the nearest integer.](b) Find the eigenvalues and eigenvectors of L.(c) Find an approximate expression for the population distribution after n years(for large n).(d) Use your answer from part (c) to find an approximation of the population vectorafter 15 years.(e) Let λ be the largest eigenvalue of L, and let u =[xy]be an eigenvectorcorresponding to λ. One can show that the proportion of individuals in theage 0 class converges to xx+y, and the proportion of individuals in the age 1 classconverges to yx+y. The fractions xx+yand yx+yare referred to as stable distributionof the two classes as t→∞ (or just the stable distribution of L). Find the stabledistribution of the Leslie matrix L from part (a).46 Chapter 4. Linear AlgebraChapter5Infinite Series5.1 Geometric Series1. [Solution on page 397]Determine whether each series is convergent or divergent. If a series is convergent,find its sum.(a)∞∑n=1(13)n−1(b)∞∑n=15(−34)n−1(c)∞∑n=14(32)n−1(d)∞∑n=19(67)n(e)∞∑n=13(5n)4n−1(f)∞∑n=172n81−n(g)∞∑n=011n+2(−10)n+1(h)∞∑n=0(−4)n+352n+1(i)∞∑n=1(34n+67 (2n+3))(j)∞∑n=13 + 4n9n2. [Solution on page 402]Determine whether each series is convergent or divergent. If a series is convergent,find its sum.(a)23+49+827+1681+32243+ · · ·(b) −52+54− 58+516− 532+ · · ·3. [Solution on page 402]Show that 1.999999 . . . = 2.4748 Chapter 5. Infinite Series5.2 Taylor Series1. [Solution on page 403]Use the definition of the Taylor series to find the first four nonzero terms of the seriesof f(x) at a.(a) f(x) = e−4x, a = 0(b) f(x) = ln(x), a = 1(c) f(x) = sin x, a = pi(d) f(x) = arcsin(2x), a = 142. [Solution on page 407]Find the Maclaurin series for f(x) (that is, find the Taylor series of f(x) at 0).(a) f(x) = e3x(b) f(x) =11− x2(c) f(x) = cos(x3)(d) f(x) = arctan(4x)(e) f(x) = x sin(2x)(f) f(x) = x2 ln(1 +x3)(g) f(x) = xe−2x(h) f(x) =ln(1 + 5x)x(i) f(x) = 5x(j) f(x) =√1 + x33. [Solution on page 411]Explain why the following approximations make sense for small values of x.(a) sinx ≈ x(b) cosx ≈ 1− x22(c) ex ≈ 1 + x(d) (1− x)n ≈ 1− nx4. [Solution on page 412]Let f(x) = 5√(1− 3x)2.(a) Explain why the following approximation makes sense for small values of x.f(x) ≈ 1− 65x(b) Use your answer from part (a) to find an approximation of 5√(0.88)25. [Solution on page 413]Consider the function f(x) = ex2.(a) Find the Maclaurin series of f(x).5.2. Taylor Series 49(b) Evaluate∫ex2dx as an infinite series.(c) Find∫ 10ex2dx as an infinite series.(d) Use the first four nonzero terms of the series from part (c) to approximate∫ 10ex2dx.50 Chapter 5. Infinite SeriesChapter6Continuous Probability Distributionsand Linear Regression6.1 Probability Density Functions and Cumulative Dis-tribution Functions1. [Solution on page 415]Determine whether each function is a probability density function on the given inter-val.(a) f(x) = 2x [0, 1](b) f(x) =x293[4, 7](c) f(x) =x3 − 412[1, 3](d) f(x) =27x4 [0, 2](e) f(x) =6305(−x2 + 3x+ 10) [−1, 4](f) f(x) =13pi(3 + sin x)[−pi2,pi2]2. [Solution on page 419]In each case, find the constant k that makes f a probability density function on theindicated interval.(a) f(x) = kx2 [2, 6](b) f(x) = k(x)12 [4, 9](c) f(x) =3√x2 + kx [0, 1]5152 Chapter 6. Continuous Probability Distributions and Linear Regression3. [Solution on page 421]Let f(x) =3125x2 defined on [0, 5].(a) Show that f is a probability density function.(b) Find the probability that X is between 1 and 2.(c) Find P (3 ≤ X ≤ 5).4. [Solution on page 422]Let f(x) =23(1 + 2x)43defined on [0,∞).(a) Show that f is a probability density function.(b) Find P (0 ≤ X ≤ 2).(c) Find P (X ≥ 13).5. [Solution on page 425]Let f(x) = 3x2e−x3defined on [0,∞).(a) Show that f is a probability density function.(b) Find P (0 ≤ X ≤ 1).(c) Find P (X > 1).(d) Find P (X ≤ 2)6. [Solution on page 427]Letf(x) =45x3 if 0 ≤ x ≤ 145x2if x > 1(a) Show that f is a probability density function.(b) Find the probability that X is between 0 and 1.(c) Find P (0 ≤ X ≤ 3)(d) Find P (X < 5)7. [Solution on page 430]Let f(x) = 1x2be the probability density function of a random variableX in (−∞,−1].(a) Find the cumulative distribution function for f(x).(b) Use your answer from part (a) to find P (X < −5).6.2. Expected Value and Variance of Continuous Random Variables 538. [Solution on page 431]Suppose the cumulative distribution function of a continuous random variable X isF (x) =0 if x ≤ 019x2 if 0 < x < 31 if x ≥ 3(a) Find the corresponding density function.(b) Find P (X ≤ 2).(c) Find P (1 ≤ X < 2).9. [Solution on page 432]The time for blood to clot, T , is a random variable with probability density functionf(t) =1(ln 16)t, 1 ≤ t ≤ 16(t in seconds).(a) What is the probability that T is between 2 and 7 seconds?(b) Find the probability that T is greater than 8 seconds.(c) Find the probability that T is at most 12 seconds.6.2 Expected Value and Variance of Continuous Ran-dom Variables1. [Solution on page 433]In each case, the probability density function of a continuous random random variableX is given as well as the interval. Find the expected value, the variance, and thestandard deviation of X.(a) f(x) =14[2, 6](b) f(x) = −13x+56[0, 2](c) f(x) =116(1 +2√x)[1, 9](d) f(x) =4x5[1,∞)54 Chapter 6. Continuous Probability Distributions and Linear Regression2. [Solution on page 439]Suppose the probability density function of a continuous random variable X isf(x) =328x−14 for x in [1, 16][For the questions below, round your answers to two decimal places.](a) Find the expected value of X.(b) Find Var(X) and σ.(c) Find the median of X. (The mean is a measure of centrality. Another measureof centrality is the median. Let X be a continuous random variable with PDFf(x) on [a, b]. The median of X is the number m such that∫ maf(x) dx = 12.)(d) Find the probability that X is greater than the mean.(e) Find the probability that X is within 1 standard deviation of the mean.3. [Solution on page 443]Find the expected value, the variance, and the standard deviation of for the proba-bility density functionf(x) =127x3 if 0 ≤ x ≤ 1127x4if x > 14. [Solution on page 445]The time for blood to clot, T , is a random variable with probability density functionf(t) =1(ln 16)t, 1 ≤ t ≤ 16(t in seconds). [For the following questions, round your answer to two decimal places.](a) Find the expected value of T .(b) Find the variance and standard deviation of T .5. [Solution on page 446]The height of a tree can be described by a random variable with probability densityfunctionh(x) =14√x+ 1, 0 ≤ x ≤ 86.3. Uniform, Exponential, and Normal Distributions 55(a) Find the mean, variance, and standard deviation of this distribution.(b) Find the probability that the height of a tree chosen randomly is greater thanthe mean height.6. [Solution on page 449]The life span of a particular cable (in years) has the probability density functionf(x) =125e−x25 for x > 0(a) Find the expected life of such a cable.(b) Find the standard deviation of this distribution.(c) What is the probability that a cable will live less than the mean number ofyears?(d) What is the median life of this distribution?6.3 Uniform, Exponential, and Normal Distributions1. [Solution on page 453]The probability density function, f(x), of a random variable X is given. Find themean, variance, and standard deviation of X.(a) f(x) = 13, for x in [2, 5] (b) f(x) = 29, for x in [−0.5, 4]2. [Solution on page 456]Let X be a continuous random variable with probability density functionf(x) = 0.02, 450 ≤ x ≤ 500(a) Find the expected value and standard deviation of X.(b) Find the probability that X is less than $486.(c) Find the probability that X is between the mean and one standard deviationabove the mean.3. [Solution on page 457]The life span of a female mosquito (in months) with an adequate food supply can bedescribed by a uniform distribution on the interval [2, 7].(a) Find the expected life span of this mosquito?56 Chapter 6. Continuous Probability Distributions and Linear Regression(b) What is the probability that the life span of one of these mosquitoes chosenrandomly is greater than 5 months?4. [Solution on page 458]The probability density function, f(x), of a random variable X is given. Find themean, variance, and standard deviation of X.(a) f(x) = 5e−5x, for x in [0,∞) (b) f(x) = 23e−23x, for x in [0,∞)5. [Solution on page 459]The lifetime of a certain organism has an exponential distribution with a mean of 48years.(a) What is the probability density function of this distribution?(b) Find the probability that a randomly selected organism lives between 45 and 50years.(c) Find the probability that a randomly selected organism lives more than 52 years.6. [Solution on page 460]The price (in thousand of dollars) of a particular commodity is exponentially dis-tributed, with a mean of $2.5. What is the probability that the price is less than$7000?7. [Solution on page 461]Suppose a random variable X has the normal distribution with µ = 55 and σ = 20.Find each probability.(a) P (X ≤ 60)(b) P (X > 25)(c) P (45 < X ≤ 80)8. [Solution on page 466]An individual’s weight from a certain population has the normal distribution with amean of 4700 g and a standard deviation of 645 g.(a) What is the probability that a randomly selected individual has a weight greaterthan 6000 g?(b) What is the probability that a randomly selected individual has a weight between3500 g and 5200 g?6.4. Linear Regression 579. [Solution on page 467]An individual’s height from a certain population is normally distributed with a meanof 1.15 m and a standard deviation of 0.3 m. Find an interval centered at the meansuch that the probability that the height of an individual belongs to the interval is50%.6.4 Linear Regression1. [Solution on page 468]Consider the following data.x 1 2 3 4 5y 5 8 10 9 14Find the least-squares line. Plot this line along with the data.2. [Solution on page 471]Consider the following data.x 3 9 13 16 24y 11 10 9.5 6 5.5Find the least-squares line. Plot this line along with the data.3. [Solution on page 473]Fit a linear regression line through the following points. Plot this line along with thedata.(−10.7,−3.4), (−5.2,−1.6), (−3,−2.8), (−1, 2), (2.5, 1), (7, 4), (10, 7.3), (14, 9)4. [Solution on page 474]Consider the following data.Length of a chromosome (x) 39 40 41 43 46 49 50 52Life span (y) 74 73 70 68 63 59 55 5358 Chapter 6. Continuous Probability Distributions and Linear Regression(a) Plot the data. Do the data show any relationship between x and y?(b) Find the line of best fit using the method of least squares.(c) Use the equation of the line obtained in part (b) to predict the life span of anindividual whose length of the chromosome is 54.5. [Solution on page 476]Consider the following data.Leaf width (in cm) 2.3 2.9 3.5 3.8 4.2 4.8 5 5.2 5.5 6Leaf length (in cm) 3.4 3.8 4.2 5 5.3 5.2 6 6.3 7 7.2(a) Plot the data. Do the data show any relationship between leaf width and leaflength?(b) Find the line of best fit using the method of least squares. Plot this line alongwith the data.(c) Use the equation of the line obtained in part (b) to estimate the length of a leafwhose width is 6.7 cm.PartIISolutions to Exercises59Chapter1Integration1.1 Antiderivatives and Indefinite Integrals1. [Exercise on page 3]AntiderivativesLet f be a function. A function F is called an antiderivative of f on aninterval I if the derivative of F (x) is equal to f(x); that is,F ′(x) = f(x) for all x in IFor example, the function F (x) = x2 is an antiderivative of f(x) = 2x sincethe derivative of x2 is 2x.(a) To determine whether F (x) = x9 is an antiderivative of f(x) = 9x8, we need tofind the derivative of F (x) and see if it is equal to f(x).F ′(x) =ddx[x9]= 9x9−1ddx[xn] = nxn−1= 9x8Since F ′(x) = f(x), it follows that F (x) is an antiderivative of f(x) .(b) To determine whether F (x) = x−5 is an antiderivative of f(x) = −5x−4, we needto find the derivative of F (x) and see if it is equal to f(x).F ′(x) =ddx[x−5] = −5x−5−1 ddx[xn] = nxn−1= −5x−66162 Chapter 1. IntegrationSince F ′(x) is not equal to f(x), it follows that F (x) is not an antiderivativeof f(x) .(c) We need to find the derivative of F (x) =√1 + x2 and see if it is equal tof(x) = 2x√1+x2.F ′(x) =ddx[√1 + x2]=ddx[(1 + x2)12] √x = x12=12(1 + x2)12−1 ddx[x2]ddx[(g(x))n] = n(g(x))n−1ddx[g(x)]=12(1 + x2)−12 (2x)ab− c = a− bcb,ddx[xn] = nxn−1=121(1 + x2)12(2x) x−m =1xm=2x2√1 + x2=x√1 + x2Since F ′(x) is not equal to f(x), it follows that F (x) is not an antiderivativeof f(x) .(d) We need to find the derivative of F (x) = ex2and see if it is equal to f(x) = xex2.F ′(x) =ddx[ex2]= ex2 ddx[x2]ddx[eg(x)]= eg(x)ddx[g(x)]= ex2(2x) = 2xex2Since F ′(x) 6= f(x), it follows that F (x) is not an antiderivative of f(x) .(e)F ′(x) =ddx[x lnx] =ddx[x] lnx+ xddx[lnx] Product Rule= (1) ln x+ x(1x)ddx[lnx] =1x= lnx+xx= lnx+ 1Since F ′(x) = f(x), it follows that F (x) is an antiderivative of f(x) .2. [Exercise on page 3]1.1. Antiderivatives and Indefinite Integrals 63Indefinite IntegralsLet f be a continuous function. If F (x) is an antiderivative of f(x), then forevery real number C, F (x) + C is also an antiderivative of f(x) sinceddx[F (x) + C] =ddx[F (x)] +ddx[C] Sum Rule= f(x) +ddx[C]F ′(x) = f(x) because F (x)is an antiderivative of f(x)= f(x) + 0 The derivative of a constant is 0= f(x)The family of all antiderivatives of f(x) is called the indefinite integral off(x) and is denoted by ∫f(x) dxIf F ′(x) = f(x), then ∫f(x) dx = F (x) + Cwhere C is a constant (that is, an arbitrary real number). The symbol∫is called the integral symbol (or integral sign). The function f(x) is called the integrand. The term dx indicates that the variable of integration is x. In otherwords, dx tells us that the integral is taken with respect to x. The arbitrary real number C is called the constant of integration.For example, since the derivative of x2 is 2x, it follows that∫2x dx = x2 + C.Warning. Do not forget to add the constant of integration C when evaluatingan indefinite integral.(a) The equality∫3x2 dx = x3 + C is correct because the derivative of x3 is 3x2.(b) The equality∫e−x dx = e−x + C is not correct because the derivative of e−x isnot equal to e−x. Indeed,ddx[e−x] = e−xddx[−x] = e−x(−1) = −e−x 6= e−x64 Chapter 1. Integration(c) To see if the equality∫(1 + 2x)e2x dx = xe2x +C is correct, we need to find thederivative of xe2x and see if it is equal to (1 + 2x)e2x.ddx[xe2x]=ddx[x]e2x + xddx[e2x]Product Rule: ddx[f(x)g(x)] =f ′(x)g(x) + f(x)g′(x)= (1)e2x + xe2xddx[2x]ddx[eg(x)]= eg(x)ddx[g(x)]= e2x + xe2x(2)= (1 + 2x)e2x Factor out e2xSince ddx[xe2x] = (1 + 2x)e2x, the following equality is correct.∫(1 + 2x)e2x dx = xe2x + C(d) Letf(x) =−x4 − 3x2 + 2x(x2 + 1)2and F (x) =x3 + 1x2 + 1We need to find the derivative of F (x) and see if it is equal to f(x). Rememberthe quotient rule:ddx[f(x)g()x]=f ′(x)g(x)− f(x)g′(x)[g(x)]2Using this, we haveF ′(x) =ddx[x3 + 1x2 + 1]=ddx[x3 + 1](x2 + 1)− (x3 + 1) ddx[x2 + 1](x2 + 1)2=3x2(x2 + 1)− (x3 + 1)(2x)(x2 + 1)23x4 + 3x2 − 2x4 − 2x(x2 + 1)2=x4 + 3x2 − 2x(x2 + 1)2Since F ′(x) 6= f(x), the equality ∫ f(x) dx = F (x) + C is not correct.3. [Exercise on page 3](a)∫5 dx1.1. Antiderivatives and Indefinite Integrals 65Indefinite Integral of a Constant FunctionThe indefinite integral of a constant function k is kx + C, where C is aconstant (an arbitrary real number). That is,∫k dx = kx+ CTo verify this rule, we can take the derivative of the right-hand side andsee if it is equal to the integrand k:ddx[kx+ C] =ddx[kx] +ddx[C]ddx[f(x) + g(x)] =ddx[f(x)] +ddx[g(x)]= kddx[x] + 0ddx[kf(x)] = kddx[f(x)] ,ddx[C] = 0= k(1) = kddx[x] = 1Using this, we have∫5 dx = 5x+ C .(b) Since pi is a constant, it follows that pi2 is also a constant. So∫pi2dx = pi2x+ C(c)∫x8 dxIndefinite Integral of xnFor every real number n such that n 6= −1, the indefinite integral of thepower function xn is xn+1n+1+ C. That is,∫xn dx =xn+1n+ 1+ C, n 6= −1.To verify this rule, we can take the derivative of the right-hand side andsee if it is equal to the integrand xn:ddx[xn+1n+ 1+ C]=ddx[xn+1n+ 1]+ddx[C]=1n+ 1ddx[xn+1]+ 0ddx[kf(x)] = kddx[f(x)]66 Chapter 1. Integration=1n+ 1[(n+ 1)x(n+1)−1] ddx[xn] = nxn−1=n+ 1n+ 1xn+1−1 = xnThe case n = −1 will be discussed later.Using the formula∫xn dx =xn+1n+ 1+ C , we get∫x8dx =x8+18 + 1+ C =x99+ C(d) ∫x14 dx =x14+114+ 1+ C∫xn dx =xn+1n+ 1+ C=x5454+ C =45x54 + Cab+ c =a+ bcb,abc=acb(e) ∫1x3dx =∫x−3 dx1xm= x−m=x−3+1−3 + 1 + C∫xn dx =xn+1n+ 1+ C=x−2−2 + C = −12x2+ C(f) ∫3√x dx =∫x13 dx p√xq = xqp=x13+113+ 1+ C∫xn dx =xn+1n+ 1+ C=x4343+ C =34x43 + Cab+ c =a+ bcb,abc=acb(g) ∫16√xdx =∫1x16dx =∫x−16 dx1.1. Antiderivatives and Indefinite Integrals 67=1−16+ 1x−16+1 + C =156x56 + C =65x56 + C(h) ∫x3√xdx =∫xx13dx=∫x1−13 dxxnxm= xn−m=∫x23 dx a− bc=ac− bc=x5353+ C∫xn dx =xn+1n+ 1+ C=35x53 + Cabc=acb(i) ∫4√x5 dx =∫x54 dx =194x94 + C =49x94 + C(j) ∫ 3√x2x4dx =∫x23x4dx =∫x23−4 dx =∫x−103 dx=1−103+ 1x−103+1 + C =1−73x−73 + C= −37× 1x73= − 37x73+ C4. [Exercise on page 4]Properties of Indefinite IntegralsLet f and g be continuous functions.(a) For any constant k, ∫kf(x) dx = k∫f(x) dx68 Chapter 1. Integration(b) The integral of the sum of two functions is the sum of integrals; that is,∫[f(x) + g(x)] dx =∫f(x) dx+∫g(x) dx(c) The integral of the difference of two functions is the difference of integrals;that is, ∫[f(x)− g(x)] dx =∫f(x) dx−∫g(x) dx(a) ∫4x dx = 4∫x dx Take out the constant 4= 4(x22)+ C∫xn dx =xn+1n+ 1+ C= 2x2 + C(b) ∫ √3x dx =∫ √3√x dx =√3∫ √x dx =√3∫x12 dx=√3x3232+ C =23√3x32 + C =2√33x32 + C(c) ∫12x5dx =∫121x5dx =12∫1x5dx Take out the constant12=12∫x−5 dx =12(x−5+1−5 + 1)+ C∫xn dx =xn+1n+ 1+ C=12(x−4−4)+ C = −18x−4 + C= − 18x4+ C x−m =1xm(d) ∫(x2 + x3) dx =∫x2dx+∫x3 dx∫(f(x) + g(x)) dx =∫f(x) dx+∫g(x) dx1.1. Antiderivatives and Indefinite Integrals 69=x33+x44+ C∫xn dx =xn+1n+ 1+ CNote. When dealing with the sum (or difference) of integrals, do not add aconstant for each integral. Add only one constant. This is because the sum (ordifference or product) of constants is still a constant.(e) ∫ (1x2− 8)dx =∫1x2dx−∫8 dx∫(f(x) − g(x))dx =∫f(x)dx− ∫ g(x)dx=∫x−2 dx−∫8 dx1x2= x−2=x−2+1−2 + 1 − 8x+ C∫xn dx =xn+1n+ 1+ C=x−1−1 − 8x+ C = −1x− 8x+ C(f) ∫(−3x2 + 5x− 1) dx=∫−3x2dx+∫5x dx−∫1dx= −3∫x2dx+ 5∫x dx−∫1 dx∫kf(x) dx = k∫f(x) dx= −3x33+ 5x22− x+ C∫xn dx =xn+1n+ 1+ C= −x3 + 52x2 − x+ C(g) ∫(x2 +√x− pi4) dx =∫x2 dx+∫ √x dx−∫pi4 dx=∫x2 dx+∫x12 dx−∫pi4 dx=x33+x3232− pi4x+ C = 13x3 +23x32 − pi4x+ C70 Chapter 1. Integration(h) ∫4x− 3√xdx =∫ (4x√x− 3√x)dxa− bc=ac− bc=∫4x√xdx−∫3√xdx∫(f(x) − g(x))dx =∫f(x)dx− ∫ g(x)dx= 4∫xx12dx− 3∫1x12dx√x = x12= 4∫x12dx− 3∫x−12dxxnxm= xn−m,1xm= x−m= 4x3232− 3x1212+ C∫xn dx =xn+1n+ 1+ C= 4(23)x32 − 3(21)x12 + Cabc=acb=83x32 − 6x 12 + C(i) ∫2−√2t+ 3√t2√tdt =∫2√tdt−∫ √2t√tdt+∫ 3√t2√tdt=∫2t12dt−∫ √2√t√tdt+∫t23t12dt= 2∫t−12 dt−∫ √2 dt+∫t16 dt= 2t1212−√2t+t7676+ C = 2(21)t12 −√2t+67t76 + C= 4√t−(√2)t+67t76 + C(j) ∫2x(3− x−3) dx =∫(6x− 2x−2) dx Distribute= 6∫x dx− 2∫x−2 dx = 6x22− 2x−1−1 + C= 3x2 + 2x−1 + C = 3x2 +2x+ CWarning. In general, the integral of the product of two functions is not equal1.1. Antiderivatives and Indefinite Integrals 71to the product of integrals; that is,∫f(x)g(x) dx 6=(∫f(x) dx)(∫g(x) dx)in general.This also applies to the quotient of two functions: In general,∫f(x)g(x)dx 6=∫f(x) dx∫g(x) dx(k) ∫(t+ 4)(2t+ 1) dt =∫(2t2 + t+ 8t+ 4) dt Distribute=∫(2t2 + 9t+ 4) dt = 2∫t2 dt+ 9∫t dt+∫4 dt= 2t33+ 9t22+ 4t+ C =23t3 +92t2 + 4t+ C(l) ∫(x+ 3)(4x− 5) dx =∫(4x2 − 5x+ 12x− 15) dx Distribute=∫(4x2 + 7x− 15) dx= 4(x33)+ 7(x22)− 15x+ C∫xn dx =xn+1n+ 1+ C=43x3 +72x2 − 15x+ C5. [Exercise on page 4]Indefinite Integrals of Exponential FunctionsDifferentiation rules for exponential functions include the following:ddx[ex] = exddx[ekx] = kekxddx[ax] = (ln a)axddx[akx] = k(ln a)akxUsing these, we derive the following rules for indefinite integrals of exponentialfunctions.72 Chapter 1. Integration∫exdx = ex + C For any real number k 6= 0,∫ekxdx =1kekx + C For a > 0, a 6= 1,∫axdx =axln a+ C For a > 0, a 6= 1, k 6= 0,∫akxdx =akxk ln a+ C(a) ∫8ex dx = 8∫ex dx∫kf(x) dx = k∫f(x) dx= 8ex + C∫ex dx = ex + C(b) The function f(x) = e−x is of the form ekx with k = −1. So, by using theformula∫ekx dx =1kekx + C, we get∫e−xdx =e−x−1 + C = −e−x + C(c) The function f(x) = e7x is of the form ekx with k = 7. So, by using the formula∫ekx dx =1kekx + C, we have∫e7xdx =17e7x + C(d) ∫(1 + x− 6e−9x) dx =∫1 dx+∫xdx− 6∫e−9x dx= x+x22− 6(1−9e−9x)+ C∫ekx dx =1kekx + C= x+x22+23e−9x + C(e) ∫25e−8x=∫251e−8xdx =25∫1e−8xdx∫kf(x) dx = k∫f(x) dx1.1. Antiderivatives and Indefinite Integrals 73=25∫e8x dx1e−b= eb=25(18e8x)+ C∫ekx dx =1kekx + C=120e8x + C(f) Using the formula∫ax dx =axln a+ C, we get∫5xdx =5xln 5+ C .(g) The function f(x) = 3−7x is of the form akx with a = 3 and k = −7. So, byusing the formula∫akx dx =akxk ln a+ C, we get∫3−7xdx =3−7x−7 ln 3 + C(h) ∫ √9 · 75x dx =∫ √9√75x dx =∫3(75x) 12 dx√x = x12=∫3 · 7 5x2 dx =∫3 · 7 52x dx (an)m = anm= 3∫752x dx∫kf(x) dx = k∫f(x) dx= 3(752x52ln 7)+ C∫akx dx =akxk ln a+ C= 3(2575x2ln 7)+ Cabc=acb=6 · 7 5x25 ln 7+ C6. [Exercise on page 4](a)Indefinite Integral of x−1Since the derivative of the function y = ln |x| is 1x, we deduce the followingintegration rule: ∫1xdx = ln |x|+ C74 Chapter 1. IntegrationIn other words, the indefinite integral of f(x) = xn, when n = −1, is thenatural logarithm of the absolute value of x plus a constant.Warning. Do not forget the absolute value symbol when taking theintegral of 1xwith respect to x.∫2xdx = 2∫1xdx∫kf(x) dx = k∫f(x) dx= 2 ln |x|+ C∫1xdx = ln |x|+ C(b) ∫59xdx =59∫1xdx∫kf(x) dx = k∫f(x) dx=59ln |x|+ C∫1xdx = ln |x|+ C(c) ∫x2(1− 4x−3) dx = ∫ (x2 − 4x−1) dx Distribute=∫x2 dx−∫4x−1 dx∫(f(x) − g(x))dx =∫f(x)dx− ∫ g(x)dx=∫x2 dx− 4∫x−1 dx∫kf(x) dx = k∫f(x) dx=∫x2 dx− 4∫1xdx x−m =1xm=x33− 4 ln |x|+ C∫xn dx =xn+1n+ 1+ C,∫1xdx = ln |x|+ C(d) ∫x5 + 3x2 − 7x+ 1x2dx=∫ (x5x2+3x2x2− 7xx2+1x2)dxa± bc=ac± bc=∫ (x3 + 3− 7x+1x2)dxxnxm= xn−m1.1. Antiderivatives and Indefinite Integrals 75=∫x3 dx+∫3 dx−∫7xdx+∫1x2dx∫(f(x) ± g(x)) dx =∫f(x) dx± ∫ g(x) dx=∫x3 dx+∫3 dx−∫7x−1 dx+∫x−2 dx 1xm= x−m=∫x3 dx+∫3 dx− 7∫x−1 dx+∫x−2 dx∫kf(x) dx = k∫f(x) dx=x44+ 3x− 7 ln |x|+ x−2+1−2 + 1 + C∫xn dx =xn+1n+ 1+ C,∫1xdx = ln |x| + C,∫kdx = kx+ C=x44+ 3x− 7 ln |x|+ x−1−1 + C=x44+ 3x− 7 ln |x| − 1x+ C7. [Exercise on page 5]∫(4x2 − 1)(3x+ 5)x3dx =∫(12x3 + 20x2 − 3x− 5)x3dx=∫ (12x3x3+20x2x3− 3xx3− 5x3)dx=∫ (12 +20x− 3x2− 5x3)dx=∫ (12 + 201x− 3x−2 − 5x−3)dx= 12x+ 20 ln |x| − 3 x−2+1−2 + 1 − 5x−3+1−3 + 1 + C= 12x+ 20 ln |x| − 3x−1−1 − 5x−2−2 + C= 12x+ 20 ln |x|+ 3x+52x2+ C8. [Exercise on page 5]Basic Trigonometric FunctionsConsider the following figure.76 Chapter 1. Integrationxyθba10 1−11−1LIn this figure, θ is the angle between the x-axis and the line L.Definition of the Six Basic Trigonometric Functions.(1) The sine of θ, denoted sin θ, is defined by sin θ = a.(2) The cosine of θ, denoted cos θ, is defined by cos θ = b.(3) The tangent of θ, denoted tan θ, is defined by tan θ = ab. Since a = sin θand b = cos θ, it follows thattan θ =sin θcos θ(4) The cotangent of θ, denoted cot θ, is defined by cot θ = ba. Since a = sin θand b = cos θ, it follows thatcot θ =cos θsin θIn other words, the cotangent of θ is the reciprocal of the tangent of θ;that is, cot θ = 1tan θ.(5) The secant of θ, denoted sec θ, is defined as sec θ = 1b. Since b = cos θ, itfollows that the secant of θ is the reciprocal of the cosine of θ. That is,sec θ =1cos θ1.1. Antiderivatives and Indefinite Integrals 77(6) The cosecant of θ, denoted csc θ, is defined as csc θ = 1a. Since a = sin θ,it follows that the cosecant of θ is the reciprocal of the sine of θ. That is,csc θ =1sin θIndefinite Integrals of (basic) Trigonometric FunctionsThe derivatives of the six basic trigonometric functions are given by the follow-ing rules.ddx[sinx] = cos xddx[cosx] = − sinxddx[tanx] = sec2 xddx[cotx] = − csc2 xddx[secx] = secx tanxddx[cscx] = − cscx cotxUsing these, we have the following integration rules.∫cosx dx = sinx+ C∫sinx dx = − cosx+ C∫sec2 x dx = tanx+ C∫csc2 x dx = − cotx+ C∫secx tanx dx = secx∫cscx cotx dx = − cscx+ C(a) ∫(3 cosx+ 5 sinx) dx=∫3 cosx dx+∫5 sinx dx∫(f(x) + g(x))dx =∫f(x)dx+∫g(x)dx= 3∫cosx dx+ 5∫sinx dx∫kf(x) dx = k∫f(x) dx= 3(sinx) + 5(− cosx) + C∫cosx dx = sin x + C∫sinx dx = − cosx + C= 3 sinx− 5 cosx+ C78 Chapter 1. Integration(b) ∫(7 cos t− 8 sin t) dt =∫7 cos t dt−∫8 sin t dt= 7∫cos t dt− 8∫sin t dt= 7(sin t)− 8(− cos t) + C∫cos t dt = sin t + C∫sin t dt = − cos t + C= 7 sin t+ 8 cos t+ C(c) ∫ (12− sec2 x) dx = ∫ 12 dx− ∫ sec2 x dx= 12x− tanx+ C∫k dx = kx+C∫sec2 x dx = tanx+ C(d) ∫ (e−2x − 9 csc2 x) dx=∫e−2x dx−∫9 csc2 x dx=1−2e−2x − 9∫csc2 x dx∫ekx dx =1kekx + C= −12e−2x − 9 (− cotx) + C∫csc2 x dx = − cotx+ C= −12e−2x + 9 cotx+ C(e) ∫(2 secx tanx− 5 cscx cotx) dx= 2∫secx tanx dx− 5∫cscx cotx dx1.1. Antiderivatives and Indefinite Integrals 79= 2 sec x− 5(− cscx) + C∫secx tanx dx = secx+ C∫cscx cotx dx = − cscx+ C= 2 secx+ 5 cscx+ C(f) ∫secx(secx− tanx) dx=∫ (sec2 x− secx tanx) dx Distribute=∫sec2 x dx−∫secx tanx dx= tanx− secx+C∫sec2 x dx = tanx+ C∫secx tanx dx = secx+ C9. [Exercise on page 5]Some Trigonometric Identities For every real number x,sin2 x+ cos2 x = 1 For every real number x,sin(2x) = 2 sinx cosx(a) ∫ (2x+ cos2 x+ sin2 x)dx =∫(2x+ 1) dx sin2 x+ cos2 x = 1= 2∫xdx+∫1 dx= 2(x22)+ 1x+ C∫xn dx =xn+1n+ 1+ C∫k dx = kx+ C= x2 + x+ C80 Chapter 1. Integration(b) ∫11− sin2 x dx =∫1cos2 xdx sin2 x+ cos2 x = 1=∫sec2 x dx secx =1cosx= tanx+ C∫sec2 x dx = tanx+ C(c) ∫ (4 + 4 tan2 t)dt =∫4(1 + tan2 t)dt= 4∫ (1 + tan2 t)dt= 4∫ (1 +sin2 tcos2 t)dt tan t =sin tcos t= 4∫cos2 t+ sin2 tcos2 tdt a+bc=ac+ bc= 4∫1cos2 tdt sin2 t+ cos2 t = 1= 4∫sec2 t dt sec t =1cos t= 4 tan t+ C∫sec2 t dt = tan t+ C(d) ∫sin(2x)cosxdx =∫2 sinx cosxcosxdx sin 2x = 2 sin x cosx=∫2 sinx dx Simplify= 2∫sinx dx = 2(− cosx) + C∫sinx dx = − cosx+ C= −2 cosx+ C10. [Exercise on page 5]1.1. Antiderivatives and Indefinite Integrals 81Inverse Trigonometric FunctionsInverse Functions. Let f be a function. Roughly speaking, the inverse func-tion of f , denoted f−1, undoes what f does. Specifically, the inverse functionf−1 (if it exists) is defined byf−1(y) = x if and only if f(x) = yFor example, consider the function y = f(x) = x2 on the interval [0,∞). Since f(2) = 4, it follows that f−1(4) = 2. Since f(3) = 9, it follows that f−1(9) = 3. Since f(4) = 16, it follows that f−1(16) = 4.Warning. f−1(x) does not mean 1f(x). In the previous example, we havef−1(4) = 2, but 1f(4)= 116. The reciprocal 1f(4)could be written as [f(4)]−1.A function f has an inverse if and only iff(x1) 6= f(x2) whenever x1 6= x2In other words, f has an inverse if and only if it does not take on the same valuetwice (geometrically, this means that no horizontal line intersects the graph off more than once).Not all functions have inverses. For example, consider the following two func-tions.y = g(x)−2 24xyy = f(x)xy The function y = g(x) does not possess an inverse because it takes on thesame value twice (both −2 and 2 have the same output, 4). This can alsobe justified by the fact that there exists a horizontal line that intersectsthe graph of g more than once (for example the line y = 4 intersects thegraph of g twice). On the other hand, the function y = f(x) has an inverse because nohorizontal line intersects the graph of f more than once.82 Chapter 1. IntegrationThe Inverse Tangent Function. Consider the graph of y = tanx on theinterval(−pi2, pi2).xy−pi20 pi2−pi piThis function has an inverse since no horizontal line intersects its graph morethan once. The inverse tangent function is denoted tan−1 or arctan, andwe havetan−1(y) = x if and only if tanx = yfor any y in (−∞,∞). For example, since tan (pi4)= 1, it follows thattan−1(1) = pi4.The Inverse Sine Function. Consider the graph of y = sinx on the interval[−pi2, pi2].xy0−pi2pi2−11Since no horizontal line intersects this graph more than once, it follows thaty = sinx, defined on the interval[−pi2, pi2], has an inverse called the inverse1.1. Antiderivatives and Indefinite Integrals 83sine function. It is denoted sin−1 or arcsin, and we havesin−1(y) = x if and only if sinx = yfor any y in [−1, 1]. For example, since sin (pi6)= 12, it follows that sin−1(12)=pi6.Indefinite Integrals of Inverse Trigonometric FunctionsIt turns out that the derivatives of the inverse tangent and inverse sine functionsare given by the following rules:ddx[tan−1(x)]=11 + x2andddx[sin−1(x)]=1√1− x2Using these, we have the following integration rules:∫11 + x2dx = tan−1 x+ C∫1√1− x2 dx = sin−1 x+ C(a) ∫31 + x2dx = 3∫11 + x2dx= 3 tan−1 x+ C∫11 + x2dx = tan−1 x+ C(b) ∫10x28x2 + 8dx =∫10x28(x2 + 1)dx =∫5x24(x2 + 1)dx=54∫x2x2 + 1dx∫kf(x) dx = k∫f(x) dx=54∫x2 + 1− 1x2 + 1dx=54∫ (x2 + 1x2 + 1− 1x2 + 1)dxa+ bc=ac+bc84 Chapter 1. Integration=54∫ (1− 1x2 + 1)dx Simplify=54(x− tan−1 x)+ C∫1 dx = x + C∫11 + x2dx = tan−1 x+ C=54x− 54tan−1 x+ C(c) ∫83√1− x2 dx =83∫1√1− x2 dx=83sin−1 x+ C∫1√1− x2 dx = sin−1 x+ C(d) ∫ √1− x21− x2 dx =∫ (√1− x2) (√1− x2)(1− x2)√1− x2 dx Rationalize=∫(1− x2)(1− x2)√1− x2 dx (√a)(√a) = a=∫1√1− x2 dx Simplify= sin−1 x+ C∫1√1− x2 dx = sin−1 x+ C11. [Solution on page 5]Table of Integrals(1) If k is a constant, then ∫k dx = kx+ C(2) For any real number n 6= −1,∫xn dx =xn+1n+ 1+ C1.1. Antiderivatives and Indefinite Integrals 85(3) For any real number k,∫kf(x) dx = k∫f(x) dx+ C(4)∫[f(x) + g(x)] dx =∫f(x) dx+∫g(x) dx(5)∫[f(x)− g(x)] dx =∫f(x) dx−∫g(x) dx(6)∫ex dx = ex + C(7) For any real number k 6= 0,∫ekx dx =1kekx + C(8) For any real number a such that a > 0 and a 6= 1,∫ax dx =axln a+ C(9) For any real number a such that a > 0 and a 6= 1, for any real numberk 6= 0, ∫akx dx =akxk(ln a)+ C(10) The indefinite integral of 1xis ln |x|+ C, that is,∫1xdx = ln |x|+ C(11)∫cosx dx = sinx+ C(12)∫sinx dx = − cosx+ C(13)∫sec2 x dx = tanx+ C(14)∫csc2 x dx = − cotx+ C(15)∫secx tanx dx = secx+ C86 Chapter 1. Integration(16)∫cscx cotx dx = − cscx+ C(17)∫11 + x2dx = arctan(x)+C (Another notation for arctan(x) is tan−1 x)(18)∫1√1− x2 dx = arcsin(x)+C (Another notation for arcsin(x) is sin−1 x)12. [Exercise on page 6](a) We want to find N(t). Remember that the rate of change of N(t) is the derivativeof N(t); that is,rate of change of N(t) = N ′(t) =ddt[N(t)]So, if N ′(t) is given, N(t) is obtained by taking the integral of N ′(t). (Rememberthat integrals and derivatives are inverse processes.)N(t) =∫N ′(t) dt =∫9e1.5t dt= 9∫e1.5t dt = 9(11.5e1.5t)+ C∫ekx dx =1kekx + C= 6e1.5t + CNow, we need to find the constant C. Since the initial population is 50 bacteria,it follows that N(0) = 50. This implies thatN(0) = 506e1.5(0) + C = 50 Substitute t with 0 into N(t)6 + C = 50 e0 = 1C = 50− 6 = 44So, N(t) = 6e1.5t + 44 .(b) Five hours later the population isN(5) = 6e1.5(5) + 44 = 6e7.5 + 44 ≈ 1089213. [Exercise on page 6]1.2. The Substitution Rule 87(a) The population P (t) at time t isP (t) =∫P ′(t) dt =∫6500√tdt= 6500∫1√tdt = 6500∫1t12dt√x = x12= 6500∫t−12 dt1xm= x−m= 6500(t1212)+ C∫xn dx =xn+1n+ 1+ C= 6500(2t12)+ C = 13000√t+ CNow, we need to find C. Since the population is 29000 when t = 1, it followsthat P (1) = 29000. This implies thatP (1) = 2900013000√1 + C = 29000C = 29000− 13000 = 16000So, P (t) = 13000√t+ 16000 .(b) We want to find t such that P (t) = 211000.P (t) = 21100013000√t+ 16000 = 21100013000√t = 195000 Subtract 16000 from both sides√t = 15 Divide both sides by 13000t = (15)2 = 225 Take the square of both sidesSo, the population will get to 211000 after 225 days.1.2 The Substitution Rule1. [Exercise on page 6](a)88 Chapter 1. IntegrationDifferentialsIf u = u(x), the differential of u, denoted du, is defined asdu = u′(x) dxFor example, if u = x, then du = 1 dx = dx. If u = x2, then du = 2x dx.Substitution MethodThe Substitution RuleLet u = g(x) be a differentiable function and let f be a continuousfunction. Then ∫f(g(x))g′(x) dx =∫f(u) duThe idea behind the substitution rule is to replace a difficult integral byan easier integral. To integrate∫f(x) dx using substitution, one can gothrough the following steps.Step 1. Choose a substitution u = u(x) that “simplifies” the integrandf(x).Step 2. Find the differential du = u′(x) dx and rewrite the entire integralonly in terms of u and du. The original integral is of the form∫f(x) dx =∫h(u) dufor some function h.Step 3. Evaluate the integral∫h(u) du.Step 4. Return to the original variable by replacing u with u(x).Note. The substitution method works when it is possible to do two things: Express the original integral only in terms of the new variable u. Evaluate the transformed integral∫h(u) du.1.2. The Substitution Rule 89A promising sign that the substitution method may work is when a func-tion (say u(x)) and its derivative u′(x) (except for a constant factor) areboth present in the original integral. For example, consider the integral∫x (x2 + 1)5dx. The function u = x2 +1 and its derivative 2x (except forthe constant factor 2) are both present in the integral. So, the substitutionu = x2 + 1 may work. (It turns out that this substitution actually works.)Let’s apply this with the integral∫(x+ 1)8 dx.Step 1. Let u = x+ 1, the quantity inside the parentheses.Step 2. The differential of u is du = (x+ 1)′ dx = 1 dx = dx. And the originalintegral becomes ∫(x+ 1)8 dx =∫u8 du.Step 3. We evaluate the transformed integral∫u8 du.∫u8 du =u8+18 + 1+ C∫xn dx =xn+1n+ 1+ C=u99+ CStep 4. Returning to the original variable by substituting u with x+ 1, we get∫(x+ 1)8 dx =∫u8 du=u99+ C=(x+ 1)99+ C(b)Step 1. We choose u = x2 + 3.Step 2. The differential is du = 2x dx so that∫2x(x2 + 3)15 dx =∫(x2 + 3)15(2x dx) =∫u15 du.Step 3. Evaluating the transformed integral, we get∫u15 du =u1616+ C∫xn dx =xn+1n+ 1+ C90 Chapter 1. IntegrationStep 4. Returning to the original variable, we get∫2x(x2 + 3)15 dx =∫u15 du=u1616+ C=(x2 + 3)1616+ C(c)Step 1. Let u = 2x− 1.Step 2. The differential is du = 2dx, so du2= dx. The integral becomes∫(2x− 1)9 dx =∫u9(du2)Steps 3 and 4.∫u9(du2)=12∫u9 du∫kf(x) dx = k∫f(x) dx=12(u1010)+ C∫xn dx =xn+1n+ 1+ C=120u10 + C=120(2x− 1)10 + C Substitute u by 2x− 1.(d)Step 1. Let u = x3 − 7.Step 2. The differential is du = 3x2 dx. Since there is no 3x2 dx in the integral(but x2 dx instead), we divide the equation du = 3x2 dx by 3 to get du3= x2 dx.The integral becomes∫x2(x3 − 7)6 dx =∫(x3 − 7)6(x2 dx) =∫u6(du3)Steps 3 and 4.∫u6(du3)=13∫u6 du∫kf(x) dx = k∫f(x) dx=13(u77)+ C∫xn dx =xn+1n+ 1+ C1.2. The Substitution Rule 91=121u7 + C=121(x3 − 7)7 + C Substitute u by x3 − 7.(e)Step 1. Let u = 1− x2.Step 2. The differential is du = −2x dx. Dividing both sides by −2, we getdu−2 = x dx. The integral becomes∫x√1− x2 dx =∫ √u(du−2)Steps 3 and 4.∫ √u(du−2)= −12∫u12 du∫kf(x) dx = k∫f(x) dx,√x = x12= −12u3232+ C∫xn dx =xn+1n+ 1+ C= −12(23)u32 + Cabc=acb= −13u32 + C= −13(1− x2) 32 + C Substitute u by 1− x2(f) We want to find∫z2√z3 + 4 dz. First, we make the substitution u = z3 + 4.Then the differential of u is du = 3z2dz. Dividing both sides of this latterequation by 3, we getdu3= z2dzThe integral becomes∫z2√z3 + 4 dz =∫ √z3 + 4(z2dz)=∫ √u(du3)=13∫ √u du =13∫u12 du=13(u3232)+ C∫xn dx =xn+1n+ 1+ C92 Chapter 1. Integration=13(23u32)+ C =29u32 + C=29(z3 + 4)32 + C Substitute u by z3 + 42. [Exercise on page 6](a) Let u = x5 + 1. Then du = 5x4 dx, so that du5= x4 dx. Therefore,∫ −3x4(x5 + 1)7dx =∫ −3(x5 + 1)7(x4 dx)=∫ −3u7(du5)= −3∫1u7(du5) ∫kf(x) dx = k∫f(x) dx= −35∫u−7 du∫kf(x) dx = k∫f(x) dx= −35(u−7+1−7 + 1)+ C∫xn dx =xn+1n+ 1+ C= −35(u−6−6)+ C=110u−6 + C=110u6+ C1xm= x−m=110(x5 + 1)6+ C Substitute u by x5 + 1(b) First, rearrange the integrand as follows:∫ −4x+ 43√(x2 − 2x+ 1)2 dx =∫ −4x+ 4(x2 − 2x+ 1) 23 dxp√xq = xqp=∫(−4x+ 4)(x2 − 2x+ 1)− 23 dx 1xm= x−mNow, let u = x2−2x+ 1. Then, du = (2x−2) dx. In the integral, we don’t have(2x−2) dx, but (−4x+4) dx instead. Multiplying the equation du = (2x−2) dxby −2, we get −2 du = (−4x + 4) dx. Hence, if we substitute x2 − 2x + 1 with1.2. The Substitution Rule 93u and (−4x+ 4) dx with du into the integral, we get∫ −4x+ 43√(x2 − 2x+ 1)2dx=∫(−4x+ 4)(x2 − 2x+ 1)− 23dx From before=∫u−23 (−2 du)= −2∫u−23 du∫kf(x) dx = k∫f(x) dx= −2u1313+ C∫xn dx =xn+1n+ 1+ C= −2(31)u13 + Cabc=acb= −6u 13 + C= −6(x2 − 2x+ 1) 13 + C Substitute u by x2 − 2x+ 1(c) We make the substitution u = x2 − 16. Then, du = 2x dx. Using this, we have∫2xx2 − 16 dx =∫1x2 − 16(2x dx)=∫1udu= ln |u|+ C∫1xdx = ln |x|+ C= ln∣∣x2 − 16∣∣+ C Substitute u by x2 − 16(d)Let u = t6 + 3. Then, du = 6t5 dt so that du6= t5 dt. We thus have∫t5t6 + 3dt =∫1t6 + 3(t5 du)=∫1u(du6)=16∫1udu∫kf(x) dx = k∫f(x) dx=16ln |u|+ C∫1xdx = ln |x|+ C94 Chapter 1. Integration=16ln∣∣t6 + 3∣∣+ C Substitute u by t6 + 3(e) Let u = 2x3 − x2, so that du = (6x2 − 2x) dx. Multiplying both sides of thislatter equation by 3, we get 3 du = (18x2 − 6x) dx. Using this, we have∫18x2 − 6x2x3 − x2 dx =∫12x3 − x2(18x2 − 6x) dx=∫1u(3 du)= 3∫1udu∫kf(x) dx = k∫f(x) dx= 3 ln |u|+ C∫1xdx = ln |x|+ C= 3 ln∣∣2x3 − x2∣∣+ C Substitute u by 2x3 − x2(f) We make the substitution u = 1 − x, the exponent on e. Then, du = − dx, so− du = dx. Therefore,∫e1−x dx =∫eu(− du)= −∫eu du∫kf(x) dx = k∫f(x) dx= −eu + C= −e1−x + C Substitute u by 1− x(g) Let u = x4. Then, du = 4x3 dx so that du4= x3 dx. Therefore,∫x3ex4dx =∫ex4(x3 dx)=∫eu(du4)=14∫eu du∫kf(x) dx = k∫f(x) dx=14eu + C∫ex dx = ex + C=14ex4+ C Substitute u by x41.2. The Substitution Rule 95(h) Let u = −x2 + 2x, the exponent on e. Then, du = (−2x + 2) dx so thatdu−2 = (x− 1) dx and∫x− 1e−x2+2xdx =∫1eu(du−2)=∫e−u(du−2)= −12∫e−u du∫kf(x) dx = k∫f(x) dx= −12(1−1e−u)+ C∫ekx dx =1kekx + C=12e−u + C=12e−(−x2+2x) + C Substitute u by −x2 + 2x=12ex2−2x + C Distribute(i) Let u = 1 + ep, the quantity in the denominator. Then, du = ep dp and∫ep1 + epdp =∫1udu= ln |u|+ C∫1xdx = ln |x|+ C= ln |1 + ep|+ C Substitute u by 1 + ep(j) Let u = e2x+e−2x. Then du = (2e2x−2e−2x) dx since ddx[eg(x)]= eg(x)ddx[g(x)].So, du2= (e2x − e−2x) dx and∫e2x − e−2xe2x + e−2xdx =∫1e2x + e−2x(e2x − e−2x) dx=∫1u(du2)=12∫1udu∫kf(x) dx = k∫f(x) dx=12ln |u|+ C∫1xdx = ln |x|+ C=12ln∣∣e2x + e−2x∣∣+ C Substitute u by e2x + e−2x96 Chapter 1. Integration(k) Let u = ln(3x). Then, du = u′(x) dx = 33xdx = 1xdx sinceddx[ln(g(x))] =ddx[g(x)]g(x). So, du = 1xdx and∫ln(3x)xdx =∫ln(3x)(1xdx)=∫u du=u22+ C∫xn dx =xn+1n+ 1+ C=[ln(3x)]22+ C Substitute u by ln(3x)(l) Let u = ln z. Then, du = 1zdz and∫(ln z)2zdz =∫u2 du=u33+ C∫xn dx =xn+1n+ 1+ C=(ln z)33+ C Substitute u by ln z(m) Let u = ln t. Then, du = 1tdt and∫1t ln tdt =∫1ln t(1tdt)=∫1udu= ln |u|+ C∫1xdx = ln |x|+ C= ln | ln t|+ C Substitute u by ln t(n) The numerator of the integrand is an exponential function with base 5. Wemake the substitution u =√x, the exponent on 5. Then, du = 12√xdx so that2 du = 1√xdx. Therefore,∫5√x√xdx =∫5√x(1√xdx)=∫5u (2 du)= 2∫5u du∫kf(x) dx = k∫f(x) dx1.2. The Substitution Rule 97= 2(5uln 5)+ C∫ax dx =axln a+ C=2ln 55u + C=2 · 5√xln 5+ C Substitute u by√x(o) Firstly, ∫x√x+ 7 dx =∫x(x+ 7)12 dx√x = x12Now, let u = x+ 7. Then, du = dx and∫x√x+ 7 dx =∫x(x+ 7)12 dx From before=∫xu12 du=∫(u− 7)u 12 du Solve u = x + 7 for xto get x = u− 7=∫(u32 − 7u 12 ) du Distribute=∫u32 du−∫7u12 du∫(f(x) ± g(x)) dx =∫f(x) dx± ∫ g(x) dx=∫u32 du− 7∫u12 du∫kf(x) dx = k∫f(x) dx=u5252− 7u3232+ C∫xn dx =xn+1n+ 1+ C=25u52 − 7(23)u32 + Cabc=acb=25u52 − 143u32 + C=25(x+ 7)52 − 143(x+ 7)32 + C Substitute u by x+ 7(p) Firstly, ∫x√1 + 2xdx =∫x(1 + 2x)12dx√x = x1298 Chapter 1. Integration=∫x(1 + 2x)−12 dx1xm= x−mNow, let u = 1 + 2x. Then, du = 2 dx, so du2= dx and∫x√1 + 2xdx =∫x(1 + 2x)−12 dx From before=∫xu−12du2=12∫xu−12 du∫kf(x) dx = k∫f(x) dx=12∫ (u− 12)u−12 duSolve u = 1 + 2x for xto get x = u−12=14∫(u− 1)u− 12 du∫kf(x) dx = k∫f(x) dx=14∫ (u12 − u− 12)du Distribute=14(∫u12 du−∫u−12 du) ∫(f(x) ± g(x)) dx =∫f(x) dx± ∫ g(x) dx=14∫u12 du− 14∫u−12 du Distribute=14u3232− 14u1212+ C∫xn dx =xn+1n+ 1+ C=14(23)u32 − 14(21)u12 + Cabc=acb=16u32 − 12u12 + C=16(1 + 2x)32 − 12(1 + 2x)12 + C Substitute u by 1 + 2x3. [Exercise on page 7](a) Let u = 4x. Then, du = 4dx so that du4= dx and∫cos(4x) dx =∫cos(u)(du4)=14∫cos(u) du∫kf(x) dx = k∫f(x) dx=14sinu+ C∫cosu du = sinu+ C1.2. The Substitution Rule 99=14sin(4x) + C Substitute u by 4x(b) Let u = 9x. Then, du = 9dx so that du9= dx and∫sin(9x) dx =∫sin(u)(du9)=19∫sin(u) du∫kf(x) dx = k∫f(x) dx=19(− cosu) + C∫sinu du = − cosu+ C= −19cos(9x) + C Substitute u by 9x(c) Let u = −x2. Then, du = −12dx so that −2du = dx and∫sec2(−x2)dx =∫sec2(u) (−2du)= −2∫sec2(u) du∫kf(x) dx = k∫f(x) dx= −2 (tanu) + C∫sec2 u du = tanu+ C= −2 tan(−x2)+ C Substitute u by −x2(d) Let u = 5x2 + 1. Then, du = 10xdx so that du10= xdx and∫−7x sin(5x2 + 1) dx =∫−7 sin(5x2 + 1)(xdx)=∫−7 sin(u)(du10)= − 710∫sin(u)∫kf(x) dx = k∫f(x) dx= − 710(− cosu) + C∫sinu du = − cosu+ C=710cos(5x2 + 1) + C Substitute u by 5x2 + 1100 Chapter 1. Integration(e) Let u = x3 + 3x. Thendu = (3x2 + 3)dxdu = 3(x2 + 1)dx Factor out 3du3= (x2 + 1)dx Divide both sides by 3So ∫(x2 + 1) csc2(x3 + 3x) dx=∫csc2(x3 + 3x)(x2 + 1)dx=∫csc2(u)(du3)=13∫csc2(u) du∫kf(x) dx = k∫f(x) dx=13(− cotu) + C∫csc2 u du = − cotu+ C= −13cotu+ C= −13cot(x3 + 3x) + C Substitute u by x3 + 3x(f) Let u = e−2x. Thendu = −2e−2xdx ddx[eg(x)]= eg(x)ddx[g(x)]du−2 = e−2xdx Divide both sides by −2So ∫e−2x sec(e−2x) tan(e−2x) dx=∫sec(e−2x) tan(e−2x)(e−2xdx)=∫sec(u) tan(u)(du−2)= −12∫secu tanu du∫kf(x) dx = k∫f(x) dx1.2. The Substitution Rule 101= −12secu+ C∫secu tanu du = secu+ C= −12sec(e−2x)+ C Substitute u by e−2x4. [Exercise on page 7](a) Let u = sinx. Then, du = cosxdx and∫cosx√sinx dx =∫ √sinx (cosxdx) =∫ √u du=∫u12 du√x = x12=u3232+ C∫xn dx =xn+1n+ 1+ C=23u32 + Cabc=acb=23(sinx)32 + C Substitute u by sinx(b) Let u = − cosx. Then, du = −(− sinx)dx = sinxdx and∫sinx e− cosx dx =∫e− cosx(sinxdx) =∫eudu= eu∫ex dx = ex + C= e− cosx Substitute u by − cosx(c) Let u = sinx. Then, du = cosxdx and∫sin2 x cosx dx =∫(sinx)2(cosxdx) =∫u2du=u33+ C∫xn dx =xn+1n+ 1+ C=sin3 x3+ C Substitute u by sinx(d) Let u = 1 + tanx. Then, du = (0 + sec2 x)dx = sec2 xdx and∫sec2 x(1 + tanx)3dx =∫1(1 + tan x)3(sec2 xdx)102 Chapter 1. Integration=∫1u3du=∫u−3 du1xm= x−m=u−3+1−3 + 1 + C∫xn dx =xn+1n+ 1+ C=u−2−2 + C= − 12u2+ C1xm= x−m=−12(1 + tan x)2+ C Substitute u by 1 + tan x(e) Let u = cotx. Then, du = − csc2 xdx. So −du = csc2 xdx and∫cot2 x csc2 x dx=∫(cotx)2(csc2 xdx)=∫u2(−du) = −∫u2 du∫kf(x) dx = k∫f(x) dx= −u33+ C∫xn dx =xn+1n+ 1+ C= −cot3 x3+ C Substitute u by cotx(f) Firstly, we rewrite the integral as∫tan(3x) dx =∫sin(3x)cos(3x)dx tanx =sinxcosxNow, let u = cos(3x). Thendu = −3 sin(3x)dx ddx[cos(g(x))] = − sin(g(x)) ddx[g(x)]So du−3 = sin(3x)dx and the integral becomes∫tan(3x) dx =∫sin(3x)cos(3x)dx =∫1cos(3x)[sin(3x)dx]1.2. The Substitution Rule 103=∫1u(du−3)= −13∫1udu∫kf(x) dx = k∫f(x) dx= −13ln |u|+ C∫1xdx = ln |x|+ C= −13ln | cos(3x)|+ C Substitute u by cos(3x)(g) Firstly, we rewrite the integral as∫cot(θ4)dθ =∫cos( θ4)sin( θ4)dθ cotx =cosxsinxNow, let u = sin( θ4). Thendu =14cos(θ4)dθddx[sin(g(x))] = cos(g(x))ddx[g(x)]So 4du = cos( θ4)dθ and the integral becomes∫cot(θ4)dθ =∫cos( θ4)sin( θ4)dθ =∫1sin( θ4)[cos(θ4)dθ]=∫1u(4 du)= 4∫1udu∫kf(x) dx = k∫f(x) dx= 4 ln |u|+ C∫1xdx = ln |x|+ C= 4 ln∣∣∣∣sin(θ4)∣∣∣∣+ C Substitute u by sin(θ4)(h)Some Trigonometric IdentitiesFor every real number x,cos2 x =1 + cos(2x)2and sin2 x =1− cos(2x)2104 Chapter 1. Integration∫cos2 x dx=∫1 + cos(2x)2dx cos2 x =1 + cos(2x)2=∫ (12+12cos(2x))dxa+ bc=ac+bc=∫12dx+12∫cos(2x) dx=12x+12(12sin(2x))+ C∫cos(2x) dx = 12sin(2x) + C byusing the substitution u = 2x=12x+14sin(2x) + C(i) Firstly, we rewrite the integral as∫cos3 x tanx dx =∫cos3 x(sinxcosx)dx tanx =sinxcosx=∫cos2 x sinx dx SimplifyNow, let u = cosx. Then du = − sinxdx, so that −du = sinxdx, and∫cos2 x sinx dx =∫(cosx)2 (sinxdx)=∫u2(−du) = −∫u2 du∫kf(x) dx = k∫f(x) dx= −u33+ C∫xn dx =xn+1n+ 1+ C= −cos3 x3+ C Substitute u by cosx(j) Firstly, we rewrite the integral as∫cosxsinx+ cscxdx =∫cosxsinx+ 1sinxdx cscx =1sinx=∫cosxsin2 x+1sinxdx a+bc=ac+ bc1.2. The Substitution Rule 105=∫sinx cosx1 + sin2 xdxabc=acbNow, let u = 1 + sin2 x = 1 + (sinx)2. Thendu =(0 + 2(sin x)1ddx[sinx])dxddx[(g(x))n] = n(g(x))n−1ddx[g(x)]du = 2 sin x cosxdxddx[sinx] = cos xdu2= sinx cosxdx Divide both sides by 2So∫sinx cosx1 + sin2 xdx =∫11 + sin2 x(sinx cosx dx)=∫1u(du2)=12∫1udu∫kf(x) dx = k∫f(x) dx=12ln |u|+ C∫1xdx = ln |x|+ C=12ln |1 + sin2 x|+ C Substitute u by 1 + sin2 x5. [Exercise on page 7](a) Firstly, we rewrite the integral as∫11 + 9x2dx =∫11 + (3x)2dxNow, let u = 3x. Then du = 3dx, so that du3= dx, and∫11 + 9x2dx =∫11 + (3x)2dx=∫11 + u2(du3)=13∫11 + u2du∫kf(x) dx = k∫f(x) dx=13tan−1(u) + C∫11 + u2du = tan−1 u+ C106 Chapter 1. Integration=13tan−1(3x) + C Substitute u by 3x(b) First, we rewrite the integral as∫1√16− 25x2 dx =∫1√16(1− 25x216) dx = ∫ 14√1− (5x4)2 dx=14∫1√1− (5x4)2 dxLet u = 5x4. Then, du = 54dx so that 4du5= dx. The integral above becomes14∫1√1− (5x4)2 dx = 14∫1√1− u2(4du5)=14(45)∫1√1− u2 du∫kf(x) dx = k∫f(x) dx=15sin−1(u) + C∫1√1− u2 du = sin−1 u+ C=15sin−1(5x4)+ C Substitute u by5x4(c) ∫e3x1 + e6xdx=∫e3x1 + (e3x)2dx (an)m = anm=∫11 + (e3x)2(e3xdx)=∫11 + u2(du3)u = e3x, du = 3e3xdx, sothat du3= e3xdx=13∫11 + u2du∫kf(x) dx = k∫f(x) dx=13tan−1(u) + C∫11 + u2du = tan−1 u+ C=13tan−1(e3x)+ C Substitute u by e3x1.2. The Substitution Rule 107(d) Firstly, we rewrite the integral as∫tanxcosx+ secxdx =∫tanxcosx+ 1cosxdx secx =1cosx=∫tanxcos2 x+1cosxdx a+bc=ac+ bc=∫cosx tanx1 + cos2 xdxabc=acb=∫cosx(sinxcosx)1 + cos2 xdx tanx =sinxcosx=∫sinx1 + (cos x)2dx SimplifyNow, let u = cosx. Then du = − sinxdx, so that −du = sinxdx, and∫sinx1 + (cos x)2dx =∫11 + u2(−du)= −∫11 + u2du∫kf(x) dx = k∫f(x) dx= − tan−1(u) + C∫11 + u2du = tan−1 u+ C= − tan−1(cosx) + C Substitute u by cosx6. [Exercise on page 8](a) The population after t months isP (t) =∫P ′(t) dt =∫400te−t24 dtTo find this integral, we make the substitution u = − t24, so thatdu = −2t4dtdu = − t2dt−2du = tdt Multiply both sides by −2Substituting these into the integral, we getP (t) =∫400te−t24 dt =∫400eu (−2du)108 Chapter 1. Integration= −800∫eu du∫kf(x) dx = k∫f(x) dx= −800eu + C∫ex dx = ex + C= −800e− t24 + CWe now find C. Since the initial population is 1500, it follows that P (0) = 1500.This implies that−800e− 024 + C = 1500−800 + C = 1500 e0 = 1C = 1500 + 800 = 2300So, the population after t months is given by the formulaP (t) = −800e− t24 + 2300(b) The size of the population after 5 months isP (5) = −800e− 524 + 2300 ≈ 2298.467. [Exercise on page 8]Firstly, we need to find N(t).N(t) =∫N ′(t)dt =∫200tt2 + 5dt= 200∫1t2 + 5(tdt)∫kf(x) dx = k∫f(x) dx= 200∫1u(du2)u = t2 + 5, du = 2tdt, sothat du2= tdt= 200(12)∫1udu∫kf(x) dx = k∫f(x) dx= 100 ln |u|+ C∫1xdx = ln |x|+ C= 100 ln |t2 + 5|+ C Substitute u by t2 + 5Now, we find C. Since N(0) = 180, it follows that1.3. The Definite Integral 109100 ln |(0)2 + 5|+ C = 180100 ln |5|+ C = 180C = 180− 100 ln |5|So, N(t) = 100 ln |t2 + 5|+ 180− 100 ln |5|. And the number of people infected after39 days isN(39) = 100 ln |392 + 5|+ 180− 100 ln |5| ≈ 7521.3 The Definite Integral1. [Exercise on page 8]Left Endpoints, Right Endpoints, and Midpoints The left endpoint of an interval [a, b] is a. For example, the left endpointof [1, 5] is 1. The right endpoint of an interval [a, b] is b. For example, the rightendpoint of [1, 5] is 5. The midpoint of an interval [a, b] isa+ b2. For example, the midpointof [1, 5] is1 + 52= 3.Approximating the Area Under a CurveLet f be a function defined on an interval [a, b] such that f(x) ≥ 0 for all x in[a, b]. Let S be the region under the graph of f , above the x-axis, and betweenx = a and x = b. The following figure (where f(x) = x2, a = 0, b = 1) is anillustration of S.11y = x2S0 xyTo find an approximation of the area of S by taking n subdivisions with left110 Chapter 1. Integrationendpoints, one can proceed as follows.Step 1. Subdivide the interval [a, b] into n subintervals, each of width∆x =b− anStep 2. Find the left endpoint of each subinterval. Let [ai−1, ai] denote theith subinterval. Thena0 = a, a1 = a0 + ∆x, a2 = a1 + ∆x, · · · , an = an−1 + ∆x = bLet xi be the left endpoint of [ai−1, ai]. Then xi = ai−1.Step 3. Evaluate the function f at xi for each i.Step 4. An approximation of the area of S by taking n subdivisionswith left endpoints is given by the sumn∑i=1f(xi)∆x = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆xThis is illustrated by the following figure.0.25 0.5 0.75 10.250.50.751y = x2n = 40 xyThe left endpoint approximation is0.25f(0) + 0.25f(0.25) + 0.25f(0.5) + 0.25f(0.75) = 0.21875Now, if each xi is the right endpoint of the subinterval [ai−1, ai] (that is,1.3. The Definite Integral 111xi = ai), the sumn∑i=1f(xi)∆x is called an approximation of the area of Sby taking n subdivisions with right endpoints. This is illustrated by thefollowing figure.0.25 0.5 0.75 10.250.50.751y = x2n = 40 xyThe right endpoint approximation is0.25f(0.25) + 0.25f(0.5) + 0.25f(0.75) + 0.25f(1) = 0.46875If each xi is the midpoint of the subinterval [ai−1, ai] (that is, xi =ai−1 + ai2),the sumn∑i=1f(xi)∆x is called an approximation of the area of S by takingn subdivisions with midpoints. This is illustrated by the following figure.0.25 0.5 0.75 10.250.50.751y = x2n = 40 xyThe midpoint approximation is0.25f(0.125) + 0.25f(0.375) + 0.25f(0.625) + 0.25f(0.875) = 0.328125112 Chapter 1. IntegrationNote: The midpoint rule is more accurate than the left endpoint and rightendpoint rules.Let’s find an approximation of the area between the function f(x) = x2 + 3, thex-axis, x = 0, and x = 1 by taking n = 4 subdivisions with left endpoints. Herea = 0 and b = 1.Step 1. We subdivide the interval [0, 1] into 4 subintervals, each of width∆x =b− an=1− 04=14We havea0 = a = 0 a3 = a2 + ∆x =12+14=34a1 = a0 + ∆x = 0 +14=14a4 = a3 + ∆x =34+14= 1 = ba2 = a1 + ∆x =14+14=12So,First Subinterval: [a0, a1] =[0,14]Second Subinterval: [a1, a2] =[14,12]Third Subinterval: [a2, a3] =[12,34]Fourth Subinterval: [a3, a4] =[34, 1]Step 2.Left Endpoint of[0,14]: x1 = 0Left Endpoint of[14,12]: x2 =14Left Endpoint of[12,34]: x3 =12Left Endpoint of[34, 1]: x4 =341.3. The Definite Integral 113Step 3. Evaluating the function at the left endpoints, we getf(x1) = f (0) = 02 + 3 = 3f(x2) = f(14)=(14)2+ 3 =116+ 3 =4916f(x3) = f(12)=(12)2+ 3 =14+ 3 =134f(x4) = f(34)=(34)2+ 3 =916+ 3 =5716Step 4. An approximation of the area between the function f(x) = x2+3, the x-axis,x = 0, and x = 1 by taking n = 4 subdivisions with left endpoints is4∑i=1f(xi)∆x = f(x1)∆x+ f(x2)∆x+ f(x3)∆x+ f(x4)∆x= 3(14)+4916(14)+134(14)+5716(14)=4864+4964+5264+5764=20664= 3.21875All this can be summarized in the following table.Intervals Left endpoints xi f(xi) f(xi)∆x[0, 14]0 3 34[14, 12]1449164964[12, 34]121341316[34, 1]3457165764Sum 20664= 3.218752. [Exercise on page 8]Here, a = 1, b = 3, and n = 4. The width of each subinterval is∆x =b− an=3− 14=24=12.114 Chapter 1. IntegrationTo subdivide the interval [1, 3] into 4 subintervals, we need to find ai for each i.a0 = a = 1 a3 = a2 + ∆x = 2 +12=52a1 = a0 + ∆x = 1 +12=32a4 = a3 +12=52+12= 3 = ba2 = a1 + ∆x =32+12= 2The intervals, the right endpoints xi, and the values of the function at xi are shownin the following table.Intervals Right endpoints xi f(xi) f(xi)∆x[1, 32]322313[32, 2]2 1214[2, 52]522515[52, 3]3 1316Sum 5760= 0.95So, an approximation of the area using right endpoints is4∑i=1f(xi)∆x = f(x1)∆x+ f(x2)∆x+ f(x3)∆x+ f(x4)∆x=23(12)+12(12)+25(12)+13(12)=13+14+15+16=2060+1560+1260+1060=5760= 0.953. [Exercise on page 8]Here n = 5, a = −2, and b = 3 so that∆x =b− an=3− (−2)5=3 + 25= 1.1.3. The Definite Integral 115The ai’s area0 = a = −2 a3 = 0 + 1 = 1a1 = −2 + 1 = −1 a4 = 1 + 1 = 2a2 = −1 + 1 = 0 a5 = 2 + 1 = 3 = bThe intervals, the midpoints xi, the values of the function at xi, and the sum5∑i=1f(xi)∆x are shown in the following table.Intervals Midpoints xi f(xi) f(xi)∆x[−2,−1] −32e−32 + 1 e−32 + 1 ≈ 1.22313[−1, 0] −12e−12 + 1 e−12 + 1 ≈ 1.60653[0, 1] 12e12 + 1 e12 + 1 ≈ 2.64872[1, 2] 32e32 + 1 e32 + 1 ≈ 5.48168[2, 3] 52e52 + 1 e52 + 1 ≈ 13.18249Sum ≈ 24.14255Thus, an approximation of the area using midpoints is 24.14255 .4. [Exercise on page 9]Riemann SumLet f be a function defined on an interval [a, b] (f(x) need not be greater thanor equal to 0). Divide the interval [a, b] into n equal subintervals[a0, a1], [a1, a2], · · · [an−1, an]The width of each subinterval is ∆x =b− an. Let xi be an arbitrary point inthe ith subinterval [ai−1, ai].The sumn∑i=1f(xi)∆x = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆xis called a Riemann sum of f on the interval [a, b].116 Chapter 1. IntegrationFrom the problem, a = 0, b = 2, and n = 4. The width is∆x =2− 04=12The ai’s area0 = a = 0 a3 = 1 +12=32a1 = 0 +12=12a4 =32+12= 2 = ba2 =12+12= 1The intervals, the right endpoints xi, the values of the function at xi, and the sum4∑i=1f(xi)∆x are shown in the following table.Intervals Right endpoints xi f(xi) f(xi)∆x[0, 12]12−34−38[12, 1]1 0 0[1, 32]325458[32, 2]2 3 32Sum 74= 1.75Therefore, the Riemann sum of f(x) = x2 − 1, 0 ≤ x ≤ 2, obtained by taking fourrectangles with right endpoints is 1.75 .5. [Exercise on page 9](a)Definite IntegralsLet f be a function defined on an interval [a, b] (f(x) need not be greaterthan or equal to 0). Divide the interval [a, b] into n equal subintervals[a0, a1], [a1, a2], · · · [an−1, an]1.3. The Definite Integral 117The width of each subinterval is ∆x =b− an. Let xi be an arbitrary pointin the ith subinterval [ai, ai+1]. The definite integral of f from a to b,denoted∫ baf(x) dx, is given by∫ baf(x) dx = limn→∞n∑i=1f(xi)∆x, (1.3.1)provided the limit exists. If the limit exists, we say that f is (Riemann)Integrable.∫is the integral symbol. f(x) is called the integrand. a and b are called the limits of integration or bounds: a is called thelower limit (or lower bound) and b is called the upper limit (orupper bound). dx simply indicates that the variable is x.The sumn∑i=1f(xi)∆x = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆xgives an approximation to the definite integral∫ baf(x) dx.Note. When f(x) ≥ 0 for all x in [a, b], the definite integral∫ baf(x) dxgives the exact area of the region under the graph of f(x), above thex-axis, and between x = a and x = b. The case f(x) ≤ 0 will be discussedin a later section.0.25 0.5 0.75 10.250.50.751y = x2xy118 Chapter 1. IntegrationThe area of the shaded region is∫ 10x2 dx.Let’s approximate the definite integral∫ 51√x− 1 dx by using four rectangles,the midpoints, and f(x) =√x− 1. In this case, a = 1, b = 5, n = 4, and∆x =5− 14= 1The ai’s area1 = a = 1 a4 = 3 + 1 = 4a2 = 1 + 1 = 2 a5 = 4 + 1 = 5 = ba3 = 2 + 1 = 3An approximation of∫ 51√x− 1 dx is given in the following table.Intervals Midpoints xi f(xi) f(xi)∆x[1, 2] 32√12√12≈ 0.70710[2, 3] 52√32√32≈ 1.22474[3, 4] 72√52√52≈ 1.58113[4, 5] 92√72√72≈ 1.87082Sum ≈ 5.38379Thus, an approximation of∫ 51√x− 1 dx is 5.38379 .(b)Here a = 3, b = 11, n = 4, and∆x =11− 34= 2So,a0 = a = 3 a3 = 9a1 = 5 a4 = 11 = ba2 = 71.3. The Definite Integral 119An approximation of∫ 113(1− x2) dx is given in the following table.Intervals Midpoints xi f(xi) f(xi)∆x[3, 5] 4 −15 −30[5, 7] 6 −35 −70[7, 9] 8 −63 −126[9, 11] 10 −99 −198Sum −424Therefore, an approximation of∫ 113(1− x2) dx is −424 .6. [Exercise on page 9]Definite Integrals as Limits of Riemann SumsIn the equation (1.3.1), xi is an arbitrary point in the ith subinterval. Takingthe xi’s as the right endpoints, we get the following theorem.Theorem 1.3.1. Let f be a function. Suppose that f is integrable on aninterval [a, b]. Then ∫ baf(x) dx = limn→∞n∑i=1f(xi)∆x,where ∆x = b−anandxi = a+ i∆x, 1 ≤ i ≤ n (1.3.2)A similar theorem can be stated by taking the xi’s as the left endpoints or themidpoints. We choose the right endpoints essentially because they often makethe computations easier.(a) We want to express∫ 135x4 dx as a limit of Riemann sums. Since the integral istaken from 5 to 13, it follows that a = 5 and b = 13. Let f(x) = x4. Then∆x =b− an=13− 5n=8nand xi = a+ i∆x = 5 +8in120 Chapter 1. IntegrationUsing Theorem 1.3.1, we have∫ 135x4 dx = limn→∞n∑i=1f(xi)∆x= limn→∞n∑i=1(xi)4 · 8nf(xi) = (xi)4= limn→∞n∑i=1(5 +8in)4· 8nxi = 5 +8in(b) For the integral∫ −1−3 x2e−x dx, we have a = −3, b = −1,∆x =−1− (−3)n=−1 + 3n=2nand xi = a+ i∆x = −3 + 2inSo, ∫ −1−3x2e−x dx = limn→∞n∑i=1f(xi)∆x= limn→∞n∑i=1(xi)2e−xi · 2n= limn→∞n∑i=1(−3 + 2in)2e−(−3+2in ) · 2n7. [Exercise on page 9](a)limn→∞n∑i=1ln(xi)xi∆x = limn→∞n∑i=1f(xi)∆x f(xi) =ln(xi)xi=∫ 31f(x) dx f(x) =lnxx, [a, b] = [1, 3]=∫ 31lnxxdx(b)limn→∞n∑i=1(2xi − 3xi cosxi)∆x = limn→∞n∑i=1f(xi)∆x f(xi) = 2xi − 3xi cosxi1.3. The Definite Integral 121=∫ pi0f(x) dxf(x) = 2x − 3x cosx,[a, b] = [0, pi]=∫ pi0(2x− 3x cosx) dx8. [Exercise on page 9](a) We want to find a, b, and f(x) such thatlimn→∞n∑i=1(in)3· 1n=∫ baf(x) dxFirstly, we haven∑i=1(in)3· 1n= limn→∞n∑i=1(xi)3∆x,where xi =inand ∆x = 1n. Rewriting xi, we getxi =in= i · 1n= i∆xOn the other hand, xi is given by the formula xi = a+ i∆x (see (1.3.2)). So,a+ i∆x = i∆xa = 0 Subtract i∆x from both sidesNow, since ∆x = 1nand since ∆x is given by the formula ∆x = b−an, it followsthatb− 0n=1nb = 1 Multiply both sides by nSo a = 0, b = 1, andn∑i=1(in)3· 1n= limn→∞n∑i=1(xi)3∆x= limn→∞n∑i=1f(xi)∆x f(xi) = (xi)3=∫ 10f(x) dx f(x) = x3=∫ 10x3 dx122 Chapter 1. Integration(b) We want to find a, b, and f(x) such thatlimn→∞n∑i=1(2inln(1 +2in))· 2n=∫ baf(x) dxFirstly, we haven∑i=1(2inln(1 +2in))· 2n= limn→∞n∑i=1xi ln(1 + xi)∆x,where xi =2inand ∆x = 2n. Rewriting xi, we getxi =2in= i · 2n= i∆xOn the other hand, xi is given by the formula xi = a+ i∆x (see (1.3.2)). So,a+ i∆x = i∆xa = 0 Subtract i∆x from both sidesNow, since ∆x = 2nand since ∆x is given by the formula ∆x = b−an, it followsthatb− 0n=2nb = 2 Multiply both sides by nSo a = 0, b = 2, andn∑i=1(2inln(1 +2in))· 2n= limn→∞n∑i=1xi ln(1 + xi)∆x xi =2in,∆x =2n= limn→∞n∑i=1f(xi)∆x f(xi) = xi ln(1 + xi)=∫ 20f(x) dxf(x) = x ln(1 + x),a = 0, b = 2=∫ 20x ln(1 + x) dx9. [Exercise on page 9](a)1.3. The Definite Integral 123By definition, the definite integral∫ 302x dx is the exact area of the region Sunder the graph of f(x) = 2x, above the x-axis, and between x = 0 and x = 3.The region S is illustrated in the following figure.f(x) = 2xSx1 2 3 4y1234560Using the formula for the area of a triangle, A = 12BH (where B is the lengthof the base and H is the height), the exact area of S isA =12BH =12(3)(6) = 9.So,∫ 302x dx = 9 .(b)The definite integral∫ 40(4−x) dx is the exact area of the region S in the followingfigure.y = 4− xx1 2 3 4y1234S0124 Chapter 1. IntegrationThus, ∫ 40(4− x) dx = 12(4)(4) = 8(c)The definite integral∫ 0−2√4− x2 dx is the exact area of the region S in thefollowing figure.y =√4− x2Sx−2 −1 1 2y−1120Notice that the area of S is the quarter of the area of the circle of radius r = 2.Well, the area of a circle with radius r = 2 ispir2 = pi(2)2 = 4piIt thus follows that the area of S is4pi4= pi so that∫ 0−2√4− x2 dx = pi10. [Exercise on page 10]By definition, the definite integral∫ 51f(x) dx is the exact area of the region S shownin the following figure.xy0 1 512SUsing the formula for the area of a triangle with base B = 5 − 1 = 4 and height1.3. The Definite Integral 125H = 2, we get ∫ 51f(x) dx =12BH =12(4)(2) = 411. [Exercise on page 10]By definition, the integral∫ 4−3 f(x)dx is the area of the region underneath the graphof f over the interval [−3, 4]; that is, the area of S1 plus the area of S2, where S1 andS2 are the regions shown in the following figure.xyy = f(x)S1S2−3 −1 0 445 Finding the area of S1. The region S1 is a trapezoid withlong base = 5, short base = 2, and height = 2Using the formula for the area of a trapezoid, we havearea(S1) =(long base + short base)height2=(5 + 2)22= 7 Finding the area of S2. The region S2 is also a trapezoid withlong base = 4, short base = 2, and height = 5So,area(S2) =(long base + short base)height2=(4 + 2)52= 15Adding up these two areas, we get the area underneath the graph of y = f(x) over126 Chapter 1. Integrationthe interval [−3, 4]:∫ 4−3f(x) dx = area(S1) + area(S2) = 7 + 15 = 221.4 The Fundamental Theorem of Calculus1. [Exercise on page 10]The Fundamental Theorem of Calculus (Part I)The Fundamental Theorem of Calculus (FTC) is one of the most powerfulresults in Calculus. It has two parts: the first part says that differentiation andintegration are inverse processes, and the second part provides a method forevaluating definite integrals.Theorem 1.4.1. [FTC (Part I)] Let f be a function and let [a, b] be aninterval. Suppose that f is continuous on [a, b], and let F be the functiondefined byF (x) =∫ xaf(t) dt a ≤ x ≤ bThen F is differentiable on (a, b) and F ′(x) = f(x) for every x in (a, b); thatis,ddx[∫ xaf(t) dt]= f(x) a < x < b(a) Let f(t) = t3. Using the FTC (part I), we getdydx=ddx[∫ x0f(t) dt]= f(x) = x3(b) Let f(t) = 5 sin t. Using the FTC (part I), we finddydx=ddx[∫ x1f(t) dt]= f(x) = 5 sinx(c) Let f(t) = 11+t4. Using the FTC (part I), we finddydx=ddx[∫ x−2f(t) dt]= f(x) =11 + x4(d) For the function y =∫ x20(1 + e3t) dt, the upper bound of integration is x2, not1.4. The Fundamental Theorem of Calculus 127x. So, we can’t use the FTC (part I). Instead, we will use a more general result.Leibniz’s Rule for IntegralsLet g(x) and h(x) be differentiable functions. Let y = f(t) be a functionthat is continuous at every t between g(x) and h(x). Thenddx[∫ h(x)g(x)f(t) dt]= f [h(x)]ddx[h(x)]− f [g(x)] ddx[g(x)]Proof. Let c be a number between g(x) and h(x). Then, using the prop-erties∫ baf(t) dt =∫ caf(t) dt+∫ bcf(t) dt and∫ baf(t) dt = −∫ abf(t) dtwe have ∫ h(x)g(x)f(t) dt =∫ cg(x)f(t) dt+∫ h(x)cf(t) dt= −∫ g(x)cf(t) dt+∫ h(x)cf(t) dt For the integral y = − ∫ g(x)cf(t) dt, let u = g(x). Then, y =− ∫ ucf(t) dt and by the chain rule,dydx=dydududx=ddu[y]ddx[u]=ddu[−∫ ucf(t) dt]ddx[g(x)]= − ddu[∫ ucf(t) dt]ddx[g(x)]= −f(u) ddx[g(x)] FTC (Part I)= −f [g(x)] ddx[g(x)] Likewise,ddx[∫ h(x)cf(t) dt]= f [h(x)]ddx[h(x)]128 Chapter 1. IntegrationSo,ddx[∫ h(x)g(x)f(t) dt]=ddx[∫ h(x)cf(t) dt−∫ g(x)cf(t) dt]=ddx[∫ h(x)cf(t) dt]− ddx[∫ g(x)cf(t) dt]= f [h(x)]ddx[h(x)]− f [g(x)] ddx[g(x)]This ends the proof.We come back to the function y =∫ x20(1 + e3t) dt. Let f(t) = 1 + e3t, g(x) = 0,and h(x) = x2. Using the Leibniz rule, we haveddx[∫ h(x)g(x)f(t) dt]= f [h(x)]ddx[h(x)]− f [g(x)] ddx[g(x)]= f(x2)ddx[x2]− f(0) ddx[0]= (1 + e3x2)(2x)− (1 + e3(0)) (0)= 2x(1 + e3x2)− 0 = 2x(1 + e3x2)(e) For the function y =∫ √x2x+17 sec t dt, we have f(t) = 7 sec t, g(x) = 2x + 1, andh(x) =√x. Using the Leibniz rule, we findddx[∫ h(x)g(x)f(t) dt]= f [h(x)]ddx[h(x)]− f [g(x)] ddx[g(x)]= f(√x) ddx[√x]− f(2x+ 1) ddx[2x+ 1]= 7 sec(√x) · 12√x− 7 sec(2x+ 1)(2) ddx[√x] =12√x=7 sec(√x)2√x− 14 sec(2x+ 1)(f)dydx=ddx[∫ x+x41−x2t5 dt]1.4. The Fundamental Theorem of Calculus 129=(x+ x4)5 ddx[x+ x4]− (1− x2)5 ddx[1− x2] Leibniz’s Rule=(x+ x4)5(1 + 4x3)− (1− x2)5 (−2x)=(1 + 4x3) (x+ x4)5+ 2x(1− x2)52. [Exercise on page 11](a)The Fundamental Theorem of Calculus (Part II)The definite integral is defined as the limit of the sum of the areas ofrectangles as the number of rectangles approaches infinity. This definitionis hard to work with. Fortunately, there is an easier way of evaluating adefinite integral with the following remarkable result which connects thedefinite integral to antidifferentiation.Theorem 1.4.2. [FTC (Part II)] Let f be a function. Suppose that fis continuous on an interval [a, b]. If F is any antiderivative of f , then∫ baf(x) dx = F (b)− F (a)Note: The difference F (b)− F (a) is denoted by F (x)∣∣∣∣ba. So,∫ baf(x) dx = F (x)∣∣∣∣ba= F (b)− F (a)Let’s evaluate∫ 813 dx. We first need to find an antiderivative or the indefiniteintegral of f(x) = 3. We get∫3 dx = 3x+ C∫k dx = kx+ CSo, an antiderivative of the function f(x) = 3 is F (x) = 3x. Using the Funda-mental Theorem of Calculus, we find∫ 813 dx = 3x∣∣∣∣81∫ baf(x) dx = F (x)∣∣∣∣ba130 Chapter 1. Integration= F (8)− F (1) F (x)∣∣∣∣ba= F (b)− F (a)= 3(8)− 3(1)= 24− 3 = 21The Constant C is not needed for definite integralsLet F (x) be an antiderivative of f(x). Then for any constant C, F (x)+Cis also an antiderivative of f(x). By the fundamental theorem of calculus,we have∫ baf(x) dx = (F (x) + C)∣∣∣∣ba= (F (b) + C)− (F (a) + C)= F (b) + C − F (a)− C C − C = 0= F (b)− F (a)= F (x)∣∣∣∣baSo,(F (x) + C)∣∣∣∣ba= F (x)∣∣∣∣baThis shows that we do not need to add the constant C when evaluating adefinite integral.(b)∫ 1−35 dx = 5x∣∣∣∣1−3∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba= 5(1)− 5(−3) F (x)∣∣∣∣ba= F (b)− F (a)= 5 + 15 = 201.4. The Fundamental Theorem of Calculus 131(c)∫ 10x3 dx =x44∣∣∣∣10∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=144− 044=14F (x)∣∣∣∣ba= F (b)− F (a)(d)∫ 941√xdx =∫ 941x12dx√x = x12=∫ 94x−12 dx1xm= x−m=x1212∣∣∣∣94∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba= 2√x∣∣∣∣94abc=acb,√x = x12= 2√9− 2√4 F (x)∣∣∣∣ba= F (b)− F (a)= 2(3)− 2(2) = 6− 4 = 2(e)∫ 413x dx = 3∫ 41x dx∫ bakf(x) dx = k∫ baf(x) dx= 3x22∣∣∣∣41∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=32x2∣∣∣∣41=32(4)2 − 32(1)2 F (x)∣∣∣∣ba= F (b)− F (a)132 Chapter 1. Integration=32(16)− 32=482− 32=452(f)Properties of Definite IntegralsLet f and g be continuous on [a, b].(i)∫ bakf(x) dx = k∫ baf(x) dx for any constant k.(ii)∫ ba[f(x) + g(x)] dx =∫ baf(x) dx+∫ bag(x) dx.(iii)∫ ba[f(x)− g(x)] dx =∫ baf(x) dx−∫ bag(x) dx.Using the first property, we have∫ 21(−x2 + 3x− 1) dx =(−x33+32x2 − x)∣∣∣∣21∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=(−233+32(2)2 − 2)−(−133+32(1)2 − 1)F (x)∣∣∣∣ba= F (b)− F (a)=(−83+ 6− 2)−(−13+32− 1)=(−83+ 4)−(−13+12)=(43)−(16)=86− 16=761.4. The Fundamental Theorem of Calculus 133Alternate Solution∫ 21(−x2 + 3x− 1) dx =∫ 21−x2 dx+∫ 213x dx−∫ 211 dx= −∫ 21x2 dx+ 3∫ 21x dx−∫ 211 dx= −x33∣∣∣∣21+ 3x22∣∣∣∣21− x∣∣∣∣21= −(83− 13)+ 3(42− 12)− (2− 1)= −(73)+ 3(32)− (1) = −73+92− 1= −146+276− 66=−14 + 27− 66=76(g) ∫ 21(1x2− 4x3)dx =∫ 21(x−2 − 4x−3) dx 1xm= x−m=∫ 21x−2 dx−∫ 214x−3 dx∫ ba[f(x)± g(x)] dx =∫ baf(x) dx±∫ bag(x) dx=∫ 21x−2 dx− 4∫ 21x−3 dx∫ bakf(x) dx = k∫ baf(x) dx=x−1−1∣∣∣∣21− 4x−2−2∣∣∣∣21∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba= −1x∣∣∣∣21+ 2x−2∣∣∣∣211xm= x−m= −1x∣∣∣∣21+ 21x2∣∣∣∣211xm= x−m= −(12− 11)+ 2(14− 11)F (x)∣∣∣∣ba= F (b)− F (a)134 Chapter 1. Integration= −(−12)+ 2(−34)=12− 32=−22= −1(h)∫ 3−2(x2 − 3) dx =(x33− 3x)∣∣∣∣3−2∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=(333− 3(3))−((−2)33− 3(−2))F (x)∣∣∣∣ba= F (b)− F (a)=(273− 9)−(−83+ 6)= (9− 9)−(−8 + 183)= 0−(103)= −103(i) ∫ 414 + 6x√xdx =∫ 41(4x12+6xx12)dxa+ bc=ac+bc,√x = x12=∫ 41(4x−12 + 6x12)dx1xm= x−m=∫ 414x−12 dx+∫ 416x12 dx∫ ba[f(x)± g(x)] dx =∫ baf(x) dx±∫ bag(x) dx= 4∫ 41x−12 dx+ 6∫ 41x12 dx∫ bakf(x) dx = k∫ baf(x) dx= 4x1212∣∣∣∣∣41+ 6x3232∣∣∣∣∣41∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba= 4(21)x12∣∣∣∣41+ 6(23)x32∣∣∣∣41abc=acb1.4. The Fundamental Theorem of Calculus 135= 8x12∣∣∣∣41+ 4(x12 )3∣∣∣∣41= 8√x∣∣∣∣41+ 4(√x)3∣∣∣∣41√x = x12= 8(√4−√1)+ 4((√4)3−(√1)3)F (x)∣∣∣∣ba= F (b)− F (a)= 8(2− 1) + 4(23 − 13)= 8(1) + 4(8− 1)= 8 + 4(7) = 8 + 28 = 36(j) ∫ 812 + t3√t2dt =∫ 81(2t23+tt23)dta+ bc=ac+bc, p√xq = xqp=∫ 81(2t−23 + t13)dt1xm= x−m= 2∫ 81t−23 dt+∫ 81t13 dt∫ ba[f(x)± g(x)] dx =∫ baf(x) dx ±∫ bag(x) dx,∫ bakf(x) dx = k∫ baf(x) dx= 2t1313∣∣∣∣∣81+t4343∣∣∣∣∣81∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba= 2(31)t13∣∣∣∣81+34t43∣∣∣∣81abc=acb= 6(813 − 1 13 ) + 34(843 − 1 43)F (x)∣∣∣∣ba= F (b)− F (a)= 6(3√8− 1)+34((813)4− 1)p√xq = xqp= 6(2− 1) + 34((2)4 − 1)= 6(1) +34(16− 1)= 6 +34(15)136 Chapter 1. Integration= 6 +454=24 + 454=694(k) ∫ 20(2x− 3)(4x2 + 1)dx =∫ 20(8x3 + 2x− 12x2 − 3) dx Distribute= 8∫ 20x3 dx+ 2∫ 20x dx∫ ba[f(x)± g(x)] dx =∫ baf(x) dx±∫ bag(x) dx− 12∫ 20x2 dx−∫ 203 dx∫ bakf(x) dx = k∫ baf(x) dx= 8x44∣∣∣∣20+ 2x22∣∣∣∣20− 12x33∣∣∣∣20− 3x∣∣∣∣20∫ baf(x) dx = F (x)∣∣∣∣ba= 2x4∣∣∣∣20+ x2∣∣∣∣20− 4x3∣∣∣∣20− 3x∣∣∣∣20= 2(24 − 04)+ (22 − 02)− 4 (23 − 03)− 3(2− 0) F (x)∣∣∣∣ba= F (b)− F (a)= 2 (16− 0) + (4− 0)− 4 (8− 0)− 3(2)= 32 + 4− 32− 6 = −2(l) ∫ 40(4− x)√x dx =∫ 40(4√x− x√x) dx Distribute=∫ 40(4x12 − xx 12)dx√x = x12= 4∫ 40x12 dx−∫ 40x32 dx∫ ba[f(x)± g(x)] dx =∫ baf(x) dx ±∫ bag(x) dx,∫ bakf(x) dx = k∫ baf(x) dx1.4. The Fundamental Theorem of Calculus 137= 4x3232∣∣∣∣∣40− x5252∣∣∣∣∣40∫ baf(x) dx = F (x)∣∣∣∣ba= 4(23)x32∣∣∣∣40− 25x52∣∣∣∣40abc=acb=83(432 − 0)− 25(452 − 0)F (x)∣∣∣∣ba= F (b)− F (a)=83(412)3− 25(412)5=83(2)3 − 25(2)5=83(8)− 25(32)=643− 645= 64(13− 15)= 64(5− 315)=12815(m) ∫ 10e−2xdx =1−2e−2x∣∣∣∣10∫ekx dx =1kekx + C,∫ baf(x) dx = F (x)∣∣∣∣ba= −12(e−2(1) − e−2(0)) F (x)∣∣∣∣ba= F (b)− F (a)= −12(e−2 − e0)= −12(e−2 − 1)(n) ∫ ln 205e3x dx = 5∫ ln 20e3x dx∫ bakf(x) dx = k∫ baf(x) dx= 513e3x∣∣∣∣ln 20∫ekx dx =1kekx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=53e3x∣∣∣∣ln 20138 Chapter 1. Integration=53(e3 ln 2 − e0) F (x)∣∣∣∣ba= F (b)− F (a)=53((eln 2)3 − 1) a ln(b) = ln(b)a=53((2)3 − 1) elnx = x=53(7) =353(o)∫ 21(x2 − 3x)dx =∫ 21x2dx− 3∫ 211xdx∫ ba[f(x)± g(x)] dx =∫ baf(x) dx ±∫ bag(x) dx,∫ bakf(x) dx = k∫ baf(x) dx=x33∣∣∣∣21− 3 ln |x|∣∣∣∣21∫ baf(x) dx = F (x)∣∣∣∣ba,∫1xdx = ln |x|+ C=(233− 133)− 3 (ln |2| − ln |1|) F (x)∣∣∣∣ba= F (b)− F (a)=(83− 13)− 3(ln 2− ln 1)=73− 3(ln 2− 0) ln(1) = 0=73− 3 ln 2(p) ∫ e1x2 − xx2dx =∫ e1(x2x2− xx2)dxa− bc=ac− bc=∫ e1(1− 1x)dx1.4. The Fundamental Theorem of Calculus 139= (x− ln |x|)∣∣∣∣e1∫k dx = kx + C,∫1xdx = ln |x| + C,∫ baf(x) dx = F (x)∣∣∣∣ba= (e− ln |e|)− (1− ln |1|) F (x)∣∣∣∣ba= F (b)− F (a)= (e− 1)− (1− 0) ln(ex) = x, ln(1) = 0= e− 1− 1 = e− 23. [Exercise on page 11](a)∫ pi60cosx dx = sinx∣∣∣∣pi60= sin(pi6)− sin(0) F (x)∣∣∣∣ba= F (b)− F (a)=12− 0 = 12sin(pi6)=12, sin(0) = 0(b) ∫ pi403 sec2 x dx = 3∫ pi40sec2 x dx∫kf(x) dx = k∫f(x) dx= 3 tan x∣∣∣∣pi40∫sec2 x dx = tanx+ C= 3 tan(pi4)− 3 tan(0) F (x)∣∣∣∣ba= F (b)− F (a)= 3(1)− 3(0) = 3 tan(pi4)= 1, tan(0) = 04. [Exercise on page 11](a)140 Chapter 1. IntegrationProperties of Definite Integrals (continued)Suppose f is continuous on [a, b].(iv)∫ aaf(x) dx = 0;(v) Let c be a real number such that f is continuous on an intervalcontaining a, b, and c. Then∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx(vi)∫ baf(x) dx = −∫ abf(x) dx.Using∫ aaf(x) dx = 0, we get∫ 22f(x) dx = 0 .(b)∫ 5−2f(x) dx =∫ 3−2f(x) dx+∫ 53f(x) dx∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx= 4 + (−10) = −6(c)Using∫ baf(x) dx = −∫ abf(x) dx, we get∫ 35f(x) dx = −∫ 53f(x) dx = −(−10) = 10(d) ∫ −2512f(x) dx= 12∫ −25f(x) dx Take out 12= 12(∫ 35f(x) dx+∫ −23f(x) dx) ∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx= 12(−∫ 53f(x) dx−∫ 3−2f(x) dx) ∫ baf(x) dx = −∫ abf(x) dx1.4. The Fundamental Theorem of Calculus 141= 12 [−(−10)− 4]∫ 53f(x) dx = −10,∫ 3−2f(x) dx = 4= 12(10− 4) = 72Alternate Solution∫ −2512f(x) dx= 12∫ −25f(x) dx Take out 12= 12[−∫ 5−2f(x) dx] ∫ baf(x) dx = −∫ abf(x) dx= 12[−(−6)] From part (b),∫ 5−2f(x) dx = −6= 12(6) = 725. [Exercise on page 12](a) From the equation∫ 86f(x)2dx = 3, it follows that 12∫ 86f(x) dx = 3 (by taking12out). Multiplying both sides of the latter equation by 2, we get∫ 86f(x) dx = 6Now, we have∫ 617f(x) dx= 7∫ 61f(x) dx∫kf(x) dx = k∫f(x) dx= 7(∫ 81f(x) dx+∫ 68f(x)dx) ∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx= 7(∫ 81f(x) dx−∫ 86f(x)dx) ∫ baf(x) dx = −∫ abf(x) dx= 7(15− 6) = 63∫ 81f(x) dx = 15,∫ 86f(x) dx = 6142 Chapter 1. Integration(b) ∫ 16(4− f(x)5)dx=∫ 164 dx−∫ 16f(x)5dx= 4x∣∣∣∣16− 15∫ 16f(x) dx∫kf(x) dx = k∫f(x) dx= 4(1)− 4(6)− 15∫ 16f(x) dx F (x)∣∣∣∣ba= F (b)− F (a)= 4− 24− 15(∫ 86f(x) dx+∫ 18f(x) dx) ∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx= −20− 15(∫ 86f(x) dx−∫ 81f(x) dx) ∫ baf(x) dx = −∫ abf(x) dx= −20− 15(6− 15)∫ 81f(x) dx = 15,∫ 86f(x) dx = 6= −20 + 95= −9156. [Exercise on page 12]Order Properties of Definite IntegralsLet a and b be real numbers such that a ≤ b. Suppose f is continuous on theinterval [a, b].(vii) If f(x) ≥ 0 for every x in [a, b], then ∫ baf(x) dx ≥ 0;(viii) If f(x) ≤ g(x) for every x in [a, b], then∫ baf(x) dx ≤∫ bag(x) dx(ix) If m and M are real numbers such that m ≤ f(x) ≤ M for every x in[a, b], thenm(b− a) ≤∫ baf(x) dx ≤M(b− a)We want to prove that 1 ≤ ∫ 10√1 + x2 dx ≤ √2. Since the bounds of integration are0 and 1, it follows that 0 ≤ x ≤ 1. Taking the square of this, we get 02 ≤ x2 ≤ 12;1.4. The Fundamental Theorem of Calculus 143that is,0 ≤ x2 ≤ 10 + 1 ≤ 1 + x2 ≤ 1 + 1 Add 1 to each side1 ≤ 1 + x2 ≤ 2√1 ≤√1 + x2 ≤√2 Take the square root of each side1 ≤√1 + x2 ≤√21(1− 0) ≤∫ 10√1 + x2 dx ≤√2(1− 0) Use property (ix)1 ≤∫ 10√1 + x2 dx ≤√2This ends the proof.7. [Exercise on page 12](a)When evaluating a definite integral by substitution, two methods are possible.Definite Integral Substitution Method 1Take the indefinite integral first and then evaluate it.Let u = x2 − 1. Then, du = 2xdx so that∫2x(x2 − 1)3 dx =∫u3du=u44∫xn dx =xn+1n+ 1+ C=(x2 − 1)44+ C Substitute u by x2 − 1Evaluating this from 0 to 2, we get∫ 202x(x2 − 1)3 dx = (x2 − 1)44∣∣∣∣20∫ baf(x) dx = F (x)∣∣∣∣ba=((2)2 − 1)44− ((0)2 − 1)44F (x)∣∣∣∣ba= F (b)− F (a)=(3)44− (−1)44144 Chapter 1. Integration=814− 14=804= 20Definite Integral Substituion Method 2 (Usually Preferable)Change the bounds during substitution. In this method, we don’t returnto the original variable. Step 1. Let u = x2 − 1. Then du = 2xdx Step 2. Find the new bounds on u.If x = 0, u = (0)2 − 1 = −1If x = 2, u = (2)2 − 1 = 3 Step 3. Using the new bounds, find∫ 202x(x2 − 1)3dx =∫ 3−2u3 du Substitute, change bounds=u44∣∣∣∣3−2∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=(3)44− (−1)44F (x)∣∣∣∣ba= F (b)− F (a)=814− 14=804= 20(b)Step 1. We make the substitution u = 4t− 1. Then du = 4 dt, so du4= dt.Step 2. We now change the bounds.If t = 0, u = 4(0)− 1 = −11.4. The Fundamental Theorem of Calculus 145If t = 1, u = 4(1)− 1 = 3Step 3. Then,∫ 10(4t− 1)50 dt =∫ 3−1u50du4Substitute, change bounds=14∫ 3−1u50du∫ bakf(x) dx = k∫ baf(x) dx=14u5151∣∣∣∣3−1∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=1204u51∣∣∣∣3−1=1204((351 − (−1)51) F (x)∣∣∣∣ba= F (b)− F (a)=1204(351 + 1)(c)We can first rewrite the integral∫ 103√26x+ 1 dx =∫ 10(26x+ 1)13 dx p√xq = xqpNow, let u = 26x + 1. Then, du = 26 dx so that du26= dx. We now change thebounds.If x = 0, u = 1If x = 1, u = 27The integral thus becomes∫ 10(26x+ 1)13 dx =∫ 271u13du26Substitute, change bounds=126∫ 271u13du∫ bakf(x) dx = k∫ baf(x) dx=126u4343∣∣∣∣∣271∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba146 Chapter 1. Integration=126(34)u43∣∣∣∣271abc=acb=3104(2743 − 1 43)F (x)∣∣∣∣ba= F (b)− F (a)=3104((2713)4− 1)=3104((3)4 − 1)=3104(81− 1)=3104(80) =3013(d)Let u = x2. Then, du = 2x dx so that du2= x dx.If x = 0, u = 02 = 0If x = 1, u = 12 = 1So, ∫ 10xex2dx =∫ 10eudu2Substitute, change bounds=12∫ 10eudu∫ bakf(x) dx = k∫ baf(x) dx=12eu∣∣∣∣10∫ex dx = ex + C,∫ baf(x) dx = F (x)∣∣∣∣ba=12(e− e0) F (x)∣∣∣∣ba= F (b)− F (a)=12(e− 1) e0 = 1(e)We make the substitution u = 1x. Then, du = − 1x2dx so that −du = 1x2dx.If x =13, u =113= 31.4. The Fundamental Theorem of Calculus 147If x =12, u =112= 2So, ∫ 12131x2e1xdx =∫ 23eu(−du) Substitute, change bounds= −∫ 23eudu∫ bakf(x) dx = k∫ baf(x) dx= −eu∣∣∣∣23∫ baf(x) dx = F (x)∣∣∣∣ba= −(e2 − e3) F (x)∣∣∣∣ba= F (b)− F (a)= e3 − e2(f)Let u = x3 + 1. Then, du = 3x2 dx so that du3= x2 dx.If x = 1, u = 13 + 1 = 2If x = 2, u = 23 + 1 = 9The integral thus becomes∫ 21x2(x3 + 1)2dx =∫ 921u2du3Substitute, change bounds=13∫ 921u2du∫ bakf(x) dx = k∫ baf(x) dx=13∫ 92u−2du1xm= x−m=13(u−1−1)∣∣∣∣92∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba= −131u∣∣∣∣92= −13(19− 12)F (x)∣∣∣∣ba= F (b)− F (a)= −13(− 718)148 Chapter 1. Integration=754(g)First, rewrite the integral∫ 801√1 + 2xdx =∫ 80(1 + 2x)−12 dx√x = x12 ,1xm= x−mNow, let u = 1 + 2x. Then, du = 2 dx so that du2= dx.If x = 0, u = 1If x = 8, u = 17Therefore,∫ 801√1 + 2xdx =∫ 80(1 + 2x)−12 dx From before=∫ 171u−12du2Substitute, change bounds=12∫ 171u−12 du∫ bakf(x) dx = k∫ baf(x) dx=12u1212∣∣∣∣∣171∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=12(21)u12∣∣∣∣171abc=acb= u12∣∣∣∣171= 1712 − 1 12 F (x)∣∣∣∣ba= F (b)− F (a)=√17− 1(h)First, ∫ 20t2√1 + t3 dt =∫ 20t2(1 + t3)12 dt√x = x121.4. The Fundamental Theorem of Calculus 149We make the substitution u = 1 + t3. Then, du = 3t2dt so that du3= t2dt.If t = 0, u = 1If t = 2, u = 9Therefore,∫ 20t2√1 + t3 dt =∫ 91u12du3Substitute, change bounds=13∫ 91u12 du∫ bakf(x) dx = k∫ baf(x) dx=13u3232∣∣∣∣∣91∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=13(23)u32∣∣∣∣91abc=acb=29(932 − 1 32)F (x)∣∣∣∣ba= F (b)− F (a)=29((912)3− 1)=29((3)3 − 1)=29(27− 1)=29(26) =529(i)Let u = x2 + 2x− 1. Then, du = (2x+ 2)dx so that du2= (x+ 1)dx.If x = 1, u = 2If x = 2, u = 4 + 4− 1 = 7Therefore,∫ 21(x+ 1)√x2 + 2x− 1 dx =∫ 721√u(du2)Substitute, change bounds150 Chapter 1. Integration=∫ 721u12(du2) √x = x12=∫ 72u−12(du2)1xm= x−m=12∫ 72u−12du∫ bakf(x) dx = k∫ baf(x) dx=12u1212∣∣∣∣∣72∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=12(21)u12∣∣∣∣72abc=acb= u12∣∣∣∣72= 712 − 2 12 F (x)∣∣∣∣ba= F (b)− F (a)=√7−√2(j)Let u = x2 + 1. Then, du = 2xdx so that du2= xdx.If x = −2, u = (−2)2 + 1 = 5If x = −1, u = (−1)2 + 1 = 2Thus,∫ −1−2xx2 + 1dx =∫ 251u(du2)Substitute, change bounds=12∫ 251udu∫ bakf(x) dx = k∫ baf(x) dx=12ln |u|∣∣∣∣25∫1xdx = ln |x|+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=12(ln |2| − ln |5|) F (x)∣∣∣∣ba= F (b)− F (a)=12(ln 2− ln 5)1.4. The Fundamental Theorem of Calculus 151(k)Let u = 5− ex. Then, du = −exdx so that −du = exdx.If x = 0, u = 5− e0 = 5− 1 = 4If x = ln 4, u = 5− eln 4 = 5− 4 = 1Thus,∫ ln 40ex5− exdx =∫ 141u(−du) Substitute, change bounds= −∫ 141udu∫ bakf(x) dx = k∫ baf(x) dx= − ln |u|∣∣∣∣14∫1xdx = ln |x|+ C,∫ baf(x) dx = F (x)∣∣∣∣ba= − (ln |1| − ln |4|) F (x)∣∣∣∣ba= F (b)− F (a)= −(0− ln 4) ln(1) = 0= ln 4(l)First, ∫ 40x√1 + 2xdx =∫ 40x(1 + 2x)−12 dx√x = x12 ,1xm= x−mLet u = 1 + 2x. Then du = 2dx so that du2= dx.If x = 0, u = 1If x = 4, u = 9Therefore,∫ 40x√1 + 2xdx =∫ 91xu−12du2=∫ 91(u− 12)u−12du2u = 1 + 2x⇔ x = u− 12=14∫ 91(u− 1)u− 12 du∫ bakf(x) dx = k∫ baf(x) dx152 Chapter 1. Integration=14∫ 91(u12 − u− 12)du Distribute=14∫ 91u12 du− 14∫ 91u−12 du∫ ba[f(x)± g(x)] dx =∫ baf(x) dx±∫ bag(x) dx=14u3232∣∣∣∣∣91− 14u1212∣∣∣∣∣91∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=14(23)u32∣∣∣∣91− 14(21)u12∣∣∣∣91abc=acb=16(932 − 1 32 )− 12(912 − 1 12)F (x)∣∣∣∣ba= F (b)− F (a)=16(27− 1)− 12(3− 1)=16(26)− 12(2)=133− 1 = 1038. [Exercise on page 12](a) Let u = 2x− pi6. Then du = 2dx so that du2= dx.If x = 0, u = −pi6If x =pi6, u = 2(pi6)− pi6=pi6So,∫ pi60sin(2x− pi6)dx =∫ pi6−pi6sin(u)(du2)Substitute, change bounds=12∫ pi6−pi6sin(u) du∫kf(x) dx = k∫f(x) dx=12(− cosu)∣∣∣∣pi6−pi6∫sinu du = − cosu+ C= −12cos(pi6)−[−12cos(−pi6)]F (x)∣∣∣∣ba= F (b)− F (a)1.4. The Fundamental Theorem of Calculus 153= −12(√32)−[−12(√32)]cos(−pi6)= cos(pi6)=√32= −√32+√32= 0(b) Let u = 1 + tanx. Then du = sec2 xdx.If x = 0, u = 1 + tan 0 = 1 + 0 = 1If x =pi4, u = 1 + tanpi4= 1 + 1 = 2So,∫ pi40√1 + tan xcos2 xdx =∫ pi40√1 + tan x(1cos2 xdx)=∫ pi40√1 + tan x(sec2 x dx)secx =1cosx=∫ 21√u du Substitute, change bounds=∫ 21u12 du√x = x12=u3232∣∣∣∣21∫xn dx =xn+1n+ 1+ C=23u32∣∣∣∣21abc=acb=23· 2 32 − 23(1)32 F (x)∣∣∣∣ba= F (b)− F (a)=23· 2 32 − 23=2(232 − 1)3(c) Let u = cosx. Then du = − sinxdx so that −du = sinxdx.If x = 0, u = cos(0) = 1If x =pi2, u = cospi2= 0154 Chapter 1. IntegrationTherefore,∫ pi20sinx cos(cosx) dx =∫ 01cos(u)(−du) Substitute, change bounds= −∫ 01cos(u) du∫kf(x) dx = k∫f(x) dx= − sin(u)∣∣∣∣01∫cosu du = sinu+ C= − sin(0)− [− sin(1)] F (x)∣∣∣∣ba= F (b)− F (a)= −0 + sin(1) sin(0) = 0= sin(1)(d) Let u = sin−1 x. Then du = 1√1−x2dx.If x = 0, u = sin−1(0) = 0If x =√32, u = sin−1(√32)=pi3So,∫ √320sin−1 x√1− x2 dx =∫ √320(sin−1 x)( 1√1− x2 dx)=∫ pi30u du Substitute, change bounds=u22∣∣∣∣pi30∫xn dx =xn+1n+ 1+ C=pi292− 022=pi218F (x)∣∣∣∣ba= F (b)− F (a)(e)∫ 1201 + x√1− x2 dx=∫ 120(1√1− x2 +x√1− x2)dxa+ bc=ac+bc1.4. The Fundamental Theorem of Calculus 155=∫ 1201√1− x2 dx+∫ 120x√1− x2 dx=∫ 1201√1− x2 dx+∫ 3411√u(du−2) Using substitution for the second integral:Let u = 1 − x2. Then du = −2x so thatdu−2 = xdx. If x = 0, u = 1 − 0 = 1. Ifx = 12, u = 1− (12)2 = 1− 14= 34=∫ 1201√1− x2 dx−12∫ 341u−12 du1√u=1u12= u−12= arcsin(x)∣∣∣∣ 120− 12u1212∣∣∣∣ 341∫1√1− x2 dx = sin−1 x+ C∫un du =un+1n+ 1+ C= arcsin(x)∣∣∣∣ 120− u 12∣∣∣∣ 341abc=acb=[arcsin(12)− arcsin(0)]−[(34) 12− (1) 12]=pi6− 0−(√32− 1)=pi6−√32+ 1(f) First, we can rewrite this integral as∫ 10x31 + x8dx =∫ 10x31 + (x4)2dxLet u = x4. Then du = 4x3dx. This implies that du4= x3dx.If x = 0, u = 04 = 0If x = 1, u = 14 = 1The integral becomes∫ 10x31 + x8dx =∫ 1011 + u2(du4)=14∫ 1011 + u2du Take out14156 Chapter 1. Integration=14arctan(u)∣∣∣∣10∫11 + u2du = tan−1 u+ C=14(arctan(1)− arctan(0))=14(pi4− 0)=pi161.5 The Area Between Two Curves1. [Exercise on page 13]Geometric Interpretation of Definite IntegralsLet a and b be real numbers such that a ≤ b, and let f be a function. Supposef is continuous on [a, b]. Suppose f(x) ≥ 0 for every x in [a, b], and let S be the region under thecurve y = f(x), above the x-axis, and between the lines x = a and x = bas illustrated in the following figure.a by = f(x)SxyThe definite integral∫ baf(x) dx can be interpreted as the area of S; thatis, ∫ baf(x) dx = Area(S) Suppose f(x) ≤ 0 for every x in [a, b]. Then the graph of f is below thex-axis as illustrated in the following figure.1.5. The Area Between Two Curves 157a by = f(x)SxyLet S be the region below the x-axis, above the graph of f , and betweenthe lines x = a and x = b. Then the integral∫ baf(x) dx can be interpretedas the opposite of the area of S; that is,∫ baf(x) dx = −Area(S) Suppose f takes on positive and negative values on [a, b]. Then the in-tegral∫ baf(x) dx is the total area above the x-axis minus the total areabelow the x-axis; that is,∫ baf(x) dx = [Total area above x-axis]− [Total area below x-axis]For example, suppose the graph of y = f(x) is the following.a by = f(x)S1S2S3xy158 Chapter 1. IntegrationThen, the geometric interpretation of the integral∫ baf(x)dx is as follows:∫ baf(x)dx = Area(S1) + Area(S3)− Area(S2)Let S1 and S2 be the shaded regions as shown below.xyS1S2AB CD4−24−2The definite integral∫ 4−2 f(x) dx is the area of S1 minus the area of S2. The region S1 is a triangle, and its area isArea(S1) =base · height2=2 · 42= 4 The region S2 is the trapezoid ABCD. And its area is given by the formula(AD +BC)DC2,where– AD is the distance from A to D (this is the long base). From the figure,the long base is AD = 4.– BC is the distance from B to C (this is the short base). From the figure,the short base is BC = 3.– DC is the distance from D to C (this is the height of the trapezoid). Fromthe figure, the height is DC = 2.1.5. The Area Between Two Curves 159So, the exact area of S2 isArea(S2) =(4 + 3)22=(7)(2)2= 7Thus, ∫ 4−2f(x) dx = Area(S1)− Area(S2) = 4− 7 = −32. [Exercise on page 13]Let S1, S2, and S3 be the shaded regions as shown below.xyS2S1 S30 1 25−2−44−2The integral∫ 5−4 f(x) dx is the total area above the x-axis minus the total area belowthe x-axis; that is,∫ 5−4dx = Area(S2)− [Area(S1) + Area(S3)]=pi(2)22− [Area(S1) + Area(S3)] The area of a circle ofradius r is pir2=pi(4)2−(2 · 22+3 · 22)The area of a triangleis base·height2= 2pi − (2 + 3) = 2pi − 53. [Exercise on page 14](a)160 Chapter 1. IntegrationArea Under a CurveLet f be a continuous function such that f(x) ≥ 0 for every x on theinterval [a, b]. Let S be the region under the curve y = f(x) and abovethe interval [a, b] as illustrated in the following figure.a by = f(x)SxyThen, the area of S is given by the following definite integral:Area(S) =∫ baf(x) dxThe area under the curve y = x2ex3over the interval [0, 1] is given byA =∫ 10x2ex3dxTo evaluate this, we make the substitution u = x3 so that du = 3x2dx ordu3= x2dx.If x = 0, u = 0If x = 1, u = 1So,A =∫ 10x2ex3dx =∫ 10eudu3Substitute, change bounds=13∫ 10eu du∫ bakf(x) dx = k∫ baf(x) dx=13eu∣∣∣∣10∫ex dx = ex + C,∫ baf(x) dx = F (x)∣∣∣∣ba1.5. The Area Between Two Curves 161=13(e1 − e0) F (x)∣∣∣∣ba= F (b)− F (a)=13(e− 1) e0 = 1(b) The area is given by A =∫ 1−245− 2x dx. To calculate this, we make the substi-tution u = 5− 2x so that du = −2dx or du−2 = dx.If x = −2, u = 5− 2(−2) = 5 + 4 = 9If x = 1, u = 5− 2 = 3The integral thus becomesA =∫ 1−245− 2x dx =∫ 394u(du−2)Substitute, change bounds=4−2∫ 391udu∫ bakf(x) dx = k∫ baf(x) dx= −2 ln |u|∣∣∣∣39∫1xdx = ln |x|+ C= −2(ln |3| − ln |9|) F (x)∣∣∣∣ba= F (b)− F (a)= −2(ln 3− ln 9)= 2(ln 9− ln 3) −(a− b) = b− a= 2(ln(32)− ln 3)= 2(2 ln 3− ln 3) ln (an) = n ln(a)= 2(ln 3) = 2 ln 3(c) The area is given byA =∫ pi40(sec2 x− cosx) dx= (tanx− sinx)∣∣∣∣pi40∫sec2 x dx = tanx + C∫cosx dx = sinx+ C162 Chapter 1. Integration=[tan(pi4)− sin(pi4)]− [tan (0)− sin (0)] F (x)∣∣∣∣ba= F (b)− F (a)=(1−√22)− (0− 0)= 1−√22(d) The area under the curve y = cot(2x) over the interval[pi12, pi4]is given byA =∫ pi4pi12cot(2x) dx =∫ pi4pi12cos(2x)sin(2x)dx cot θ =cos θsin θTo find this integral, we make the substitution u = sin(2x). Then du =2 cos(2x)dx, so that du2= cos(2x)dx.If x =pi12, u = sin(2 · pi12)= sin(pi6)=12If x =pi4, u = sin(2 · pi4)= sin(pi2)= 1The integral becomesA =∫ pi4pi12cos(2x)sin(2x)dx=∫ 1121u(du2)Substitute, change bounds=12∫ 1121udu∫kf(x) dx = k∫f(x) dx=12ln |u|∣∣∣∣112∫1xdx = ln |x|+ C=12ln |1| − 12ln |12| F (x)∣∣∣∣ba= F (b)− F (a)=12(0)− 12ln(12)ln(1) = 0, If a ≥ 0, |a| = a= 0− 12[− ln 2] ln(1a)= − ln a=12ln 21.5. The Area Between Two Curves 1634. [Exercise on page 14]The Area Between Two CurvesLet f and g be continuous functions such that f(x) ≥ g(x) for all x in theinterval [a, b]. Then, the area between the curves y = f(x) and y = g(x) froma to b is given byArea =∫ ba[f(x)− g(x)] dxThe condition f(x) ≥ g(x) on [a, b] means the graph of f is above the graph ofg on the interval [a, b].a by = f(x)y = g(x)0 xyThe area of the shaded region is equal to the area under the curve y = f(x)(given by∫ baf(x) dx) minus the area under the curve y = g(x) (given by∫ bag(x) dx). That is,Area =∫ baf(x) dx−∫ bag(x) dx=∫ ba[f(x)− g(x)] dxThe area of the region bounded above by y = 2x+ 5 and below by y = x3 on [0, 2] isArea =∫ 20[(2x+ 5)− x3]dx Area =∫ ba[f(x)− g(x)] dx=∫ 20(2x+ 5− x3) dx=∫ 202x dx+∫ 205 dx−∫ 20x3 dx∫ ba[f(x)± g(x)] dx =∫ baf(x) dx±∫ bag(x) dx164 Chapter 1. Integration= x2∣∣∣∣20+ 5x∣∣∣∣20− x44∣∣∣∣20∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba= (22 − 02) + 5(2− 0)−(244− 044)F (x)∣∣∣∣ba= F (b)− F (a)= 4 + 10− 4 = 105. [Exercise on page 14]Let f(x) = x2 and g(x) = x + 2. To find the area between the curves of thesefunctions, we can proceed as follows.Step 1. Find the interval(s).We need to find the x-coordinates of the points of intersection between the two curves.To this end, we set f(x) = g(x) and solve for x. The equation x2 = x+2 is equivalentto x2−x−2 = 0. Factoring the lefthand side, we get (x+1)(x−2) = 0. This impliesthat x = −1 or x = 2. So, the interval is [−1, 2].Step 2. Determine whether f is above g on the interval(s) found in thefirst step.Method 1. Sketch the graphs and see which one is on top.−2 −1 1 2 3 4 5345 y = x+ 2y = x2xyFrom the figure, we can conclude that y = x + 2 = g(x) is above y = x2 = f(x) onthe interval [−1, 2].1.5. The Area Between Two Curves 165Method 2. Choose an arbitrary number in the interval [−1, 2]. For example 0. Then,substitute this into f(x) and g(x):f(0) = 02 = 0 and g(0) = 0 + 2 = 2Since g(0) > f(0), the graph of g is above the graph of f on the interval [−1, 2].Step 3. Set up the integral and find the area.Since y = g(x) = x+ 2 is above y = f(x) = x2 on the interval [−1, 2], the area isArea =∫ 2−1[g(x)− f(x)] dx=∫ 2−1[(x+ 2)− x2] dx=(x22+ 2x− x33)∣∣∣∣2−1∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=(222+ 2(2)− 233)−((−1)22+ 2(−1)− (−1)33)F (x)∣∣∣∣ba= F (b)− F (a)=(2 + 4− 83)−(12− 2 + 13)=(6− 83)−(−32+13)=(103)−(−76)a− bc=ac− bc,ab+cd=ad+ bcbd=103+76=206+76=276=926. [Exercise on page 14](a)Let f(x) = x2 − 8 and g(x) = −x2 + 10.Finding the interval. Set f(x) = g(x). Then,x2 − 8 = −x2 + 10166 Chapter 1. Integration2x2 − 8 = 102x2 = 18x2 = 9x = ±√9x = ±3so that the x-coordinates of the intersection between f and g are −3 and 3. Theinterval is thus [−3, 3].We now need to determine whether f is above g. Let’s choose an arbitrarynumber in the interval [−3, 3], for example 0. Substituting this into the equationsfor f and g, we getf(0) = 02 − 8 = −8 and g(0) = −(0)2 + 10 = 10Since g(0) > f(0), the graph of y = g(x) is above the graph of y = f(x).So, the area between the curves y = x2 − 8 and y = −x2 + 10 isArea =∫ 3−3[g(x)− f(x)] dx=∫ 3−3[(−x2 + 10)− (x2 − 8)] dx=∫ 3−3(−2x2 + 18) dx=(−2x33+ 18x)∣∣∣∣3−3∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=(−23(3)3 + 18(3))−(−23(−3)3 + 18(−3))F (x)∣∣∣∣ba= F (b)− F (a)=(−543+ 54)−(543− 54)= (−18 + 54)− (18− 54)= (36)− (−36)1.5. The Area Between Two Curves 167= 36 + 36 = 72(b)Let f(x) = 4x− x2 and g(x) = x2.Set 4x− x2 = x2. Then,4x− x2 = x2−2x2 + 4x = 0−2x(x− 2) = 0x = 0 or x = 2So, the interval is [0, 2].Choose an arbitrary number in the interval [0, 2], say 1. Substituting this intof(x) and g(x), we getf(1) = 4(1)− 12 = 4− 1 = 3 and g(1) = 12 = 1Since f(1) > g(1), it follows that y = f(x) is above y = g(x).So, the area isArea =∫ 20[f(x)− g(x)] dx=∫ 204x− x2 − x2 dx=∫ 20(4x− 2x2) dx=(4x22− 2x33)∣∣∣∣20∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=(2x2 − 23x3)∣∣∣∣20=(2(2)2 − 23(2)3)−(2(0)2 − 23(0)3)F (x)∣∣∣∣ba= F (b)− F (a)=(8− 163)− (0)168 Chapter 1. Integration=243− 163=83(c)Let f(x) =x32and g(x) =x2.Setx32=x2. Then,x32=x2x3 = x Multiply both sides by 2x3 − x = 0 Subtract x from both sidesx(x2 − 1) = 0 Factor out xx(x− 1)(x+ 1) = 0 a2 − b2 = (a− b)(a+ b)x = 0 or x = 1 or x = −1So, the x-coordinates of the points of intersection between the graphs arex1 = −1, x2 = 0, and x3 = 1. Since we have three points of intersection, thereare two intervals: [−1, 0] and [0, 1].On each interval, we need to determine whether f is above g. On the interval [−1, 0], choose an arbitrary number, say −12. Substitutingthis into f(x) and g(x), we getf(−12)=12(−12)3= − 116and g(−12)=12(−12)= −14Since f(−12)< g(−12), it follows that f is above g on the interval [−1, 0]. On the interval [0, 1], we choose 12. Substituting this into f(x) and g(x), wegetf(12)=116and g(12)=14So, g is above f on the interval [0, 1].The area is thusArea =∫ 0−1[f(x)− g(x)] dx+∫ 10[g(x)− f(x)] dx=∫ 0−1(x32− x2)dx+∫ 10(x2− x32)dx1.5. The Area Between Two Curves 169=12(x44− x22)∣∣∣∣0−1+12(x22− x44)∣∣∣∣10∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=12(0−(14− 12))+12((12− 14)− 0)F (x)∣∣∣∣ba= F (b)− F (a)=12(−14+12)+12(12− 14)=12(14)+12(14)=18+18=147. [Exercise on page 15]The equation of the x-axis is y = 0. So, we need to find the area betweeny = 2x2 − 7x+ 3 and y = 0. Let f(x) = 2x2 − 7x+ 3 and g(x) = 0.Set 2x2 − 7x+ 3 = 0. Then,2x2 − 7x+ 3 = 0(2x− 1)(x− 3) = 0x =12or x = 3So, the interval is[12, 3].If we choose 1 in[12, 3]and substitute into f(x) and g(x), we have f(1) = −2 andg(1) = 0. Since g(1) > f(1), it follows that g is above f .So, the area isArea =∫ 312[g(x)− f(x)] dx=∫ 312[0− (2x2 − 7x+ 3)] dx=∫ 312(−2x2 + 7x− 3) dx170 Chapter 1. Integration=(−2x33+ 7x22− 3x)∣∣∣∣312∫xn dx =xn+1n+ 1+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=(−23(3)3 +72(3)2 − 3(3))−(−23(12)3+72(12)2− 3(12))=(−18 + 632− 9)−(− 112+78− 32)=(−362+632− 182)−(− 224+2124− 3624)=(92)−(−1724)=125248. [Exercise on page 15](a) Let f(x) = x and g(x) = x2 − 2x. From the question, x is in the interval [0, 4].We need to know if the graphs of f and g intersect in the interval [0, 4]. Setf(x) = g(x). Then,x = x2 − 2xx2 − 2x = xx2 − 3x = 0 Subtract x from both sidesx(x− 3) = 0 Factor out xx = 0 or x = 3Since 3 is between 0 and 4, there are two intervals: [0, 3] and [3, 4].Now, we need to determine if f is above g in each interval. On the interval [0, 3], choose 1. Substituting this into f(x) and g(x), we getf(1) = 1 and g(1) = 12 − 2(1) = 1− 2 = −1Since f(1) > g(1), it follows that f is above g on the interval [0, 3]. On the interval [3, 4], choose 3.5. Substituting this into f(x) and g(x), wegetf(3.5) = 3.5 and g(3.5) = (3.5)2 − 2(3.5) = 12.25− 7 = 5.25Since g(3.5) > f(3.5), it follows that g is above f on the interval [3, 4].1.5. The Area Between Two Curves 171So, the area between f and g from 0 to 4 isArea =∫ 30[f(x)− g(x)] dx+∫ 43[g(x)− f(x)] dx=∫ 30[x− (x2 − 2x)] dx+ ∫ 43[(x2 − 2x)− x] dx=∫ 30(x− x2 + 2x) dx+ ∫ 43(x2 − 2x− x) dx=∫ 30(−x2 + 3x) dx+ ∫ 43(x2 − 3x) dx=(−x33+ 3x22)∣∣∣∣30+(x33− 3x22)∣∣∣∣43=[(−333+32(3)2)− (0)]+[(433− 32(4)2)−(333− 32(3)2)]= −273+272+643− 482− 273+272= −546+816+1286− 1446− 546+816=386=193(b)Let f(x) = ex and g(x) = e−x. From the question, x is between −2 and 1, thatis, −2 ≤ x ≤ 1. To get the interval(s), we need to determine whether the graphsof f and g intersect on [−2, 1].Set ex = e−x. Then,ex = e−xex =1exa−m =1ame2x = 1 Multiply both sides by ex2x = ln 1 ln(ey) = y2x = 0 ln 1 = 0x = 0 Divide both sides by 2Since 0 is between −2 and 1, there are two intervals: [−2, 0] and [0, 1].172 Chapter 1. IntegrationWe now need to determine if f is above g on each interval.On the interval [−2, 0], we choose −1. Since f(−1) = e−1 < g(−1) = e1, itfollows that f is below g on [−2, 0].On the interval [0, 1], we choose 0.5. Since f(0.5) = e0.5 > g(0.5) = e−0.5, itfollows that f is above g on [0, 1].The area is thusArea =∫ 0−2[g(x)− f(x)] dx+∫ 10[f(x)− g(x)] dx=∫ 0−2(e−x − ex) dx+∫ 10(ex − e−x) dx= (−e−x − ex)∣∣∣∣0−2+ (ex + e−x)∣∣∣∣10∫ex dx = ex + C,∫ekx dx =1kekx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=[(−e0 − e0)− (−e2 − e−2)]+ [(e+ e−1)− (e0 + e0)]= (−1− 1) + e2 + e−2 + e+ e−1 − (1 + 1) e0 = 1= −2 + e2 + e−2 + e+ e−1 − 2= e2 + e+ e−1 + e−2 − 4(c)Let f(x) =1x+ 1and let g(x) =13.Set1x+ 1=13. Then,1x+ 1=133x+ 1= 13 = x+ 1x = 2Since 2 is not strictly between −12and 2, there is only one interval here:[−12, 2].Choose 0 in[−12, 2]. Since f(0) = 1 > 13= g(0), it follows that f is above g on[−12, 2].1.5. The Area Between Two Curves 173The area is thusArea =∫ 2− 12[f(x)− g(x)] dx=∫ 2− 12(1x+ 1− 13)dx=(ln |x+ 1| − 13x)∣∣∣∣2− 12∫k dx = kx + C,∫1x+ kdx = ln |x+k|+C,∫ baf(x) dx = F (x)∣∣∣∣ba=(ln |2 + 1| − 13(2))−(ln∣∣∣∣−12 + 1∣∣∣∣− 13(−12))F (x)∣∣∣∣ba= F (b)− F (a)=(ln |3| − 23)−(ln∣∣∣∣12∣∣∣∣+ 16)= ln(3)− 23− ln(12)− 16= ln(3) + ln(2)− 56ln(1x)= − lnx(d)Let f(x) = x4 − ln(x+ 3) and let g(x) = x3 − ln(x+ 3).Set x4 − ln(x+ 3) = x3 − ln(x+ 3). Then,x4 − ln(x+ 3) = x3 − ln(x+ 3)x4 = x3x4 − x3 = 0x3(x− 1) = 0x = 0 or x = 1So, the interval is [0, 1].Choose 0.5. Substituting this into f(x) and g(x), we getf(0.5) = (0.5)4 − ln(3.5) ≈ 0.0625− 1.2527 = −1.1902 and g(0.5) ≈ −1.1277Since f(0.5) < g(0.5), it follows that f is below g.174 Chapter 1. IntegrationSo, the area isArea =∫ 10[g(x)− f(x)] dx=∫ 10[(x3 − ln(x+ 3))− (x4 − ln(x+ 3))] dx=∫ 10(x3 − x4) dx=(x44− x55)∣∣∣∣10∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫ baf(x) dx = F (x)∣∣∣∣ba=(14− 15)− (0) = 120F (x)∣∣∣∣ba= F (b)− F (a)9. [Exercise on page 15](a) Let f(x) = e−x and g(x) = −x.To find the intersection points, we need to solve the equation e−x = −x. Butthis equation is difficult to solve exactly (actually, it is impossible). So, we willuse a different approach than usual: a graphical one. The idea is to sketch thegraphs of f and g and determine which one is on top (as well as the interval(s)over which we need to integrate) graphically. Drawing the graphs of f and g,we get−2 1y = e−xy = −x0 xyFrom the figure, the two graphs does not intersect. So, the interval is the onegiven in the problem, that is, [−2, 1]. Again from the figure, the graph of y = e−x1.5. The Area Between Two Curves 175is on top. So, the area between the two curves is∫ 1−2[e−x − (−x)] dx =∫ 1−2(e−x + x) dx=(e−x−1 +x22)∣∣∣∣1−2∫ekx dx =1kekx + C,∫xn dx =xn+1n+ 1+ C=(−e−1 + 122)−(−e−(−2) + (−2)22)F (x)∣∣∣∣ba= F (b)− F (a)= −e−1 + 12+ e2 − 2= e2 − e−1 − 32(b) Let f(x) = cosx and g(x) = x. The region between the graphs of f and g overthe interval [−pi,−pi4] is shown in the following figure.−pi −pi2pi2y = cosxy = xxyFrom the graph, the area between the two curves is∫ −pi4−pi(top− bottom) dx =∫ −pi4−pi(cosx− x) dx=(sinx− x22)∣∣∣∣−pi4−pi∫cosx dx = sinx+ C ,∫xn dx =xn+1n+ 1+ C=[sin(−pi4)−(−pi4)22]−[sin (−pi)− (−pi)22]176 Chapter 1. Integration=(−√22− pi232)−(0− pi22)= −√22− pi232+pi22= −√22− pi232+16pi232=15pi232−√2210. [Solution on page 15]Since the leaf is bounded above by f(x) = x − x2 and below by g(x) = x2 − x, andsince x is between 0 and 1, the area isArea =∫ 10[f(x)− g(x)] dx=∫ 10[(x− x2)− (x2 − x)] dx=∫ 10(x− x2 − x2 + x) dx = ∫ 10(−2x2 + 2x) dx=(−2x33+ 2x22)∣∣∣∣10∫xn dx =xn+1n+ 1+ C=(−23(1)2 + (1)2)− (0) F (x)∣∣∣∣ba= F (b)− F (a)= −23+ 1 =−2 + 33=131.6 Total Change and Average Values1. [Exercise on page 15]Total Change or Net Change or Cumulative ChangeLet y = F (x) be a function such that F ′(x) is continuous for all x in the interval[a, b]. Then, the total (or net or cumulative) change in F (x) as x goes from1.6. Total Change and Average Values 177a to b is given by the definite integral∫ baF ′(x) dx; that is,(total change in F (x) from a to b) =∫ baF ′(x) dx = F (b)− F (a)In other words, the integral of the instantaneous rate of change is the totalchange.(a) The total change in F (x) is∫ 31F ′(x) dx =∫ 31x3 dx=x44∣∣∣∣31=344− 144F (x)∣∣∣∣ba= F (b)− F (a)=814− 14=804= 20(b) The total change in F (x) is∫ 20F ′(x) dx =∫ 204xe3x2−5 dxTo find this integral, we make the substitution u = 3x2 − 5. Then du = 6xdx,so that du6= xdx.If x = 0, u = 3(0)2 − 5 = −5If x = 2, u = 3(2)2 − 5 = 7So, ∫ 204xe3x2−5 dx = 4∫ 20e3x2−5(xdx)∫kf(x) dx = k∫f(x) dx= 4∫ 7−5eu(du6)Substitute, change bounds=46∫ 7−5eu du∫kf(x) dx = k∫f(x) dx178 Chapter 1. Integration=23eu∣∣∣∣7−5∫ex dx = ex + C=23(e7 − e−5)2. [Exercise on page 16]Since the population is growing and since∫ 94N ′(t) dt = N(t)∣∣∣∣94= N(9)−N(4),it follows that the integral∫ 94N ′(t) dt is the total number of bacteria added to thepopulation between times 4 and 9.3. [Exercise on page 16]Let h(t) be the height at time t. Since dhdtis the rate of change of h(t), it follows that∫ 72dhdtdt =∫ 72h′(t) dt = h(t)∣∣∣∣72= h(7)− h(2)So the integral∫ 72dhdtdt is the total change in height during the time period from 2 to7.4. [Exercise on page 16]The total number of individuals added to the population between times t = 3 andt = 5 is ∫ 53N ′(t) dt =∫ 5315(3t+ 1)2 dtTo evaluate this integral, we make the substitution u = 3t + 1. Then du = 3dt, sothat du3= dt.If x = 3, u = 3(3) + 1 = 10If x = 5, u = 3(5) + 1 = 16So,∫ 5315(3t+ 1)2 dt =15∫ 53(3t+ 1)2 dt∫kf(x) dx = k∫f(x) dx1.6. Total Change and Average Values 179=15∫ 1610u2(du3)Substitute, change bounds=15· 13∫ 1610u2 du∫kf(x) dx = k∫f(x) dx=115u33∣∣∣∣1610∫xn dx =xn+1n+ 1+ C=145(163 − 103) F (x)∣∣∣∣ba= F (b)− F (a)=145(4096− 1000) = 68.8 ≈ 705. [Exercise on page 16]Let dydtbe the rate of rainfall in mm/h.(a) Since dydtis the rate of rainfall, the integral∫ 240dydtdtis the net rainfall between times 0 and 24 .(b) The cumulative rainfall between times 0 and 24 is∫ 240dydtdt =∫ 2404(1− cos( pi12t))= 4∫ 240(1− cos( pi12t))dt∫kf(x) dx = k∫f(x) dx= 4∫ 2401 dt− 4∫ 240cos( pi12t)dt= 4t∣∣∣∣240− 4∫ 240cos( pi12t)dt∫k dx = kx+ C= 4t∣∣∣∣240− 4(12pisin( pi12t))∣∣∣∣240∫cos(kx) dx =1ksin(kx) + C= 4(24− 0)− 48pi(sin( pi12· 24)− sin(0))F (x)∣∣∣∣ba= F (b)− F (a)= 96− 48pi(sin(2pi)− 0) sin(0) = 0= 96− 48pi(0− 0) sin(2pi) = 0180 Chapter 1. Integration= 96− 0 = 96 mm/h6. [Exercise on page 16](a) The net displacement of the particle between times t = 3 and t = 8 is∫ 83v(t) dt =∫ 83(t2 − 7t+ 10) dt=(t33− 7t22+ 10t)∣∣∣∣83∫xn dx =xn+1n+ 1+ C=(833− 7822+ 10(8))−(333− 7322+ 10(3))F (x)∣∣∣∣ba= F (b)− F (a)=5123− 4482+ 80− 273+632− 30=10246− 13446+4806− 546+1896− 1806=1024− 1344 + 480− 54 + 189− 1806=1156m(b) The total distance traveled by the particle between times t = 3s and t = 8s isthe integral of the absolute value of v(t) over the interval [3, 8], that is,Total distance =∫ 83|v(t)| dt =∫ 83|t2 − 7t+ 10| dtWe first need to remove the absolute value symbol to find this integral. To dothis, we need to find the intervals where the velocity v(t) is positive and theintervals where it is negative. This can be done through the following steps: Set v(t) = 0 and solve for t.t2 − 7t+ 10 = 0(t− 2)(t− 5) = 0t = 2 or t = 5 Find the sign of v(t). Plotting the solutions found in the previous step onthe real line, we get1.6. Total Change and Average Values 1810 ∞2 5The numbers 2 and 5 determine three open intervals: (0, 2), (2, 5), and(5,∞). To get the sign of v(t) on an interval I, we choose a number inI and substitute it into v(t). If the result is positive, then v(t) is positiveon I. If the result is negative, then v(t) is negative on I. The followingdiagram gives the sign of v(t).0 ∞2 5+choose 1v(1) = 4 > 0−choose 3v(3) = −2 < 0+choose 6v(6) = 4 > 0From the sign analysis of v(t), one can split the integral∫ 83|v(t)| dt as follows:∫ 83|v(t)| dt =∫ 53|v(t)| dt+∫ 85|v(t)| dtSince v(t) is negative on (3, 5), it follows that |v(t)| = −v(t) if 3 < t < 5. Sincev(t) is positive on (5, 8), it follows that |v(t)| = v(t) if 5 < t < 8. So,∫ 83|v(t)| dt =∫ 53−v(t) dt+∫ 83v(t)=∫ 53−(t2 − 7t+ 10) dt+∫ 85(t2 − 7t+ 10) dt=∫ 53(−t2 + 7t− 10) dt+∫ 85(t2 − 7t+ 10)=(−t33+ 7t22− 10t)∣∣∣∣53+(t33− 7t22+ 10t)∣∣∣∣85=(−533+ 7522− 10(5))−(−333+ 7322− 10(3))+(833− 7822+ 10(8))−(533− 7522+ 10(5))=(−1253+1752− 50)−(−273+632− 30)+(5123− 4482+ 80)−(1253− 1752+ 50)=(−2506+5256− 3006)−(−546+1896− 1806)182 Chapter 1. Integration+(10246− 13446+4806)−(2506− 5256+3006)= −256−(−456)+1606−(256)=−25 + 45 + 160− 256=1556m7. [Exercise on page 16]Average ValuesLet f be a continuous function on [a, b]. The average value of f on [a, b],denoted f , is given byf =1b− a∫ baf(x) dxThis can be interpreted geometrically as follows. Multiplying both sides of theequation above by b− a, we getf · (b− a) =∫ baf(x) dx The lefthand side of this latter equation is the area of the rectangle withwidth b− a and height f . The righthand side is the area under the curve y = f(x) (if f(x) ≥ 0 on[a, b]).So, if the graph of f is above the x-axis, the average value f can be thought ofas the height of the rectangle with width b − a whose area is the same as thearea under the curve y = f(x) over [a, b].(a) The average value of f(x) =1− x√xon the interval [1, 9] isf =19− 1∫ 41f(x) dx =18∫ 81f(x) dxWe have∫ 911− x√xdx =∫ 91(1√x− x√x)dxa+ bc=ac+bc=∫ 91(x−12 − x 12)dx√x = x12 ,1xm= x−m1.6. Total Change and Average Values 183=(x1212− x3232)∣∣∣∣∣91∫xn dx =xn+1n+ 1+ C=(2x12 − 23x32)∣∣∣∣91abc=acb=(2(9)12 − 23(9)32)−(2(1)12 − 23(1)32)F (x)∣∣∣∣ba= F (b)− F (a)= (6− 18)−(2− 23)= −12− 43= −363− 43= −403So, the average value of f(x) on the interval [1, 9] isf =19− 1(∫ 91f(x) dx)=18(−403)= −53(b) The average value of f(x) = x√x2 + 16 on the interval [0, 3] isf =13∫ 30x√x2 + 16 dxWe use substitution. Let u = x2 + 16. Then, du = 2x dx so thatdu2= x dx.If x = 0, u = 16 If x = 3, u = 25The integral becomes∫ 30x√x2 + 16 dx =∫ 2516√u(du2)Substitute, change bounds=12∫ 2516u12 du∫kf(x) dx = k∫f(x) dx,√x = x12=1223u32∣∣∣∣2516∫xn dx =xn+1n+ 1+ C=13u32∣∣∣∣2516=13((25)32 − (16) 32)F (x)∣∣∣∣ba= F (b)− F (a)184 Chapter 1. Integration=13(125− 64) = 613Thus, the average value of f(x) = x√x2 + 16 on the interval [0, 3] isf =13− 0(∫ 30f(x) dx)=13(613)=619(c) First, we find the integral∫ 2pi2pi3f(t) dt.∫ 2pi2pi3f(t) dt =∫ 2pi2pi3csc2(t4)dt=∫ 2pi2pi3csc2(14t)dt= −114cot(t4)∣∣∣∣2pi2pi3Using the substituion u =kx, one can easily show that∫csc2(kx) dx = − 1kcot(kx) + C= −4 cot(t4)∣∣∣∣2pi2pi3abc=acb= −4[cot(2pi4)− cot( 2pi34)]= −4[cot(pi2)− cot(pi6)]= −4(cos pi2sin pi2− cospi6sin pi6)cotx =cosxsinx= −4(01−√3212)= −4(0−√3) abcd=adbc= 4√3The average value of f on the interval[2pi3, 2pi]isf =12pi − 2pi3∫ 2pi2pi3f(t) dt=14pi3∫ 2pi2pi3f(t) dt1.6. Total Change and Average Values 185=34pi(4√3)abc=acb=3√3pi8. [Exercise on page 17]The average snowfall in one year isf =11− 0∫ 10(2.3 + 2.3 cos(2pit)) dt=∫ 102.3 dt+ 2.3∫ 10cos(2pit) dt= 2.3t∣∣∣∣10+ 2.3∫ 10cos(2pit) dt∫k dx = kx+ C= 2.3t∣∣∣∣10+ 2.3(12pisin(2pit))∣∣∣∣10∫cos(kx) dx =1ksin(kx) + C= 2.3(1− 0) + 2.32pi(sin(2pi)− sin(0)) F (x)∣∣∣∣ba= F (b)− F (a)= 2.3 +2.32pi(0− 0) sin(2pi) = 0, sin(0) = 0= 2.3 + 0 = 2.3 cm/month9. [Exercise on page 17]First, we find the integral∫ 240T (t) dt.∫ 240T (t) dt =∫ 240[70 + sin( pi12t)]dt=∫ 24070 dt+∫ 240sin( pi12t)dt= 70t∣∣∣∣240+∫ 240sin( pi12t)dt∫k dx = kx+ C= 70t∣∣∣∣240+(−12picos( pi12t))∣∣∣∣240∫sin(kx) dx = −1kcos(kx) + C= 70(24− 0)− 12pi(cos( pi12· 24)− cos(0))F (x)∣∣∣∣ba= F (b)− F (a)= 1680− 12pi(cos(2pi)− 1) cos(0) = 1186 Chapter 1. Integration= 1680− 12pi(1− 1) cos(2pi) = 1= 1680− 0 = 1680Now, the average temperature in one day isT =124− 0∫ 240T (t) dt =124(1680) = 70Chapter2Techniques of Integration andComputational Methods2.1 Integration by Parts1. [Exercise on page 19](a)Formula for Integration by PartsLet f and g be differentiable functions. Then by the product rule, we haveddx[f(x)g(x)] = f ′(x)g(x) + f(x)g′(x)Taking the integral of both sides with respect to x, we get∫ddx[f(x)g(x)] dx =∫f ′(x)g(x) dx+∫f(x)g′(x) dxUsing the fact that integration and differentiation are inverse processes,the latter equality becomesf(x)g(x) =∫f ′(x)g(x) dx+∫f(x)g′(x) dxSubtracting∫f ′(x)g(x) dx from both sides gives∫f(x)g′(x) dx = f(x)g(x)−∫f ′(x)g(x) dx (2.1.1)If u = f(x) and v = g(x), then du = f ′(x)dx and dv = g′(x)dx. The187188 Chapter 2. Techniques of Integration and Computational Methodsequation (2.1.1) becomes∫udv = uv−∫v du (2.1.2)Equation 2.1.2 is the formula for integration by parts we will be using.Note. Suppose we want to take the integral∫f(x)g(x) dx using integra-tion parts. Then, there are two possibilities: Either we choose u = f(x) and dv = g(x)dx, or we choose u = g(x) and dv = f(x)dx.If we make a choice and the integral∫vdu, in the righthand side of (2.1.2),appears to be more difficult than the original integral, then that’s not theright choice. Actually, the idea is to choose u and dv so that the integral∫vdu is easier than the original integral.Integrating a Polynomial times an Exponential by PartsSuppose the integral is of the form∫Polynomial · Exponential dx.If we want to integrate this by parts, the right choice will beu = Polynomial and dv = Exponential dxLet’s integrate∫2xex dx.Choose u = 2x and dv = exdxThen, du = 2dx and v =∫dv =∫ex dx = exAside: When taking the integral∫ex dx to get v, we don’t need to introduce2.1. Integration by Parts 189the constant C. We will add C in the last step.∫2xex dx =∫u dv Since u = 2x and dv = exdx= uv −∫v du∫u dv = uv −∫v du= (2x)ex −∫ex(2dx) Substitute u and v= 2xex − 2∫ex dx∫kf(x) dx = k∫f(x) dx= 2xex − 2ex + C∫ex dx = ex + C= (2x− 2)ex + C(b)First of all, recall the basic integral∫ekx+l dx =1kekx+l + C k 6= 0.Choose u = x and dv = e4xdxThen, du = dx and v =∫e4x dx =14e4xTherefore,∫xe4x dx = uv −∫v du= x(14e4x)−∫14e4xdx Substitute u and v=14xe4x − 14∫e4x dx∫kf(x) dx = k∫f(x) dx=14xe4x − 14(14e4x)+ C∫ekx dx =1kekx + C=14xe4x − 116e4x + C(c)Choose u = 3x and dv = e−5xdxThen, du = 3dx and v =∫e−5x dx = −15e−5x190 Chapter 2. Techniques of Integration and Computational MethodsSubstituting this into the formula for integration by parts, we get∫3xe−5x dx = uv −∫v du= (3x)(−15e−5x)−∫ (−15e−5x)(3dx) Substitute u and v= −35xe−5x +35∫e−5x dx∫kf(x) dx = k∫f(x) dx= −35xe−5x +35(−15e−5x)+ C∫ekx dx =1kekx + C= −35xe−5x − 325e−5x + C(d)Choose u = −2x and dv = e−3x+1dxThen, du = −2dx and v =∫e−3x+1 dx = −13e−3x+1Now, substitute into the formula for integration by parts and complete the in-tegration.∫−2xe−3x+1 dx= uv −∫v du= (−2x)(−13e−3x+1)−∫ (−13e−3x+1)(−2dx) Substitute u and v=23xe−3x+1 − 23∫e−3x+1 dx∫k dx = kx+ C=23xe−3x+1 − 23(−13e−3x+1)+ C∫ekx+l dx =1kekx+l + C=23xe−3x+1 +29e−3x+1 + C(e)So,∫(−7x+ 9)e−2x+3 dx2.1. Integration by Parts 191Choose u = −7x+ 9 and dv = e−2x+3dxThen, du = −7dx and v =∫e−2x+3 dx = −12e−2x+3= uv −∫v du= (−7x+ 9)(−12e−2x+3)−∫ (−12e−2x+3)(−7dx) Substitute u and v= −12(−7x+ 9)e−2x+3 − 72∫e−2x+3 dx∫kf(x) dx = k∫f(x) dx= −12(−7x+ 9)e−2x+3 − 72(−12e−2x+3)+ C∫ekx+l dx =1kekx+l + C= −12(−7x+ 9)e−2x+3 + 74e−2x+3 + C(f)Integration by Parts When lnx is in the IntegralSuppose we have an integral of the form∫ln(g(x)) · h(x) dxTo find this using integration by parts, we chooseu = ln(g(x)) and dv = h(x)dx.Recall the chain rule for logarithmic functions: ddx[ln(gx)] = g′(x)g(x). So,if u = ln(g(x)), then du =g′(x)g(x)dx.For example, if u = ln(3x), then du =(3x)′3xdx =33xdx =1xdx.Let’s find∫x ln(2x) dx.192 Chapter 2. Techniques of Integration and Computational MethodsChoose u = ln(2x) and dv = xdxThen, du =22xdx =1xdx and v =∫x dx =x22So,∫x ln(2x) dx = uv −∫v du= ln(2x)(x22)−∫ (x22)(1xdx)Substitute u and v= ln(2x)(x22)−∫ (x2)(11dx)=12x2 ln(2x)− 12∫x dx∫kf(x) dx = k∫f(x) dx=12x2 ln(2x)− 12(x22)+ C∫xn dx =xn+1n+ 1+ C=12x2 ln(2x)− 14x2 + C(g)Choose u = lnx and dv = x2dxThen, du =1xdx and v =∫x2 dx =x33So,∫x2 lnx dx =∫u dv= uv −∫v du= (lnx)(x33)−∫ (x33)(1xdx)Substitute u and v=13x3 lnx− 13∫x2 dx∫kf(x) dx = k∫f(x) dx=13x3 lnx− 13(x33)+ C∫xn dx =xn+1n+ 1+ C=13x3 lnx− 19x3 + C2.1. Integration by Parts 193(h)We can rewrite this as∫1 · lnx dx.Choose u = lnx and dv = 1dxThen, du =1xdx and v =∫1 dx = xSo, ∫lnx dx =∫u dv= uv −∫v du= (lnx)(x)−∫(x)(1xdx)Substitute u and v= x lnx−∫1dx= x lnx− x+ C∫k dx = kx+ C(i)Choose u = ln(7x) and dv = 8xdxThen, du =77x=1xdx and v =∫8x dx = 4x2So,∫8x ln(7x) dx =∫u dv= uv −∫v du= (ln(7x))(4x2)−∫4x2(1xdx)Substitute u and v= 4x2 ln(7x)− 4∫x dx∫kf(x) dx = k∫f(x) dx= 4x2 ln(7x)− 4(x22)+ C∫xn dx =xn+1n+ 1+ C= 4x2 ln(7x)− 2x2 + C194 Chapter 2. Techniques of Integration and Computational Methods(j)Choose u = lnx and dv = 1x3dxThen, du =1xdx and v =∫1x3dx =∫x−3 dx =x−2−2 = −12x2So,∫lnxx3dx =∫u dv= uv −∫v du= (lnx)(− 12x2)−∫ (− 12x2)(1xdx)Substitute u and v= − lnx2x2+12∫1x3dx∫kf(x) dx = k∫f(x) dx= − lnx2x2+12∫x−3 dx1xm= x−m= − lnx2x2+12(x−2−2)+ C∫xn dx =xn+1n+ 1+ C= − lnx2x2+12(− 12x2)+ C1xm= x−m= − lnx2x2− 14x2+ C(k)First, we have∫3x+ 2exdx =∫(3x+ 2)e−x dx.Choose u = 3x+ 2 and dv = e−xdxThen, du = 3dx and v =∫e−x dx =1−1e−x = −e−xSo,∫3x+ 2exdx =∫(3x+ 2)e−x dx= uv −∫v du= (3x+ 2)(−e−x)−∫(−e−x)(3dx) Substitute u and v2.1. Integration by Parts 195= −(3x+ 2)e−x + 3∫e−x dx∫kf(x) dx = k∫f(x) dx= −(3x+ 2)e−x − 3e−x + C∫ekx dx =1kekx + C= [(−3x− 2)− 3]e−x= (−3x− 5)e−x + C=−3x− 5ex+ C e−x =1ex(l)Integrating a Polynomial times a Trigonometric Function by PartsLet T (x) be a function of the form sin(kx+ l) or cos(kx+ l), where k andl are real numbers. Suppose we want to find an integral of the form∫Polynomial · T (x) dxTo evaluate this integral, we use integration by parts by choosingu = Polynomial and dv = T (x) dxFor the integral∫3x sinx dx,Choose u = 3x and dv = sinxdxThen, du = 3dx and v =∫sinx dx = − cosxSo,∫3x sinx dx = uv −∫vdu Formula for Integration by Parts= 3x(− cosx)−∫− cosx(3 dx) Substitute u and v= −3x cosx+ 3∫cosx dx∫k dx = kx+ C= −3x cosx+ 3 sinx+ C∫cosx dx = sinx+ C196 Chapter 2. Techniques of Integration and Computational Methods(m)Choose u = x and dv = cos(2x)dxThen, du = dx and v =∫cos(2x) dx =12sin(2x)So,∫x cos(2x)dx = uv −∫vdu=12x sin(2x)−∫12sin(2x)dx Substitute u and v=12x sin(2x)− 12∫sin(2x)dx∫k dx = kx+ C=12x sin(2x)− 12(−12cos(2x)) ∫cos(kx) dx =1ksin(kx) + C=12x sin(2x) +14cos(2x) + C(n)Choose u = 4x− 2 and dv = sin(−5x+ 7)dxThen, du = 4dx and v =∫sin(−5x+ 7) dx = 15cos(−5x+ 7)To see why∫sin(−5x + 7) dx = 15cos(−5x + 7), we make the substitution z =−5x+ 7. Then dz = −5dx, so that dz−5 = dx and∫sin(−5x+ 7) dx =∫sin z(dz−5)= −15∫sin z dz= −15(− cos z)∫sin z dz = − cos z + C=15cos(−5x+ 7) Substitute z with −5x+ 7(The constant C will be added in the last step.)2.1. Integration by Parts 197Now, the integral∫(4x− 2) sin(−5x+ 7) dx is∫(4x− 2) sin(−5x+ 7) dx= uv −∫v du= (4x− 2)(15cos(−5x+ 7))−∫15cos(−5x+ 7)(4 dx)=15(4x− 2) cos(−5x+ 7)− 45∫cos(−5x+ 7) dx=15(4x− 2) cos(−5x+ 7)− 45(1−5 sin(−5x+ 7))+ C=15(4x− 2) cos(−5x+ 7) + 425sin(−5x+ 7) + C2. [Exercise on page 19](a)Choose u = x2 and dv = exdxThen, du = 2xdx and v =∫ex dx = exSo, ∫x2ex dx = uv −∫v du= x2ex −∫ex(2xdx) Substitute u and v= x2ex − 2∫xex dx∫kf(x) dx = k∫f(x) dxTo find∫xex dx, we use integration by parts again.Choose u = x and dv = exdxThen, du = dx and v =∫ex dx = exSo, ∫xex dx = xex −∫ex dx Substitute u and v198 Chapter 2. Techniques of Integration and Computational Methods= xex − ex∫ex dx = ex + CSubstituting this into what we got above, we have∫x2ex dx = x2ex − 2∫xex dx= x2ex − 2 (xex − ex) + C From before= x2ex − 2xex + 2ex + C= (x2 − 2x+ 2)ex + C(b)Choose u = 3x2 and dv = e−3xdxThen, du = 6xdx and v =∫e−3x dx = −13e−3xSo,∫3x2e−3x dx = (3x2)(−13e−3x)−∫ (−13e−3x)(6xdx) Substitute u and v= −x2e−3x + 2∫xe−3x dx∫kf(x) dx = k∫f(x) dx= −x2e−3x + 2(−13xe−3x −∫−13e−3x dx)Integrate by parts again= −x2e−3x + 2(−13xe−3x +13∫e−3x dx) ∫kf(x) dx = k∫f(x) dx= −x2e−3x + 2(−13xe−3x +13(−13e−3x)) ∫ekx dx =1kekx + C= −x2e−3x + 2(−13xe−3x − 19e−3x)+ C= −x2e−3x − 23xe−3x − 29e−3x + C=(−x2 − 23x− 29)e−3x + C(c)Choose u = x2 and dv = cosxdxThen, du = 2xdx and v =∫cosx dx = sinx2.1. Integration by Parts 199The integral becomes∫x2 cosxdx = x2 sinx−∫2x sinxdx (2.1.3)To evaluate the integral in the righthand side,∫2x sinx dx, we use integrationby parts again:Choose w = 2x and dz = sinxdxThen, dw = 2dx and z =∫sinx dx = − cosxTherefore,∫2x sinx dx= wz −∫z dw Formula for integration by parts= 2x(− cosx)−∫(− cosx)(2dx) Substitute w and z= −2x cosx+ 2∫cosx dx∫k dx = kx+ C= −2x cosx+ 2 sinx∫cosx dx = sinx+ CPutting this into (2.1.3), we get∫x2 cosxdx = x2 sinx−∫2x sinx dx= x2 sinx− (−2x cosx+ 2 sinx) + C= x2 sinx+ 2x cosx− 2 sinx+ C3. [Exercise on page 19](a)First we need to find the indefinite integral∫(x+ 1)e2x dx.Choose u = x+ 1 and dv = e2xdxThen, du = dx and v =∫e2x dx =12e2x200 Chapter 2. Techniques of Integration and Computational Methods∫(x+ 1)e2x dx = (x+ 1)12e2x −∫12e2xdx Substitute u and v=12(x+ 1)e2x − 12∫e2x dx∫kf(x) dx = k∫f(x) dx=12(x+ 1)e2x − 12(12e2x)+ C∫ekx dx =1kekx + C=12(x+ 1)e2x − 14e2x + CWe now find the definite integral.∫ 1−1(x+ 1)e2x dx=(12(x+ 1)e2x − 14e2x)∣∣∣∣1−1∫ baf(x) dx = F (x)∣∣∣∣ba=(12(1 + 1)e2 − 14e2)−(12(−1 + 1)e−2 − 14e−2)F (x)∣∣∣∣ba= F (b)− F (a)=(e2 − 14e2)−(0− 14e−2)=34e2 +14e−2(b)Choose u = −5x2 + 4x+ 1 and dv = e−2xdxThen, du = (−10x+ 4)dx and v =∫e−2x dx = −12e−2xSo,∫(−5x2 + 4x+ 1)e−2x dx= (−5x2 + 4x+ 1)(−12e−2x)−∫ (−12e−2x)((−10x+ 4)dx) Substitute u and v= −12(−5x2 + 4x+ 1)e−2x +∫(−5x+ 2)e−2x dxTo find∫(−5x+ 2)e−2x dx, we use integration by parts.2.1. Integration by Parts 201Choose u = −5x+ 2 and dv = e−2xdxThen, du = −5dx and v =∫e−2x dx = −12e−2xSo,∫(−5x+ 2)e−2x dx= (−5x+ 2)(−12e−2x)−∫ (−12e−2x)(−5dx) Substitute u and v= −12(−5x+ 2)e−2x − 52∫e−2x dx∫kf(x) dx = k∫f(x) dx= −12(−5x+ 2)e−2x − 52(−12e−2x) ∫ekx dx =1kekx + C= −12(−5x+ 2)e−2x + 54e−2xSubstituting this into what we got above, we have∫(−5x2+4x+1)e−2x dx = −12(−5x2+4x+1)e−2x− 12(−5x+2)e−2x+ 54e−2x+CWe now find the definite integral. LetF (x) = −12(−5x2 + 4x+ 1)e−2x − 12(−5x+ 2)e−2x + 54e−2xThen,∫ 21(−5x2 + 4x+ 1)e−2x dx = F (2)− F (1)∫ baf(x) dx = F (b)− F (a)whereF (2) = −12(−5(4) + 8 + 1)e−4 − 12(−10 + 2)e−4 + 54e−4= −12(−11)e−4 − 12(−8)e−4 + 54e−4=(112+ 4 +54)e−4 =434e−4andF (1) = −12(−5 + 4 + 1)e−2 − 12(−5 + 2)e−2 + 54e−2202 Chapter 2. Techniques of Integration and Computational Methods= 0 +32e−2 +54e−2=(32+54)e−2 =114e−2Altogether, ∫ 21(−5x2 + 4x+ 1)e−2x dx = F (2)− F (1)=434e−4 − 114e−2(c)First we find the indefinite integral∫ √x lnx dx.Choose u = lnx and dv =√xdxThen, du =1xdx and v =∫ √x dx =∫x12 dx =23x32So,∫ 41√x lnx dx =23x32 lnx−∫ (23x32)(1xdx)Substitute u and v=23x32 lnx− 23∫x12 dx∫kf(x) dx = k∫f(x) dx=23x32 lnx− 23(23x32)+ C∫xn dx =xn+1n+ 1+ C=23x32 lnx− 49x32 + CThe definite integral is∫ 41√x lnx dx=(23x32 lnx− 49x32)∣∣∣∣41∫ baf(x) dx = F (x)∣∣∣∣ba=(23(4)32 ln 4− 49(4)32)−(23(1)32 ln 1− 49(1)32)F (x)∣∣∣∣ba= F (b)− F (a)=(163ln 4− 329)−(0− 49)ln(1) = 0=163ln 4− 329+492.1. Integration by Parts 203=163ln 4− 289(d)First, we need to find the associated indefinite integral∫tan−1(2x) dx =∫1 · tan−1(2x) dxChoose u = tan−1(2x) and dv = 1dxThen, du =ddx[2x]1+(2x)2dx = 21+4x2dx and v =∫1 dx = xTo get du, we used the chain rule:ddx[tan−1(g(x))]=ddx[g(x)]1 + [g(x)]2The indefinite integral above becomes∫tan−1(2x) dx= uv −∫v du= x tan−1(2x)−∫2x1 + 4x2dx= x tan−1(2x)− 2∫x1 + 4x2dx∫k dx = kx+ C= x tan−1(2x)− 2(18ln |1 + 4x2|)+ CUse substitution: w = 1 + 4x2, sothat dw = 8xdx and dw8= xdx= x tan−1(2x)− 14ln |1 + 4x2|+ CSo an antiderivative of f(x) = tan−1(2x) is F (x) = x tan−1(2x)− 14ln |1 + 4x2|.Evaluating F (x) at the bounds of integration, and using the fact that tan−1(1) = pi4,we getF(12)=12tan−1(1)− 14ln |1 + 4(14)| = 12(pi4)− 14ln(2) =pi8− 14ln(2)and F (0) = 0− 14ln |1| = 0− 0 = 0.204 Chapter 2. Techniques of Integration and Computational MethodsSo, ∫ 120tan−1(2x) dx = F(12)− F (0) = pi8− 14ln(2)− 0 = pi8− ln 24(e) Recall the following differentiation rules:ddx[sin−1 x]=1√1− x2 andddx[sin−1(g(x))]=ddx[g(x)]√1− [g(x)]2Observing that sin−1 x = 1 · sin−1 x, we have∫sin−1 x dx =∫1 · sin−1 x dxChoose u = sin−1(x) and dv = 1dxThen, du = 1√1−x2dx and v =∫1 dx = xIntegrating by parts, we get∫sin−1 x dx= uv −∫v du= x sin−1 x−∫x√1− x2 dx= x sin−1 x−∫1√w(dw−2)Use substitution: w = 1 − x2, sothat dw = −2xdx and dw−2 = xdx= x sin−1 x+12∫w−12 dw1√w=1w12= w−12= x sin−1 x+12w1212+ C∫xn dx =xn+1n+ 1+ C= x sin−1 x+12(21w12)+ Cabc=acb= x sin−1 x+√w + C Simplify= x sin−1 x+√1− x2 + C Substitute w with 1− x24. [Exercise on page 20](a)2.1. Integration by Parts 205Choose u = ex and dv = cosxdxThen, du = exdx and v =∫cosx dx = sinxIntegrating by parts, we get∫ex cosx dx = ex sinx−∫ex sinx dx (2.1.4)To find the integral in the righthand side,∫ex sinx dx, we use integration byparts a second time.Choose u = ex and dv = sinxdxThen, du = exdx and v =∫sinx dx = − cosxTherefore, ∫ex sinxdx = ex(− cosx)−∫(− cosx)ex dx∫ex sinxdx = −ex cosx+∫ex cosx dxSubstituting this into (2.1.4), we get∫ex cosx dx = ex sinx−(−ex cosx+∫ex cosx dx)∫ex cosx dx = ex sinx+ ex cosx−∫ex cosx dx2∫ex cosx dx = ex sinx+ ex cosx Add∫ex cosx dx to both sides∫ex cosx dx =12(ex sinx+ ex cosx) + C Divide both sides by 2(b)Choose u = e2x and dv = sin(−3x)dxThen, du = 2e2xdx and v =∫sin(−3x) dx = 13cos(−3x)Integrating by parts, we get∫e2x sin(−3x) dx = e2x(13cos(−3x))−∫ (13cos(−3x))(2e2x dx) (2.1.5)∫e2x sin(−3x) dx = 13e2x cos(−3x)− 23∫e2x cos(−3x) dx (2.1.6)To find the integral in the righthand side,∫e2x cos(−3x) dx, we use integrationby parts a second time.206 Chapter 2. Techniques of Integration and Computational MethodsChoose u = e2x and dv = cos(−3x)dxThen, du = 2e2xdx and v =∫cos(−3x) dx = −13sin(−3x)Therefore,∫e2x cos(−3x)dx = e2x(−13sin(−3x))−∫ (−13sin(−3x))(2e2x dx)∫e2x cos(−3x)dx = −13e2x sin(−3x) + 23∫e2x sin(−3x) dxSubstituting this into (2.1.6), we get∫e2x sin(−3x) dx = 13e2x cos(−3x)− 23∫e2x cos(−3x) dx∫e2x sin(−3x) dx = 13e2x cos(−3x)− 23(−13e2x sin(−3x) + 23∫e2x sin(−3x) dx)∫e2x sin(−3x) dx = 13e2x cos(−3x) + 29e2x sin(−3x)− 49∫e2x sin(−3x) dxAdding 49∫e2x sin(−3x) dx to both sides of the latter equation, we get∫e2x sin(−3x) dx+ 49∫e2x sin(−3x) dx = 13e2x cos(−3x) + 29e2x sin(−3x)(1 +49)∫e2x sin(−3x) dx = 13e2x cos(−3x) + 29e2x sin(−3x)139∫e2x sin(−3x) dx = 13e2x cos(−3x) + 29e2x sin(−3x)∫e2x sin(−3x) dx = 913(13e2x cos(−3x) + 29e2x sin(−3x))∫e2x sin(−3x) dx = 313e2x cos(−3x) + 213e2x sin(−3x) + C5. [Exercise on page 20]Area =∫ 94ln(√x)dx =∫ 941 · ln (√x) dxTo find this, we use integration by parts.Choose u = ln(√x) and dv = 1dxThen, du =(√x)′√xdx =12√x√xdx =12xdx and v =∫1 dx = x2.1. Integration by Parts 207So, ∫1 · ln (√x) dx = x ln(√x)− ∫ x( 12xdx)Substitute u and v= x ln(√x)−∫12dx= x ln(√x)− 12x+ C∫k dx = kx+ CWe now find the definite integral.Area =∫ 94ln(√x)dx=(x ln(√x)− 12x)∣∣∣∣94∫ baf(x) dx = F (x)∣∣∣∣ba=(9 ln(√9)− 12(9))−(4 ln(√4)− 12(4))F (x)∣∣∣∣ba= F (b)− F (a)=(9 ln 3− 92)− (4 ln 2− 2)= 9 ln 3− 92− 4 ln 2 + 2= 9 ln 3− 4 ln 2− 526. [Exercise on page 20]The total population isTotal Population =∫ 110N ′(x) dx =∫ 110(x+ 2) ln(x+ 2) dxChoose u = ln(x+ 2) and dv = (x+ 2)dxThen, du =1x+ 2dx and v =∫(x+ 2) dx =12(x+ 2)2Aside: Instead of usingx22+ 2x as an antiderivative of x+ 2, we use12(x+ 2)2, whichis another antiderivative of x + 2. The reason we choose12(x + 2)2 is to simplifymatters as we will see in the next few lines. An alternate solution to this problemwould be to first use the substitution method (by letting u = x + 2), then integrate208 Chapter 2. Techniques of Integration and Computational Methodsby parts.∫(x+ 2) ln(x+ 2) dx=12(x+ 2)2 ln(x+ 2)−∫12(x+ 2)21x+ 2dx Substitute u and v=12(x+ 2)2 ln(x+ 2)− 12∫(x+ 2) dx∫k dx = kx+ C=12(x+ 2)2 ln(x+ 2)− 12(12(x+ 2)2)+ C∫(x+ k) dx =12(x+ k)2 + C=12(x+ 2)2 ln(x+ 2)− 14(x+ 2)2 + CWe can now find the definite integral.Total Population=(12(x+ 2)2 ln(x+ 2)− 14(x+ 2)2)∣∣∣∣110∫ baf(x) dx = F (x)∣∣∣∣ba=(12(13)2 ln(13)− 14(13)2)−(12(2)2 ln 2− 14(2)2)F (x)∣∣∣∣ba= F (b)− F (a)=1692ln 13− 1694− (2 ln 2− 1)=1692ln 13− 2 ln 2− 1694+ 1=1692ln 13− 2 ln 2− 1654≈ 216.7382− 1.3862− 41.25≈ 1742.2 Partial FractionsPartial Fraction Decomposition• A rational function is a function of the form f(x) = P (x)Q(x), where P (x) andQ(x) are both polynomials. For example x+1x2+3, 1x3+x−1 are rational functions.• Consider the fractions 1xand 1x−1 . Taking these to a common denominator, we2.2. Partial Fractions 209get1x+1x− 1 =(x− 1) + xx(x− 1) =x− 1 + xx(x− 1) =2x− 1x(x− 1)So,2x− 1x2 − x =1x+1x− 1The sum 1x+ 1x−1 is called partial fraction decomposition of the rationalfunction f(x) = 2x−1x2−x . And the fractions1xand 1x−1 are called partial frac-tions.• Finding a partial fraction decomposition of a rational function f(x) = P (x)Q(x)amounts to writing f(x) as a sum of simpler fractions.A natural question one can ask is to know whether any rational function admits apartial fraction decomposition. The answer to this question is given by the theorembelow. Before we state the theorem, recall that the degree of a polynomial P (x),denoted degP , is the largest exponent of x.Theorem 2.2.1. Let f(x) = P (x)Q(x)be a rational function. Assume degP < degQ.Then there exist rational functions F1, F2, · · · , Fn such thatP (x)Q(x)= F1(x) + F2(x) + · · ·+ Fn(x),and each Fi(x) is of the formA(ax+ b)rorAx+B(ax2 + bx+ c)swith b2 − 4ac < 0where r and s are nonnegative integers and a, b, c, A,B are constants.The inequality b2 − 4ac < 0 means that the polynomial ax2 + bx + c is irreductible;that is, not factorable.• The sum F1(x) + F2(x) + · · · + Fn(x) from Theorem 2.2.1 is called partialfraction decomposition of P (x)Q(x).• Each Fi(x) is called a partial fraction.A rational function P (x)Q(x)is called proper if deg(P ) < deg(Q). And improper ifdeg(P ) ≥ deg(Q). To find the partial fraction decomposition of a proper rationalfunction f(x) = P (x)Q(x), one can proceed as follows:210 Chapter 2. Techniques of Integration and Computational MethodsStep 1. Factor the denominator Q(x) as far as possible.Step 2. Express P (x)Q(x)as a sum of partial fractions of the form A(ax+b)rand/orAx+B(ax2+bx+c)s.Step 3. Find the constants A,B,C, · · · .(For improper rational fractions, one needs to perform long division first. This willbe illustrated in Exercise 3.)Depending on the factored form of Q(x), there are four cases.Case I. Q(x) is a product of distinct linear factors. In that case, Q(x) canbe written as Q(x) = (a1x+ b1)(a2x+ b2) · · · (akx+ bk), where no factor isrepeated. Then the partial fraction decomposition of P (x)Q(x)has the formP (x)Q(x)=A1a1x+ b1+A2a2x+ b2+ · · ·+ Akakx+ bk,where A1, A2, · · · , Ak are constants. For example, the partial fraction de-composition of 4x(x−1)(x+3)(x−5) is of the form4x(x− 1)(x+ 3)(x− 5) =Ax− 1 +Bx+ 3+Cx− 5 ,where A,B, and C are constants.Case II. Q(x) is a product of linear factors where at least one is repeated.If Q(x) has a factor of the form (ax+ b)r, then the partial fraction decom-position must include a sum of the formA1ax+ b+A2(ax+ b)2+ · · · Ar(ax+ b)rwhere A1, A2, · · · , Ar are constants. For example, the partial fraction de-composition of 1(x−1)(x+4)3 is1(x− 1)(x+ 4)3 =Ax− 1 +Bx+ 4+C(x+ 4)2+D(x+ 4)3,where A,B,C and D are constants.Case III. Q(x) has distinct irreductible quadratic factors. If Q(x) has anon repeated irreductible factor ax2 + bx + c, then the partial fraction2.2. Partial Fractions 211decomposition must include a term of the form Ax+Bax2+bx+c, where A andB are constants. For example, the partial fraction decomposition of4x2−5(x−2)(x2+3)(x2−x+1) is4x2 − 5(x− 2)(x2 + 3)(x2 − x+ 1) =Ax− 2 +Bx+ Cx2 + 3+Dx+ Ex2 − x+ 1Case IV. Q(x) has a repeated irreductible quadratic factor. If Q(x) has afactor of the form (ax2 + bx + c)s, where b2 − 4ac < 0, then the partialfraction decomposition must include a sum of the formA1x+B1ax2 + bx+ c+A2x+B2(ax2 + bx+ c)2+ · · ·+ Asx+Bs(ax2 + bx+ c)s,where A1, A2, · · · , As, and B1, B2, · · · , Bs are constants. For example, thepartial fraction decomposition of 1(x−5)2(x2+1)3 is1(x− 5)2(x2 + 1)3 =Ax− 5 +B(x− 5)2 +Cx+Dx2 + 1+Ex+ F(x2 + 1)2+Gx+H(x2 + 1)3Note. Cases III and IV are optional and illustrated in Exercise 4.1. [Exercise on page 20](a) We want to decompose the rational function f(x) = 2x+6x2−3x into partial fractions.• Here P (x) = 2x+ 6 and Q(x) = x2 − 3x. The degree of P (x) is 1, and thedegree of Q(x) is 2. So degP < degQ, which means that f(x) is a properrational function.• Factoring Q(x), we get Q(x) = x2 − 3x = x(x− 3).• Since Q(x) = x(x−3) is a product of distinct linear factors (which falls intoCase I), the partial fraction decomposition of f(x) has the form2x+ 6x(x− 3) =Ax+Bx− 3where A and B are constants.• Finding the constants A and B. We will present two methods:Method 1. (See the box below for another method.)212 Chapter 2. Techniques of Integration and Computational Methods2x+ 6x(x− 3) =Ax+Bx− 32x+ 6x(x− 3)x(x− 3) =Axx(x− 3) + Bx− 3x(x− 3)Multiply both sides bythe common denomi-nator x(x− 3)2x+ 6 = A(x− 3) +Bx Simplify2x+ 6 = Ax− 3A+Bx Expand the righthand side2x+ 6 = (A+B)x− 3AWrite the righthandside in the standardform for polynomialsRecall that two polynomials are equal if and only if they have the samecoefficients.- In the polynomial 2x + 6, the coefficient of x is 2, while the coefficientof x in (A+B)x− 3A is A+B. So, A+B = 2.- In the polynomial 2x + 6, the constant term is 6, while the constantterm in (A+B)x− 3A is −3A. So, −3A = 6.This gives the following system.A + B = 2−3A = 6From the second equation, we have A = 6−3 = −2. Substituting this intothe first equation, we get −2 + B = 2, which implies that B = 2 + 2 = 4.Hence, A = −2 and B = 4.So the partial fraction decomposition of 2x+6x2−3x is2x+ 6x2 − 3x =−2x+4x− 3 (2.2.1)Alternate Way of Finding the Constants A and BMethod 2. We start off with the partial fraction decomposition2x+ 6x(x− 3) =Ax+Bx− 32.2. Partial Fractions 213Multiplying both sides by the common denominator, we get2x+ 6x(x− 3)x(x− 3) =Axx(x− 3) + Bx− 3x(x− 3)Simplifying this, we get2x+ 6 = A(x− 3) +BxThe idea here is to substitute x with values that simplify the righthandside of the latter equation. Using the numbers that make the denominatorQ(x) = x(x− 3) zero (that is, the numbers 0 and 3), we can easily find Aand B: For x = 0, the equation 2x+ 6 = A(x− 3) +Bx becomes2(0) + 6 = A(0− 3) +B(0)6 = −3A−2 = A Divide both sides by −3 For x = 3, the equation 2x+ 6 = A(x− 3) +Bx becomes2(3) + 6 = A(3− 3) +B(3)12 = 3B4 = B Divide both sides by 3Note. Method 2 is more straightforward but only works well in Case I(that is, when the denominator is a product of distinct linear factors). Itcan also be used in other cases but more subtly. Because Method 1 workswell in all the cases, we will use it more often.(b) • Here P (x) = −2x+ 5, degP = 1 and Q(x) = x(x− 1)2, degQ = 3. Clearly,degP < degQ.• Since Q(x) is a product of linear factors x(x− 1)2 = x(x− 1)(x− 1), where(x−1) is repeated (this falls into Case II), the partial fraction decompositionof f(x) has the form−2x+ 5x(x− 1)2 =Ax+Bx− 1 +C(x− 1)2 , (2.2.2)214 Chapter 2. Techniques of Integration and Computational Methodswhere A,B, and C are constants. Note that the partial fraction Axis due tothe factor x, while the sum Bx−1 +C(x−1)2 is due to the factor (x− 1)2.• Finding the constants A,B and C. Multiplying both sides of (2.2.2) by thecommon denominator x(x−1)2, we get −2x+5 = A(x−1)2+Bx(x−1)+Cx.Distributing the righthand side, we get−2x+ 5 = Ax2 − 2Ax+ A+Bx2 −Bx+ Cx−2x+ 5 = (A+B)x2 + (−2A−B + C)x+ A0x2 − 2x+ 5 = (A+B)x2 + (−2A−B + C)x+ AUsing the fact that two polynomials are equal if and only if they have thesame coefficients, we get the following system.A + B = 0 coefficients of x2−2A − B + C = −2 coefficients of xA = 5 constant termsFrom the third equation, we have A = 5. Substituting this into the firstequation, we get 5+B = 0, which implies that B = −5. Substituting A andB into the second equation, we get −2(5)− (−5) + C = −2, which impliesthat C = 3.• Hence, A = 5, B = −5, C = 3, and the partial fraction decomposition off(x) is−2x+ 5x(x− 1)2 =5x− 5x− 1 +3(x− 1)2 (2.2.3)2. [Exercise on page 20](a)∫2x+ 6x2 − 3x dxSolution. From (2.2.1), we have 2x+6x2−3x =−2x+ 4x−3 . So the integral∫2x+6x2−3x dxbecomes∫2x+ 6x2 − 3x dx =∫ (−2x+4x− 3)dx= −2∫1xdx+ 4∫1x− 3 dx= −2∫1xdx+ 4∫1uduFor the second integral, letu = x− 3. Then du = dx2.2. Partial Fractions 215= −2 ln |x|+ 4 ln |u|+ C∫1xdx = ln |x|+ C= −2 ln |x|+ 4 ln |x− 3|+ C Substitute u with x− 3(b) First, we need to find the partial fraction decomposition of f(x) = 6xx2−x−2 .Factoring Q(x), we get Q(x) = (x + 1)(x − 2). Since Q(x) is a product ofdistinct linear factors, the partial fraction decomposition of f(x) has the form6xx2 − x− 2 =Ax+ 1+Bx− 2 ,where A and B are constants that we need to determine.6x(x+ 1)(x− 2) =Ax+ 1+Bx− 26x = A(x− 2) +B(x+ 1)Multiply both sides bythe common denomi-nator (x+ 1)(x− 2)6x = Ax− 2A+Bx+B Expand the righthand side6x = (A+B)x− 2A+BWrite the righthandside in the standardform for polynomials6x+ 0 = (A+B)x− 2A+BThis implies thatA + B = 6 coefficients of x−2A + B = 0 constant termsFrom the first equation, B = 6−A. Substituting this into the second equation,we get −2A + 6 − A = 0, which is equivalent to −3A = −6. So, A = 2 andB = 6− 2 = 4. (Note that there are many methods to solve a system of linearequations. The one we used here is called the substitution method.)So the partial fraction decomposition of f(x) is6xx2 − x− 2 =2x+ 1+4x− 2216 Chapter 2. Techniques of Integration and Computational MethodsWe can now find the integral∫6xx2 − x− 2 dx:∫6xx2 − x− 2 dx=∫ (2x+ 1+4x− 2)dx= 2∫1x+ 1dx+ 4∫1x− 2 dx= 2∫1udu+ 4∫1wdwUse substitution. First integral:u = x + 1. Second integral: w =x− 2= 2 ln |u|+ 4 ln |w|+ C∫1xdx = ln |x|+ C= 2 ln |x+ 1|+ 4 ln |x− 2|+ C Substitute u with x+1 and w withx− 2(c) First, we need to find the partial fraction decomposition of f(x) = x2+2x+2x3−3x2+2x .Factoring Q(x), we getQ(x) = x(x2 − 3x+ 2) = x(x− 1)(x− 2)Since Q(x) is a product of distinct linear factors, the partial fraction decompo-sition of f(x) has the formx2 + 2x+ 2x3 − 3x2 + 2x =Ax+Bx− 1 +Cx− 2 ,where A,B, and C are constants that we need to determine.Multiplying both sides of the latter equation by x(x−1)(x−2), and distributingthe righthand side of the resulting equality, we getx2 + 2x+ 2 = (A+B + C)x2 + (−3A− 2B − C)x+ 2AThis implies thatA + B + C = 1 coefficients of x2−3A − 2B − C = 2 coefficients of x2A = 2 constant termsFrom the third equation, we have A = 1. Substituting this into the other2.2. Partial Fractions 217equations, we get the system1 + B + C = 1−3 − 2B − C = 2which is equivalent toB + C = 0−2B − C = 5Isolating B in the first equation of this latter system, we get B = −C. Sub-stituting this into the second equation, we get −2(−C)− C = 5, which impliesthat C = 5, and B = −C = −5.So, the partial fraction decomposition of f(x) isx2 + 2x+ 2x3 − 3x2 + 2x =1x− 5x− 1 +5x− 2Now, we can integrate∫x2 + 2x+ 2x3 − 3x2 + 2x dx.∫x2 + 2x+ 2x3 − 3x2 + 2x dx =∫ (1x− 5x− 1 +5x− 2)dx=∫1xdx− 5∫1x− 1 dx+ 5∫1x− 2 dx= ln |x| − 5 ln |x− 1|+ 5 ln |x− 2|+ C(d) To find the definite integral∫ 1012x2 + 5x− 12 dx, we first need to find the in-definite integral ∫ 1012x2 + 5x− 12 dxLet P (x) = 1 and Q(x) = 2x2 + 5x − 12. Factoring Q(x), we get Q(x) =(2x − 3)(x + 4). Since Q(x) is a product of distinct linear factors, the partialfraction decomposition of P (x)Q(x)is of the form1(2x− 3)(x+ 4) =A2x− 3 +Bx+ 4Multiplying both sides of this latter equation by (2x− 3)(x+ 4), we get1 = A(x+ 4) +B(2x− 3)218 Chapter 2. Techniques of Integration and Computational Methods Substituting x with −4, the latter equation becomes1 = A(−4 + 4) +B(2(−4)− 3)1 = −11B− 111= B Divide both sides by −11 Substituting x with 32into the equation 1 = A(x+ 4) +B(2x− 3), we get1 = A(32+ 4)+B(2(32)− 3)1 =112A211= A Multiply both sides by211So, the partial fraction decomposition of P (x)Q(x)is12x2 + 5x− 12 =211(2x− 3) +− 111x+ 4=211(2x− 3) −111(x+ 4)abc=abcIntegrating this, we have∫12x2 + 5x− 12 dx=∫ (211(2x− 3) −1x+ 4)dx=211∫12x− 3 dx−111∫1x+ 4dx=211∫1u(du2)− 111∫1x+ 4dxUse substitution for the first in-tegral: Let u = 2x − 3. Thendu = 2dx so that du2= dx=111∫1udu− 111∫1x+ 4dx=111ln |u| − 111ln |x+ 4|+ C∫1udu = ln |u|+ C=111ln |2x− 3| − 111ln |x+ 4|+ C Substitute u with 2x− 32.2. Partial Fractions 219Now, we can find the definite integral∫ 1012x2+5x−12 dx.∫ 1012x2 + 5x− 12 dx=(111ln |2x− 3| − 111ln |x+ 4|)∣∣∣∣10=(111ln |2− 3| − 111ln |1 + 4|)−(111ln |0− 3| − 111ln |0 + 4|)=(111ln | − 1| − 111ln |5|)−(111ln | − 3| − 111ln |4|)= − 111ln 5−(111ln 3− 111ln 4)ln | − 1| = ln(1) = 0, | − 3| = 3= − 111ln 5− 111ln 3 +111ln 4(e) Using the partial fraction decomposition from (2.2.3), we get∫ −2x+ 5x(x− 1)2 dx=∫ (5x− 5x− 1 +3(x− 1)2)dx= 5∫1xdx− 5∫1x− 1 dx+ 3∫1(x− 1)2 dx= 5 ln |x| − 5∫1udu+ 3∫1u2duUse substitution: u = x−1,so that du = dx= 5 ln |x| − 5 ln |u|+ 3∫u−2 du∫1xdx = ln |x|+ C= 5 ln |x| − 5 ln |u|+ 3u−1−1 + C∫un du =un+1n+ 1+ C= 5 ln |x| − 5 ln |u| − 3u+ C= 5 ln |x| − 5 ln |x− 1| − 3x− 1 + C Substitute u with x− 1(f) First, we need to decompose f(x) = 2x2−3x−14x3+4x2+xinto partial fractions.• Factoring Q(x), we get Q(x) = x(4x2 + 4a+ 1) = x(2x+ 1)2.• Since Q(x) is a product of linear factors, some of which are repeated, the220 Chapter 2. Techniques of Integration and Computational Methodspartial fraction decomposition of f(x) has the form2x2 − 3x− 14x3 + 4x2 + x=Ax+B2x+ 1+C(2x+ 1)2• Finding the constants. Multiplying both sides of this latter equation byx(2x+ 1)2, and distributing the righthand side of the resulting equality, weget2x2 − 3x− 1 = (4A+ 2B)x2 + (4A+B + C)x+ ATherefore,4A + 2B = 2 coefficients of x24A + B + C = −3 coefficients of xA = −1 constant termsSolving this system, we get A = −1, B = 3, and C = −2.• So the partial fraction decomposition of f(x) is2x2 − 3x− 14x3 + 4x2 + x=−1x+32x+ 1− 2(2x+ 1)2We now find the integral.∫2x2 − 3x− 14x3 + 4x2 + xdx=∫ (−1x+32x+ 1− 2(2x+ 1)2)dx= −∫1xdx+ 3∫12x+ 1dx− 2∫1(2x+ 1)2dx= −∫1xdx+ 3∫1u(du2)− 2∫1u2(du2)Use substitution: u = 2x+ 1,so that du = 2dx and du2= dx= −∫1xdx+32∫1udu− 22∫u−2 du= − ln |x|+ 32ln |u| −(u−1−1)+ C∫un du =un+1n+ 1+ C= − ln |x|+ 32ln |u|+ 1u+ C= − ln |x|+ 32ln |2x+ 1|+ 12x+ 1+ C Substitute u with 2x+ 12.2. Partial Fractions 2213. [Exercise on page 21]LetP (x) = 2x3 − 3x2 + 3x+ 2 and Q(x) = x2 − x− 2Then f(x) = P (x)Q(x). Here the degree of P (degP = 3) is not less than the degree of Q(degQ = 2). In this case, we need to use long division.Improper Rational FunctionsRecall that a rational function P (x)Q(x)is improper if deg(P ) ≥ deg(Q). To findthe integral∫P (x)Q(x)dx when degP ≥ degQ, one can proceed as follows.• Perform long division to divide the dividend P (x) by the divisor Q(x)until a remainder R(x) is obtained such that degR < degQ. If S(x) isthe quotient, thenP (x) = Q(x)S(x) +R(x)P (x)Q(x)= S(x) +R(x)Q(x)Divide both sides by Q(x)• The integral becomes∫P (x)Q(x)dx =∫S(x) dx+∫R(x)Q(x)dx,where∫S(x) dx can be evaluated easily because S(x) is a polynomial,and∫R(x)Q(x)dx can be evaluated using the method of partial fractions.Performing long division, we have222 Chapter 2. Techniques of Integration and Computational Methods2x− 1x2 − x− 2)2x3 −3x2 +3x +22x3 −2x2 −4x−x2 +7x +2−x2 +x +26xSo, the remainder is R(x) = 6x and the quotient is S(x) = 2x − 1. Since, P (x)Q(x)=S(x) + R(x)Q(x), it follows that2x3 − 3x2 + 3x+ 2x2 − x− 2 = 2x− 1 +6xx2 − x− 2So, the integral becomes∫2x3 − 3x2 + 3x+ 2x2 − x− 2 =∫(2x− 1)dx+∫6xx2 − x− 2dx= x2 − x+∫6xx2 − x− 2dxFrom Exercise 2b, we have∫6xx2 − x− 2dx = 2 ln |x+ 1|+ 4 ln |x− 2|+ CThus, ∫2x3 − 3x2 + 3x+ 2x2 − x− 2 dx = x2 − x+ 2 ln |x+ 1|+ 4 ln |x− 2|+ C4. [Optional] [Exercise on page 21](a) First, we need to decompose 5x3+5xinto partial fractions.• Here P (x) = 5, degP = 0 and Q(x) = x3−5x, degQ = 3. So degP < degQ.• Factoring Q(x), we get Q(x) = x(x2 + 5).• Since Q(x) contains an irreductible quadratic factor (x2 + 5), with no rep-etition (this falls into Case III), the partial fraction decomposition of f(x)2.2. Partial Fractions 223has the form5x(x2 + 5)=Ax+Bx+ Cx2 + 5, (2.2.4)where A,B, and C are constants. Note that the partial fraction Axis due tothe factor x, while Bx+Cx2+5is due to the irreductible quadratic factor x2 + 5.• Finding the constants. Multiplying both sides of (2.2.4) by the commondenominator x(x2 + 5), we get 5 = A(x2 + 5) + (Bx+C)x. Distributing theright-hand side, we get 5 = Ax2 + 5A+Bx2 + Cx, which is equivalent to0x2 + 0x+ 5 = (A+B)x2 + Cx+ 5AIdentifying the coefficients, we get the following system of three equations.A + B = 0 coefficients of x2C = 0 coefficients of x5A = 5 constant termsFrom the third equation, we have A = 55= 1. Substituting this into the firstequation, we get 1 + B = 0, which implies that B = −1. From the secondequation, C = 0.• Hence, A = 1, B = −1, C = 0, and the partial fraction decomposition off(x) isf(x) =5x3 + 5x=1x+−1x+ 0x2 + 5=1x− xx2 + 5Now, we can find the integral:∫5x3 + 5xdx=∫ (1x− xx2 + 5)dx=∫1xdx−∫xx2 + 5dx= ln |x| −∫xx2 + 5dx∫1xdx = ln |x|+ C= ln |x| −∫1u(du2)Use substitution: u = x2+5, sothat du = 2xdx and du2= xdx= ln |x| − 12∫1udu∫k dx = kx+ C224 Chapter 2. Techniques of Integration and Computational Methods= ln |x| − 12ln |u|+ C∫1udu = ln |u|+ C= ln |x| − 12ln |x2 + 5|+ C Substitute u with x2 + 5(b) First, we need to decompose 2x3+x2+4x+3(x2+3)2into partial fractions.Solution.• Here P (x) = 2x3 + x2 + 4x+ 3, degP = 3, and Q(x) = (x2 + 3)2, degQ = 4.So degP < degQ.• Since Q(x) contains a repeated irreductible quadratic factor (this falls intoCase IV), the partial fraction decomposition of f(x) has the form (see )2x3 + x2 + 4x+ 3(x2 + 3)2=Ax+Bx2 + 3+Cx+D(x2 + 3)2, (2.2.5)where A,B,C, and D are constants.• We need to find the constants. Multiplying both sides of (2.2.5) by thecommon denominator (x2 + 3)2, we get2x3 + x2 + 4x+ 3 = (Ax+B)(x2 + 3) + Cx+DDistributing the right-hand side, we get2x3 + x2 + 4x+ 3 = Ax3 + 3Ax+Bx2 + 3B + Cx+D,which is equivalent to2x3 + x2 + 4x+ 3 = Ax3 +Bx2 + (3A+ C)x+ 3B +DIdentifying the coefficients, we get the following system of four equations.A = 2 coefficients of x3B = 1 coefficients of x23A+ C = 4 coefficients of x3B +D = 3 constant termsFrom the first two equations, we have A = 2 and B = 1. Substituting A intothe third equation, we get 3(2)+C = 4, which implies that C = 4−6 = −2.And substituting B into the fourth equation, we get 3(1) + D = 3, whichimplies that D = 0.2.2. Partial Fractions 225• Hence, A = 2, B = 1, C = −2, D = 0, and the partial fraction decomposi-tion of f(x) is2x3 + x2 + 4x+ 3(x2 + 3)2=2x+ 1x2 + 3− 2x(x2 + 3)2(2.2.6)Using this, we get∫2x3 + x2 + 4x+ 3(x2 + 3)2dx=∫ (2x+ 1x2 + 3− 2x(x2 + 3)2)dx=∫2x+ 1x2 + 3dx−∫2x(x2 + 3)2dx=∫ (2xx2 + 3+1x2 + 3)dx−∫2x(x2 + 3)2dx=∫2xx2 + 3dx+∫1x2 + 3dx−∫2x(x2 + 3)2dx=∫1udu+∫1x2 + 3dx−∫1u2duUse substitution: u = x2 + 3, sothat du = 2xdx. The integral inthe middle will be treated belowwith a different substitution= ln |u|+∫1x2 + 3dx−(−1u) ∫1udu = ln |u|+ C= ln |x2 + 3|+∫1x2 + 3dx+1x2 + 3Substitute u with x2 + 3Before we continue, we need to find the integral∫1x2 + 3dx. First, we rewriteit as ∫1x2 + 3dx =∫13(1 + x23)dx =13∫11 +(x√3)2dxLet w = x√3. Then dw = 1√3dx, so that dx =√3 dw. The integral becomes∫1x2 + 3dx =13∫11 +(x√3)2 dx=13∫11 + w2(√3 dw)=√33∫11 + w2dw∫k dx = kx+ C226 Chapter 2. Techniques of Integration and Computational Methods=√33arctan(w)∫11 + u2dw = arctan(w) + C=√33arctan(x√3)Substitute w withx√3Returning to the original integral, we get∫2x3 + x2 + 4x+ 3(x2 + 3)2dx = ln |x2 + 3|+∫1x2 + 3dx+1x2 + 3= ln |x2 + 3|+√33arctan(x√3)+1x2 + 3+ C2.3 Practicing Integration and Integration by Tables1. [Exercise on page 21]Strategy for IntegrationThis exercise book covers three techniques of integration: substitution, inte-gration by parts, and partial fractions. A natural question one can ask is thefollowing: Which technique to use for a given integral? It is very helpful to recognize an integral. If you can recognize an integralor the type of integral, it will only take you a few seconds to choose theright technique. Doing many integrals helps to recognize different types of integrals. If you are not able to recognize an integral, you could do the following.– Try to simplify the integrand.– Try the substitution method.– Try integration by parts.– Try partial fractions.Useful Tips When Integral Has an Exponential/LogarithmSuppose we have an integral of the form∫P (x)eg(x) dx2.3. Practicing Integration and Integration by Tables 227where P (x) is a polynomial. If the derivative of g(x) is P (x) (apart from a constant factor), use sub-stitution by letting u = g(x). Otherwise, try integration by parts by lettingu = P (x) and dv = eg(x)dxSuppose we want to take an integral of the form∫h(x)[ln(g(x))]n dx If the derivative of ln(g(x)) is h(x) (apart from a constant factor), thenuse substitution by letting u = ln(g(x)). Otherwise, try integration by parts by lettingu = [ln(g(x))]n and dv = h(x)dx(a) Let’s find∫x2ex3dx. Notice that the derivative of x3 is 3x2 and 3x2 is in theintegral (apart from the constant factor 3 ). So, we will use substitution. Letu = x3. Then, du = 3x2 dx so thatdu3= x2 dx. Therefore,∫x2ex3dx =∫eu(du3)Substitute=13∫eu du∫kf(x) dx = k∫f(x) dx=13eu + C∫ex dx = ex + C=13ex3+ C Substitute u by x3(b) Here, the integral has the same form as before, but the derivative of 3x (whichis 3), is not equal to x2, apart from a constant factor. So, we will try integrationby parts.Choose u = x2 and dv = e3x dxThen, du = 2x dx and v =∫e3x dx =13e3x228 Chapter 2. Techniques of Integration and Computational Methods∫x2e3x dx= uv −∫vdu= x2(13e3x)−∫ (13e3x)(2xdx) Substitute u and v=13x2e3x − 23∫xe3x dx∫kf(x) dx = k∫f(x) dx=13x2e3x − 23[x(13e3x)−∫ (13e3x)(dx)]Integrate by parts again=13x2e3x − 23(13xe3x − 13∫e3x dx) ∫kf(x) dx = k∫f(x) dx=13x2e3x − 23(13xe3x − 13(13e3x))+ C∫ekx dx =1kekx + C=13x2e3x − 23(13xe3x − 19e3x)+ C=13x2e3x − 29xe3x +227e3x + C(c) One can rewrite this as∫1x(lnx)12 dx. Since the derivative of lnx is1xand1xis in the integral, we will use the substitution method.Let u = lnx. Then, du =1xdx. So,∫ √lnxxdx =∫1x(lnx)12 dx√x = x12=∫u12 du Substitute=u3232+ C∫xn dx =xn+1n+ 1+ C=23u32 + Cabc=acb=23(lnx)32 + C Substitute u by lnx(d) Let’s try the substitution u = x + 1. Then, du = dx. Solving the equation2.3. Practicing Integration and Integration by Tables 229u = x+ 1 for x, we get x = u− 1. So,∫x2√x+ 1 dx=∫(u− 1)2√u du Substitute=∫(u2 − 2u+ 1)u 12 du √x = x 12=∫ (u52 − 2u 32 + u 12)du Distribute=u7272− 2u5252+u3232+ C∫xn dx =xn+1n+ 1+ C=27u72 − 45u52 +23u32 + Cabc=acb=27(x+ 1)72 − 45(x+ 1)52 +23(x+ 1)32 + C Substitute u by x+ 1(e) The derivative of ln(4x) is44x=1x, and we don’t have1xin the integral. So, itwon’t be a good idea to try substitution. Let’s try integration by parts instead.Choose u = ln(4x) and dv = (−3x+ 2)dxThen, du = 1xdx and v =∫(−3x+ 2) dx = −32x2 + 2x∫(−3x+ 2) ln(4x) dx= ln(4x)(−32x2 + 2x)−∫ (−32x2 + 2x)(1xdx)Substitute u and v= −32x2 ln(4x) + 2x ln(4x)−∫ (−32x+ 2)dx Distribute= −32x2 ln(4x) + 2x ln(4x)−(−32x22+ 2x)+ C∫xn dx =xn+1n+ 1+ C,∫k dx = kx+ C= −32x2 ln(4x) + 2x ln(4x) +34x2 − 2x+ C(f) At first glance, this integral looks similar to the integral from Exercise 1c. How-ever, they are quite different. In Exercise 1c, we used substitution because the230 Chapter 2. Techniques of Integration and Computational Methodsderivative of lnx was in the integral. Here, the derivative of lnx is not in theintegral. So, we will try integration by parts.Choose u = lnx and dv = 1√xdxThen, du = 1xdx and v =∫1√xdx =∫x−12 dx = 2x12So, ∫lnx√xdx= (lnx)(2x12 )−∫(2x12 )(1xdx)Substitute u and v= 2√x lnx− 2∫x−12 dx∫kf(x) dx = k∫f(x) dx,x12x1= x12−1 = x−12= 2√x lnx− 2(2x12)+ C∫xn dx =xn+1n+ 1+ C,abc=acb= 2√x lnx− 4√x+ CWe now find the definite integral.∫ e1lnx√xdx=(2√x lnx− 4√x)∣∣∣∣e1∫ baf(x) dx = F (x)∣∣∣∣ba=(2√e ln e− 4√e)− (2√1 ln 1− 4√1) F (x)∣∣∣∣ba= F (b)− F (a)= 2√e− 4√e− (0− 4)= −2√e+ 4(g) Since the derivative of the denominator is the numerator (apart from a constantfactor), we will use the substitution method. Let u = 3x4 + 7. Then, du =12x3 dx so thatdu12= x3 dx.If x = 0, u = 7 If x = 1, u = 102.3. Practicing Integration and Integration by Tables 231Therefore,∫ 10x33x4 + 7dx=∫ 1071u(du12)Substitute=112∫ 1071udu∫kf(x) dx = k∫f(x) dx=112ln |u|∣∣∣∣107∫1xdx = ln |x|+ C,∫ baf(x) dx = F (x)∣∣∣∣ba=112(ln |10| − ln |7|) F (x)∣∣∣∣ba= F (b)− F (a)(h) First we rewrite the integral as∫tan (√x)√xdx =∫sin(√x)√x cos(√x)dx tan θ =sin θcos θLet’s try the substitution u = cos(√x). Using the chain rule, we getdu =ddx[cos(√x)] dx = − sin(√x) ddx[√x] dxdu = − sin(√x)(12√x)dxddx[√x] =12√x−2du = sin(√x)√xdx Multiply both sides by −2So,∫sin(√x)√x cos(√x)dx =∫1cos(√x)(sin(√x)√xdx)=∫1u(−2 du) Substitute= −2∫1udu∫kf(x) dx = k∫f(x) dx= −2 ln |u|+ C∫1udu = ln |u|+ C= −2 ln | cos(√x)|+ C Substitute u by cos(√x)2. [Exercise on page 21]232 Chapter 2. Techniques of Integration and Computational Methods(a) Here the integrand is 1x2−4 . Since the derivative of the denominator is not equalto the numerator (except for a constant factor), it is not a good idea to use thesubstitution u = x2 − 4. Observe that the integrand f(x) = 1x2−4 is a rationalfunction. So, we can try partial fractions.First, we need to find A and B such that1(x− 2)(x+ 2) =Ax− 2 +Bx+ 2Multiplying both sides of this latter equation by (x− 2)(x+ 2), we get1 = A(x+ 2) +B(x− 2)1 = Ax+ 2A+Bx− 2B Distribute the righthand side1 = (A+B)x+ 2A− 2B0x+ 1 = (A+B)x+ 2A− 2BThis implies thatA+B = 0 and 2A− 2B = 1From the first equation, we have A = −B. Substituting this into the secondequation, we get 2(−B)−2B = 1 or −4B = 1. This latter equation implies thatB = −14. Therefore, A = −B = 14.Alternate Method to Find A and B1(x− 2)(x+ 2) =Ax− 2 +Bx+ 21 = A(x+ 2) +B(x− 2) Multiply both sides by(x− 2)(x+ 2)This method consists of substituting x with numbers into the latter equa-tion. To make things easier, we substitute x with the roots of the denom-inator; that is, −2 and 2. If x = −2, the lefthand side of the equation 1 = A(x+ 2) +B(x− 2)becomes 1. And the righthand side becomes A(−2+2)+B(−2−2) =2.3. Practicing Integration and Integration by Tables 233−4B. So,1 = −4B−14= B Divide both sides by −4 If x = 2, the lefthand side of the equation 1 = A(x+ 2) + B(x− 2)becomes 1. And the righthand side becomes A(2+2)+B(2−2) = 4A.So,1 = 4A14= A Divide both sides by 4Using the values of A and B found above, we have the following partial fractiondecomposition:1(x− 2)(x+ 2) =14x− 2 +−14x+ 2Now, we can find the integral∫1x2−4 dx.∫1x2 − 4 dx =∫1(x− 2)(x+ 2) dx=∫ ( 14x− 2 +−14x+ 2)dx=14∫1x− 2 dx−14∫1x+ 2dx=14∫1udu− 14∫1wdwUse substitution: u = x−2,du = dx. And w = x + 2,dw = dx=14ln |u| − 14ln |w|+ C∫1xdx = ln |x|+ C=14ln |x− 2| − 14ln |x+ 2|+ C Return to the original variable(b) Let f(x) = P (x)Q(x)with P (x) = x and Q(x) = x2 − 4. Since the derivative of thedenominator, 2x, is equal to the numerator (apart from the constant factor 2),we will use substitution. Let u = x2 − 4x. Then du = 2xdx, so that du2= xdx.234 Chapter 2. Techniques of Integration and Computational MethodsThe integral becomes∫xx2 − 4 dx =∫1u(du2)=12∫1udu∫kf(x) dx = k∫f(x) dx=12ln |u|+ C∫1udu = ln |u|+ C=12ln |x2 − 4|+ CNote. It is possible to find this integral using partial fractions. (Using substi-tution is much easier.)(c) Let f(x) = P (x)Q(x)with P (x) = 1 and Q(x) = x2 + 4. The derivative of thedenominator is not equal to the numerator (apart from a constant). So, we won’tuse the substitution u = Q(x) as we did in Exercise 2b. The partial fractionsand integration by parts methods won’t work either. So, which technique touse?Recall the following basic integration formula:∫11 + u2du = tan−1(u) + CHere, we want to take the integral of 14+x2. To get a 1 in place of 4, we factor 4in the denominator to obtain14 + x2=14(1 + x24) = 14(1 +(x2)2)Now, we make the substitution u = x2. Then du = 12dx, so that 2du = dx and∫14 + x2dx =∫14(1 +(x2)2) dx=14∫11 +(x2)2 dx ∫ kf(x) dx = k ∫ f(x) dx=14∫11 + u2(2du) u =x2, 2du = dx=24∫11 + u2du∫kf(x) dx = k∫f(x) dx2.3. Practicing Integration and Integration by Tables 235=12tan−1(u) + C∫11 + u2du = tan−1 u+ C=12tan−1(x2)+ C(d) For the integral∫x4 sin2(x5) dx, the obvious substitution is u = x5. Let’s see ifthis works. Taking the differential of u, we getdu = 5x4 dx, so thatdu5= x4dxThe integral becomes∫x4 sin2(x5) dx=∫sin2(x5)(x4 dx)=∫sin2(u)(du5)=15∫sin2(u) du∫kf(x) dx = k∫f(x) dx=15∫1− cos(2u)2duUse the trigonometrig iden-tity sin2(u) = 1−cos(2u)2=15∫ (12− 12cos(2u))dua− bc=ac− bc=15[12u− 12(12sin(2u))]+ C∫cos(ku) du =1ksin(ku) + C=15(12u− 14sin(2u))+ C=110u− 120sin(2u) + C Distribute=110x5 − 120sin(2x5) + C Substitute u by x5(e) The obvious substitution is u =√x. The differential of u is thusdu =12√xdx =12udxSo, du =12udx. Multiplying both sides by 2u, we get 2u du = dx. The integral236 Chapter 2. Techniques of Integration and Computational Methodsbecomes ∫e√x dx =∫eu(2udu) Substitute=∫2ueu duTo find this integral, we use integration by parts.Choose w = 2u and dv = euduThen, dw = 2du and v =∫eudu = euSo, ∫2ueu du = wv −∫vdw= 2ueu −∫2eudu Substitute w and v= 2ueu − 2eu + CAltogether,∫e√x dx =∫2ueu du= 2ueu − 2eu + C= 2√xe√x − 2e√x + C Substitute u by√x= 2(−1 +√x) e√x + C(f) The obvious substitution is u = x4 + 1. Taking the differential of u, we getdu = 4x3 dx, which implies that du4= x3 dx. So,∫x7 cos(x4 + 1)dx=∫x4 · x3 cos(x4 + 1) dx x7 = x4 · x3=∫x4 cos(u)(du4)u = x4 + 1,du4= x3dx=∫(u− 1) cos(u)(du4)From u = x4 + 1, we havex4 = u− 12.3. Practicing Integration and Integration by Tables 237=14∫(u− 1) cos(u) du∫kf(x) dx = k∫f(x) dx=14[zt−∫t dz] Use integration by parts: choosez = u− 1 and dt = cos(u). Thendz = 1du and t = sin(u)=14[(u− 1) sin(u)−∫sin(u) du]Substitute z and t=14[(u− 1) sin(u)− (− cos(u))] + C∫sinu du = − cosu+ C=14(u− 1) sin(u) + 14cos(u) + C Distribute=14x4 sin(x4 + 1) +14cos(x4 + 1) + C Substitute u by x4 + 1(g) Let u = cosx. Then du = − sinx dx, so that −du = sinx dx and∫sinxcos2 x+ cosxdx =∫1(cosx)2 + cosx(sinx dx)=∫1u2 + u(−du) u = cosx, −du = sinx dx=∫ −1u(u+ 1)duTo find this latter integral, we use partial fractions. Since the denominator is theproduct of distinct linear factors, the partial fraction decomposition of −1u(u+1)isof the form −1u(u+ 1)=Au+Bu+ 1Multiplying both sides by u(u+ 1), we get−1 = A(u+ 1) +BuTo find the constants A and B, we substitute in values of u. To make thingseasier, we substitute the roots/zeros of the denominator u(u+ 1); that is, 0 and−1. If u = 0, the equation −1 = A(u+ 1) +Bu becomes−1 = A(0 + 1) +B(0)−1 = A238 Chapter 2. Techniques of Integration and Computational Methods If u = −1, the equation −1 = A(u+ 1) +Bu becomes−1 = A(−1 + 1) +B(−1)−1 = −B1 = BSo, A = −1 and B = 1. And the partial fraction decomposition of −1u(u+1)is−1u(u+ 1)=−1u+1u+ 1Now, we can find the integral∫ −1u(u+1)du:∫ −1u(u+ 1)du =∫ (−1u+1u+ 1)du= −∫1udu+∫1u+ 1du= − ln |u|+ ln |u+ 1|+ C∫1udu = ln |u|+ C= − ln | cosx|+ ln | cosx+ 1|+ C Substitute u by cosx(h) Let u =√x+ 10. (This is not an obvious substitution!) Thendu =12√x+ 10dxddx[√x+ 10]=12√x+ 10du =12udx u =√x+ 102u du = dx Multiply both sides by 2uSo,∫13√x+ 10 + xdx =∫13u+ x(2u du)=∫13u+ (u2 − 10)(2u du)From u =√x+ 10, we haveu2 = x+ 10 or u2 − 10 = x=∫2uu2 + 3u− 10 du2.3. Practicing Integration and Integration by Tables 239=∫2u(u− 2)(u+ 5) duTo find this latter integral, we use partial fractions. Let f(u) = 2u(u−2)(u+5) .Since the denominator of f(u) is a product of distinct linear factors, the partialfraction decomposition of f(u) is of the form2u(u− 2)(u+ 5) =Au− 2 +Bu+ 5Multiplying both sides by (u− 2)(u+ 5), we get2u = A(u+ 5) +B(u− 2) If u = −5, the equation 2u = A(u+ 5) +B(u− 2) becomes2(−5) = A(−5 + 5) +B(−5− 2)−10 = −7B107= B If u = 2, the equation 2u = A(u+ 5) +B(u− 2) becomes2(2) = A(2 + 5) +B(2− 2)4 = 7A47= ASo, the partial fraction decomposition of f(u) is2u(u− 2)(u+ 5) =47u− 2 +107u+ 5Thus,∫2u(u− 2)(u+ 5) du=∫ ( 47u− 2 +107u+ 5)du=47∫1u− 2 du+107∫1u+ 5du=47ln |u− 2|+ 107ln |u+ 5|+ C Use substitution240 Chapter 2. Techniques of Integration and Computational Methods=47ln |√x+ 10− 2|+ 107ln |√x+ 10 + 5|+ C Substitute u by √x+ 103. [Exercise on page 22]Table of Integrals(1) If k is a constant, then ∫k dx = kx+ C(2) For any real number n 6= −1,∫xn dx =xn+1n+ 1+ C(3) For any real number k,∫kf(x) dx = k∫f(x) dx+ C(4)∫[f(x) + g(x)] dx =∫f(x) dx+∫g(x) dx(5)∫[f(x)− g(x)] dx =∫f(x) dx−∫g(x) dx(6)∫ex dx = ex + C(7) For any real number k 6= 0,∫ekx dx =1kekx + C(8) For any real number a such that a > 0 and a 6= 1,∫ax dx =axln a+ C(9) For any real number a such that a > 0 and a 6= 1, for any real numberk 6= 0, ∫akx dx =akxk(ln a)+ C2.3. Practicing Integration and Integration by Tables 241(10) The indefinite integral of 1xis ln |x|+ C, that is,∫1xdx = ln |x|+ C(11)∫cosx dx = sinx+ C(12)∫sinx dx = − cosx+ C(13)∫sec2 x dx = tanx+ C(14)∫csc2 x dx = − cotx+ C(15)∫secx tanx dx = secx+ C(16)∫cscx cotx dx = − cscx+ C(17)∫11 + x2dx = tan−1 x+ C (Another notation for tan−1 x is arctan x)(18)∫1√1− x2 dx = sin−1 x+ C (Another notation for sin−1 x is arcsin x)(19)∫ √x2 + a2 dx =x2√x2 + a2 +a22· ln |x+√x2 + a2|+ C(20)∫1√x2 + a2dx = ln∣∣∣x+√x2 + a2∣∣∣+ C(21)∫1√x2 − a2 dx = ln∣∣∣x+√x2 − a2∣∣∣+ C(22)∫1x√a2 − x2 dx = −1a· ln∣∣∣∣a+√a2 − x2x∣∣∣∣+ C (a 6= 0)(23)∫1x√a2 + x2dx = −1a· ln∣∣∣∣∣a+√a2 + x2x∣∣∣∣∣+ C (a 6= 0)(a) Consider the table of integrals above. Rule 19 says that∫ √x2 + a2 dx =x2√x2 + a2 +a22· ln |x+√x2 + a2|+ C242 Chapter 2. Techniques of Integration and Computational MethodsUsing this with a = 3, we get∫ √x2 + 9 dx =∫ √x2 + 32 dx=x2√x2 + 32 +322· ln |x+√x2 + 32|+ C=x2√x2 + 9 +92· ln |x+√x2 + 9|+ C(b) ∫10√2x2 + 50dx =∫10√2(x2 + 25)dx=∫10√2√x2 + 25dx=10√2∫1√x2 + 25dx∫kf(x) dx = k∫f(x) dx=10√2∫1√x2 + 52dx=10√2ln∣∣∣x+√x2 + 52∣∣∣+ C Rule 20=10√22ln∣∣∣x+√x2 + 52∣∣∣+ C 10√2=10√2√2√2=10√22= 5√2 ln∣∣∣x+√x2 + 25∣∣∣+ C(c) ∫ −4x√64− x2 dx = −4∫1x√82 − x2 dx∫kf(x) dx = k∫f(x) dx= −4(−18)ln∣∣∣∣8 +√82 − x2x∣∣∣∣+ C Rule 22=12ln∣∣∣∣8 +√64− x2x∣∣∣∣+ C(d) First, we can rewrite the integral as∫5√3x2 − 4 dx =∫5√(√3x)2 − 4 dx2.4. Improper Integrals 243Let u =√3x. Then du =√3 dx. This implies that du√3= dx. The integralbecomes∫5√3x2 − 4 dx =∫5√(√3x)2 − 4 dx=∫5u2 − 4(du√3)=5√3∫1√u2 − 22 du Take out5√3=5√3ln∣∣∣u+√u2 − 22∣∣∣+ C Rule 21=5√3ln∣∣∣∣∣√3x+√(√3x)2− 22∣∣∣∣∣+ C Substitute u with √3x=5√3ln∣∣∣√3x+√3x2 − 4∣∣∣+ C2.4 Improper Integrals1. [Exercise on page 22](a)Improper Integrals of the Form∫ ∞af(x)dxLet a be a real number, and let f be a function. The improper integral∫ ∞af(x)dx is defined to be the limit of∫ taf(x)dx as t approaches ∞.That is, ∫ ∞af(x)dx = limt→∞∫ taf(x)dxIf the limit exists (as a finite number), the improper integral∫ ∞af(x)dxis said to be convergent. Otherwise, it is said to be divergent.244 Chapter 2. Techniques of Integration and Computational MethodsFor Review – Limits of a Constant, 1tn, and tnLet n be a positive real number, let k be a constant, and let p be anonnegative integer. Then, we have the following basic limits.limt→±∞(k) = k (2.4.1)limt→±∞(1tn)= 0 (2.4.2)limt→−∞tp = ∞ if p is even−∞ if p is odd limt→∞ (tn) =∞ (2.4.3)By definition, we have∫ ∞11x4dx = limt→∞∫ t11x4dxCalculating the integral∫ t11x4dx, we get∫ t11x4dx =∫ t1x−4dx=x−4+1−4 + 1∣∣∣∣t1∫xn dx =xn+1n+ 1+ C=x−3−3∣∣∣∣t1F (x)∣∣∣∣ba= F (b)− F (a)= − 13x3∣∣∣∣t1=(− 13t3)−(− 13(1)3)= − 13t3+13Now, we find the limit.limt→∞∫ t11x4dx = limt→∞(− 13t3+13)From above= limt→∞(− 13t3)+ limt→∞(13)2.4. Improper Integrals 245= −13limt→∞(1t3)+ limt→∞(13)= −13(0) +13limt→±∞(k) = k, limt→±∞(1tn)= 0= 0 +13=13In conclusion, since the limit is a finite number, it follows that the improper int-geral∫ ∞11x4dx is convergent . (It converges to 13, and one can write∫ ∞11x4dx =13.)(b)By definition,∫ ∞45√xdx = limt→∞∫ t45√xdx.The integral is∫ t45√xdx = 5∫ t41x12dx∫kf(x) dx = k∫f(x) dx,√x = x12= 5∫ t4x−12 dx1xm= x−m= 5x1212∣∣∣∣∣t4∫xn dx =xn+1n+ 1+ C= 10x12∣∣∣∣t4abc=acb= 10(t12 − 4 12)F (x)∣∣∣∣ba= F (b)− F (a)= 10(t12 − 2)Now, we find the limit.limt→∞∫ t45√xdx = limt→∞(10(t12 − 2))From above= 10 limt→∞(t12 − 2)= 10(limt→∞(t12)− limt→∞(2))= 10 (∞− 2) limt→∞(tn) =∞, limt→±∞(k) = k246 Chapter 2. Techniques of Integration and Computational Methods= 10(∞) = ∞Since the limit is not a (finite) number, the improper integral∫ ∞45√xdx isdivergent .(c)For Review – Limits of Logarithmic FunctionsWe have the following basic limits.limt→0+ln t = −∞ and limt→∞ln t =∞. (2.4.4)We also have the following rule. Let g(t) be a function.If limt→±∞g(t) =∞, then limt→±∞ln[g(t)] =∞. (2.4.5)By definition,∫ ∞15xdx = limt→∞∫ t15xdx.So, we find∫ t15xdx = 5∫ t11xdx∫kf(x) dx = k∫f(x) dx= 5 ln |x|∣∣∣∣t1∫1xdx = ln |x|+ C= 5 (ln |t| − ln |1|) F (x)∣∣∣∣ba= F (b)− F (a)= 5 (ln |t| − 0)= 5 ln |t|Hence, the limit islimt→∞(5 ln |t|) = 5 limt→∞ln |t|= 5(∞) limt→∞ln |t| =∞= ∞2.4. Improper Integrals 247Since the limit is not a finite number, the improper integral∫ 515xdx is divergent .(d)By definition,∫ ∞212x− 3 dx = limt→∞∫ t212x− 3 dx.To evaluate∫ t212x− 3 dx, let u = 2x− 3. Then, du = 2dx so that dx =du2. Forour new bounds, we findIf x = 2, u = 2(2)− 3 = 1 and if x = t, u = 2t− 3.The integral becomes∫ t212x− 3 dx =∫ 2t−311u(du2)Substitute, change bounds=12∫ 2t−311udu∫kf(x) dx = k∫f(x) dx=12ln |u|∣∣∣∣2t−31∫1xdx = ln |x|+ C=12ln |2t− 3| − 12ln |1| F (x)∣∣∣∣ba= F (b)− F (a)=12ln |2t− 3| − 0=12ln |2t− 3|Finding the limit, we havelimt→∞∫ t212x− 3 dx = limt→∞12ln |2t− 3| From above=12limt→∞ln |2t− 3|=12(∞) limt→∞ln |2t− 3| =∞= ∞Since the limit is not a finite number, it follows that the improper integral∫ ∞212x− 3 dx is divergent .(e)248 Chapter 2. Techniques of Integration and Computational MethodsBy definition, we have∫ ∞31x(lnx)4dx = limt→∞∫ t31x(lnx)4dx.To evaluate the integral∫ t31x(lnx)4dx, substitute u = ln x. Then, du = 1xdx.For our new bounds, we findIf x = 3, u = ln 3 and if x = t, u = ln tThe integral becomes∫ t31x(lnx)4dx =∫ t31(lnx)4(1xdx)=∫ ln tln 31u4du Substitute, change bounds=∫ ln tln 3u−4 du1xm= x−m= − 13u3∣∣∣∣ln tln 3∫xn dx =xn+1n+ 1+ C=(− 13(ln t)3)−(− 13(ln 3)3)F (x)∣∣∣∣ba= F (b)− F (a)= − 13(ln t)3+13(ln 3)3Finding the limit, we havelimt→∞∫ t31x(lnx)4dx = limt→∞(− 13(ln t)3+13(ln 3)3)From above= limt→∞(− 13(ln t)3)+ limt→∞(13(ln 3)3)= −13limt→∞(1(ln t)3)+13(ln 3)3limt→±∞(k) = k= −131(limt→∞ln t)3 + 13(ln 3)3= −13(1(∞)3)+13(ln 3)3limt→∞ln t =∞= −13(1∞)+13(ln 3)32.4. Improper Integrals 249= −13(0) +13(ln 3)3=13(ln 3)31∞ = 0Since the limit is a finite number, we can conclude that the improper integral∫ ∞31x(lnx)4dx converges .(f) ∫ ∞0cos(2x) dx = limt→∞∫ t0cos(2x) dx= limt→∞(12sin(2x)∣∣∣∣t0) ∫cos(kx) dx =1ksin(kx) + C= limt→∞(12sin(2t)− 12sin(0))F (x)∣∣∣∣ba= F (b)− F (a)= limt→∞(12sin(2t))sin(0) = 0Since the limit of sin(2t) as t approaches ∞ does not exist (because sin(2t)bounces between −1 and 1, no matter how far out you look), it follows that theimproper integral∫∞0cos(2x) dx is divergent .2. [Exercise on page 22](a)Improper Integrals of the Form∫ b−∞f(x)dxLet b be a real number, and let f be a function. The improper integral∫ b−∞ f(x)dx is defined to be the limit of∫ btf(x)dx as t approaches −∞.That is, ∫ b−∞f(x)dx = limt→−∞∫ btf(x)dxIf the limit exists (that is, it is a finite number), the improper integral∫ b−∞ f(x)dx is said to be convergent. Otherwise, it is said to be diver-gent.250 Chapter 2. Techniques of Integration and Computational MethodsFor Review - Limits of Exponential Functionslimt→−∞et = 0 and limt→∞et =∞. (2.4.6)If limt→∞g(t) = −∞, then limt→∞eg(t) = 0. (2.4.7)If limt→∞g(t) =∞, then limt→∞eg(t) =∞. (2.4.8)If limt→−∞g(t) = −∞, then limt→−∞eg(t) = 0. (2.4.9)If limt→−∞g(t) =∞, then limt→−∞eg(t) =∞. (2.4.10)Here is good way to memorize these limits. Suppose we want to find thelimit of eg(t) as t approaches ±∞. If the limit of the exponent, g(t), is −∞, then the limit of eg(t) is 0. If the limit of the exponent, g(t), is ∞, then the limit of eg(t) is ∞.If the base is not e, one can use the following formula to change it into e.ax = ex ln a (2.4.11)One should also know the following useful limits. Let k be a positive realnumber. Then,limt→∞te−kt = 0 and limt→−∞tekt = 0. (2.4.12)By definition, we have∫ 0−∞ 3x dx = limt→−∞∫ 0t3x dx.Calculating the integral∫ 0t3x dx, we find∫ 0t3x dx =3xln 3∣∣∣∣0t∫ax dx =axln a+ C=30ln 3− 3tln 3F (x)∣∣∣∣ba= F (b)− F (a)=1ln 3− 3tln 32.4. Improper Integrals 251Finding the limit, we havelimt→−∞∫ 0t3x dx = limt→−∞(1ln 3− 3tln 3)From above= limt→−∞(1ln 3)− 1ln 3limt→−∞3t=1ln 3− 1ln 3limt→−∞et ln 3 ax = ex ln a=1ln 3− 1ln 3(0) limt→−∞et ln 3 = 0=1ln 3− 0 = 1ln 3Because the limit is a finite number, the integral is convergent .(b) By definition,∫ 0−∞ 3xe2x dx = limt→−∞∫ 0t3xe2x dx.To finding the integral∫ 0t3xe2x dx, let us first find the indefinite integral∫3xe2x dxby parts.Choose u = 3x and dv = e2xThen, du = 3dx and v =∫e2xdx =12e2xTherefore,∫3xe2x dx = uv −∫vdu=32xe2x −∫32e2x dx Substitute u and v=32xe2x − 32[12e2x]+ C∫ekx dx =1kekx + C=32xe2x − 34e2x + CSo, an antiderivative of f(x) = 2xe3x is F (x) = 32xe2x − 34e2x. Using the Funda-mental Theorem of Calculus (part II), we get∫ 0t3xe2x dx = F (0)− F (t)∫ baf(x) dx = F (b)− F (a)=(0− 34e0)−(32te2t − 34e2t)252 Chapter 2. Techniques of Integration and Computational Methods= −34− 32te2t +34e2tFinding the limit, we havelimt→−∞∫ 0t3xe2x dx = limt→−∞(−34)− limt→−∞(32te2t)+ limt→−∞(34e2t)= −34− 32limt→−∞(te2t) +34limt→−∞(e2t)= −34− 32(0) +34(0)limt→−∞(te2t) = 0limt→−∞e2t = 0= −34Since the limit is a finite number, we can conclude that the integral is convergent .(c) By definition,∫ 1−∞xx2+7dx = limt→−∞∫ 1txx2+7dx.To find the integral∫ 1txx2+7dx, substitute u = x2 + 7. Then, du = 2xdx so thatxdx = du2. For our new bounds, we findIf x = t, u = t2 + 7 and if x = 1, u = 12 + 7 = 8The integral becomes∫ 1txx2 + 7dx =∫ 1t1x2 + 7(xdx)=∫ 8t2+71u(du2)Substitute, change bounds=12ln |u|∣∣∣∣8t2+7∫1xdx = ln |x|+ C=12ln |8| − 12ln |t2 + 7| F (x)∣∣∣∣ba= F (b)− F (a)=12ln(8)− 12ln(t2 + 7) t2 + 7 ≥ 0Finding the limit, we havelimt→−∞∫ 1txx2 + 7dx =12ln(8)− 12limt→−∞ln(t2 + 7) From above2.4. Improper Integrals 253=12ln(8)− 12(∞) limt→−∞(t2 + 7) =∞=12ln(8)−∞ = −∞Since the limit is not a finite number, it follows that the integral diverges .(d) By definition,∫ 0−∞−2x+ 53x2 − 13x+ 4 dx = limt→−∞∫ 0t−2x+ 53x2 − 13x+ 4 dxTo find the integral∫ 0t−2x+53x2−13x+4 dx, we will use partial fractions. First, we findthe corresponding indefinite integral Let f(x) = −2x+53x2−13x+4 . Factoring the denominator of f(x), we get3x2 − 13x+ 4 = (3x− 1)(x− 4) Since the denominator is a product of distinct linear factors, the partialfraction decomposition of f(x) has the form−2x+ 5(3x− 1)(x− 4) =A3x− 1 +Bx− 4 To find A and B, we multiply both sides of the latter equation by thecommon denominator (3x− 1)(x− 4) to obtain−2x+ 5 = A(x− 4) +B(3x− 1)– Substituting x with 4 into the latter equation, we get−2(4) + 5 = A(4− 4) +B(3(4)− 1)−3 = 11B− 311= B– If x is 13, the equation −2x+ 5 = A(x− 4) +B(3x− 1) becomes−2(13)+ 5 = A(13− 4)+B(3(13)− 1)−23+ 5 = −113A+B(1− 1)254 Chapter 2. Techniques of Integration and Computational Methods133= −113A−1311= ASo,∫ −2x+ 5(3x− 1)(x− 4) dx=∫ ( −13113x− 1 +− 311x− 4)dx= −1311∫13x− 1 dx−311∫1x− 4 dx= −1311∫1u(du3)− 311∫1wdwUse substitution: Let u =3x − 1. Then du = 3dx,du3= dx. Let w = x − 4.Then dw = dx= −1333∫1udu− 311∫1wdw∫kf(x) dx = k∫f(x) dx= −1333ln |u| − 311ln |w|+ C∫1udu = ln |u|+ C= −1333ln |3x− 1| − 311ln |x− 4|+ C Substitute u by 3x−1, Sub-stitute w by x− 4Thus, an antiderivative of −2x+53x2−13x+4 isF (x) = −1333ln |3x− 1| − 311ln |x− 4|Using the Fundamental Theorem of Calculus (part II), we get∫ 0t−2x+ 53x2 − 13x+ 4 dx= F (0)− F (t)=[−1333ln | − 1| − 311ln | − 4|]−[−1333ln |3t− 1| − 311ln |t− 4|]=[−1333ln(1)− 311ln(4)]−[−1333ln |3t− 1| − 311ln |t− 4|] | − 1| = 1| − 4| = 4= − 311ln(4)−(−1333ln |3t− 1| − 311ln |t− 4|)ln(1) = 02.4. Improper Integrals 255= − 311ln(4) +133(13 ln |3t− 1|+ 9 ln |t− 4|)Sincelimt→−∞ln |3t− 1| = ln | limt→−∞(3t− 1)| = ln | −∞| =∞andlimt→−∞ln |t− 4| = ln | limt→−∞(t− 4)| = ln | −∞| =∞,it follows thatlimt→−∞[− 311ln(4) +133(13 ln |3t− 1|+ 9 ln |t− 4|)]= − 311ln(4) +133[13(∞) + 9(∞)]= − 311ln(4) +∞ = ∞Since the limit limt→−∞∫ 0tf(x) dx is not a number, it follows that the improperintegral∫ 0−∞−2x+ 53x2 − 13x+ 4 dx diverges .(e)Improper Integrals of the Form∫ ∞−∞f(x)dxLet f be a function. The improper integral∫∞−∞ f(x)dx is defined to bethe sum of improper integrals∫ a−∞ f(x)dx and∫∞af(x)dx, where a is anyreal number. That is,∫ ∞−∞f(x)dx =∫ a−∞f(x)dx+∫ ∞af(x)dx.We say that∫∞−∞ f(x)dx converges if both∫ a−∞ f(x)dx and∫∞af(x)dxare convergent, and diverges otherwise.It is convenient to choose a = 0. We can then write the integral as∫ ∞−∞x3e−x4dx =∫ 0−∞x3e−x4dx+∫ ∞0x3e−x4dx256 Chapter 2. Techniques of Integration and Computational MethodsDetermining whether∫ 0−∞x3e−x4dx converges or diverges, we find∫ 0−∞x3e−x4dx= limt→−∞∫ 0tx3e−x4dx= limt→−∞∫ 0−t4eu(du−4) Use substitution: Let u = −x4. Thendu = −4x3dx, so that du−4 = x3dx.And change the bounds= limt→−∞(−14∫ 0−t4eu du) ∫kf(x) dx = k∫f(x) dx= limt→−∞(−14eu∣∣∣∣0−t4)= limt→−∞(−14e0 +14e−t4)F (x)∣∣∣∣ba= F (b)− F (a)= −14+14(0)Since limt→−∞(−t4) = −∞, it followsthat limt→−∞e−t4= 0= −14+ 0 = −14Since the limit is a finite number, it follows that the improper integral∫ 0−∞ x3e−x4dxconverges.Likewise, one can show that∫∞0x3e−x4dx =14.Since both integrals∫ 0−∞ x3e−x4dx and∫∞0x3e−x4dx are convergent, we canconclude that the integral∫∞−∞ x3e−x4dx is convergent . (In fact, it convergesto −14+ 14= 0. Note that no matter what a you choose, you will get to the sameconclusion.)(f) Choose a = 0. Then, by definition,∫ ∞−∞7e−8x dx =∫ 0−∞7e−8x dx+∫ ∞07e−8x dx2.4. Improper Integrals 257Determining whether the first integral is convergent or not, we find∫ 0−∞7e−8x dx = limt→−∞∫ 0t7e−8x dx= limt→−∞(−78e−8x∣∣∣∣0t) ∫ekx dx =1kekx + C= limt→−∞(−78e0 +78e−8t)F (x)∣∣∣∣ba= F (b)− F (a)= −78+78limt→−∞(e−8t)limt→±∞(k) = k= −78+78(∞) limt→−∞e−8t =∞= −78+ (∞) = ∞Since the limit is not a finite number, the improper integral∫ 0−∞ 7e−8x dx isdivergent. Therefore, the integral∫∞−∞ 7e−8x dx is divergent . (We don’t need tofind the improper integral∫∞07e−8x dx because the first one is already divergent.)(g) • Again, it is convenient to choose a = 0. So∫ ∞−∞51 + 16x2dx =∫ 0−∞51 + 16x2dx+∫ ∞051 + 16x2dx• Determining whether ∫ 0−∞ 51+16x2 dx converges or diverges.- By definition, ∫ 0−∞51 + 16x2dx = limt→−∞∫ 0t51 + 16x2dx- Finding the integral∫ 0t51+16x2dx. First, we have∫ 0t51 + 16x2dx =∫ 0t51 + (4x)2dxLet u = 4x. Then du = 4dx, so that du4= dx.If x = t, u = 4(t) = 4tIf x = 0, u = 4(0) = 0258 Chapter 2. Techniques of Integration and Computational MethodsThe integral becomes∫ 0t51 + 16x2dx =∫ 04t51 + u2(du4)=[54arctan(u)]04t∫11 + u2du = tan−1 u+ C=54arctan(0)− 54arctan(4t) F (x)∣∣∣∣ba= F (b)− F (a)= −54arctan(4t) arctan(0) = 0- Finding the limit. Since limt→−∞arctan(t) = −pi2, it follows thatlimt→−∞∫ 0t51 + 16x2dx = limt→−∞(−54arctan(4t))= −54(−pi2)=5pi8- Since the limit is finite, it follows that the improper integral∫ 0−∞51+16x2dxconverges.• Similarly, one can show that the improper integral ∫∞051+16x2dx converges(to 5pi8).• Since both integrals ∫ 0−∞ 51+16x2 dx and ∫∞0 51+16x2 dx are convergent, it fol-lows that∫∞−∞51+16x2dx converges (to 5pi8+ 5pi8=5pi4).(h) By definition, ∫ ∞−∞sin(pix) dx =∫ a−∞sin(pit) dx+∫ ∞asin(pix) dxwhere a is any real number. Choosing a = 0, we can write the integral as∫ ∞−∞sin(pix) dx =∫ 0−∞sin(pit) dx+∫ ∞asin(pix) dxThe first improper integral on the righthand side is∫ 0−∞sin(pix) dx= limt→−∞[∫ 0tsin(pix) dx]= limt→−∞[− 1picos(pix)∣∣∣∣0t] ∫sin(kx) dx = −1kcos(kx) + C2.4. Improper Integrals 259= limt→−∞[− 1picos(0) +1picos(pit)]F (x)∣∣∣∣ba= F (b)− F (a)= limt→−∞(− 1pi+1picos(pit))cos(0) = 1= − 1pi+1pilimt→−∞cos(pit)Since limt→−∞cos(pit) does not exist (because cos(pit) bounces between −1 and 1,no matter how far out you look), it follows that the limit limt→−∞[∫ 0tsin(pix) dx]does not exist. This implies that the improper integral∫ 0−∞ sin(pix) dx diverges.Thus, the improper integral∫∞−∞ sin(pix) dx is divergent .3. [Exercise on page 22]The total change in y is∫ ∞01500e−0.04t dt= limb→∞∫ b01500e−0.04t dt= limb→∞1500−0.04e−0.04t∣∣∣∣b0∫ekx dx =1kekx + C,∫ baf(x) dx = F (x)∣∣∣∣ba= limb→∞− 15000.04(e−0.04b − 1) F (x)∣∣∣∣ba= F (b)− F (a)= −15000.04(limb→∞e−0.04b − 1)= −15000.04(0− 1) limb→∞e−0.04b = 0=15000.04= 375004. [Exercise on page 23]The total reaction to the drug over all the time since it was administrated is∫ ∞0r′(t) dt = limz→∞∫ z03t2e−0.5t dtFirst, we need to find the indefinite integral∫3t2e−0.5t dt.260 Chapter 2. Techniques of Integration and Computational Methods∫3t2e−0.5t dt= uv −∫v duBy parts: choose u = 3t2 anddv = e−0.5t. Then du = 6tdt andv = 1−0.5e−0.5t = −2e−0.5t= −6t2e−0.5t −∫−12te−0.5t dt Substitute u and v= −6t2e−0.5t −(xy −∫y dx) By parts again: choose x = −12tand dy = e−0.5t. Then dx =−12dt and y = −2e−0.5t= −6t2e−0.5t −(24te−0.5t −∫24e−0.5t dt)Substitute x and y= −6t2e−0.5t −(24te−0.5t − 24e−0.5t−0.5) ∫ekx dx =1kekx + C= −6t2e−0.5t − (24te−0.5t + 48e−0.5t)+ C= −6t2e−0.5t − 24te−0.5t − 48e−0.5t + CSo, an antiderivative of 3t2e−0.5t isF (t) = −6t2e−0.5t − 24te−0.5t − 48e−0.5tApplying the Fundamental Theorem of Calculus, we get∫ z03t2e−0.5t dt = F (z)− F (0)=(−6z2e−0.5z − 24ze−0.5z − 48e−0.5z)− (0 + 0− 48e0)= −6z2e−0.5z − 24ze−0.5z − 48e−0.5z + 48 e0 = 1Now, taking the limit as z approaches ∞, we getlimz→∞∫ z03t2e−0.5t dt= limz→∞(−6z2e−0.5z − 24ze−0.5z − 48e−0.5z + 48)= −6 limz→∞(z2e−0.5z)− 24 limz→∞(ze−0.5z)− 48 limz→∞(e−0.5z) + limz→∞(48)2.5. Numerical Integration 261= −6(0)− 24(0)− 48(0) + 48 = 48 limz→∞ zke−0.5z = 0 for all realnumbers kHence, the total reaction of the drug over all the time since it was administrated is∫ ∞03t2e−0.5t dt = 482.5 Numerical Integration1. [Exercise on page 23]Trapezoidal RuleLet f be a function and [a, b] be an interval. Suppose f is continuous on [a, b].Also suppose that we subdivide [a, b] into n equal subintervals[a0, a1], [a1, a2], · · · [an−1, an]We havex0 = a and xi+1 = xi + ∆x for 0 ≤ i ≤ n− 1where ∆x =b− anis the width of each subinterval. So,x0 = a, x3 = x2 + ∆xx1 = a+ ∆x...x2 = x1 + ∆x xn = xn−1 + ∆x = bThe trapezoidal rule says that the definite integral∫ baf(x) dx can be ap-proximated by the expression ∆x[12f(x0) + f(x1) + · · ·+ f(xn−1) + 12f(xn)];that is, ∫ baf(x) dx ≈ ∆x[12f(x0) + f(x1) + · · ·+ f(xn−1) + 12f(xn)]262 Chapter 2. Techniques of Integration and Computational MethodsLet f(x) =√3x2 + 5. For the integral∫ 10√3x2 + 5 dx, a = 0 and b = 1. Moreover,from the question, n = 4.The width is ∆x =b− an=1− 04= 0.25.So,x0 = a = 0 x3 = 0.5 + 0.25 = 0.75x1 = 0 + 0.25 = 0.25 x4 = 0.75 + 0.25 = 1 = bx2 = 0.25 + 0.25 = 0.5Evaluating the function at each xi, we getf(x0) =√3(0)2 + 5 ≈ 2.23607 f(x3) =√3(0.75)2 + 5 ≈ 2.58602f(x1) =√3(0.25)2 + 5 ≈ 2.27761 f(x4) =√3(1)2 + 5 ≈ 2.82843f(x2) =√3(0.5)2 + 5 ≈ 2.39792By the trapezoidal rule,∫ 10√3x2 + 5 dx ≈ ∆x[12f(x0) + f(x1) + f(x2) + f(x3) +12f(x4)]= 0.25[12(2.23607) + 2.27761 + 2.39792 + 2.58602 +12(2.82843)]= 0.25(9.7938) = 2.44845Thus,∫ 10√3x2 + 5 dx ≈ 2.44845 .2. [Exercise on page 23]Simpson’s RuleLet f be a function and [a, b] be an interval. Suppose f is continuous on[a, b]. Let n be an even integer. Suppose that we subdivide [a, b] into n equalsubintervals[a0, a1], [a1, a2], · · · [an−1, an]2.5. Numerical Integration 263We havex0 = a and xi+1 = xi + ∆x for 0 ≤ i ≤ n− 1where ∆x =b− anis the width of each subinterval. So,x0 = a, x3 = x2 + ∆xx1 = a+ ∆x...x2 = x1 + ∆x xn = xn−1 + ∆x = bSimpson’s rule says that∫ baf(x) dx ≈ ∆x3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + · · ·+ 2f(xn−2) + 4f(xn−1) + f(xn)]Note the pattern of coefficients:1, 4, 2, 4, 2, 4, 2, 4, 2, · · · , 4, 2, 4, 1Examples:If n = 2, ∆x =b− a2and∫ baf(x) dx ≈ ∆x3[f(x0) + 4f(x1) + f(x2)]If n = 4, ∆x =b− a4and∫ baf(x) dx ≈ ∆x3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + f(x4)]If n = 6, ∆x =b− a6and∫ baf(x) dx ≈ ∆x3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + 4f(x5) + f(x6)]Consider the integral∫ 10√3x2 + 5 dx. In Exercise 1, we found xi and f(xi), 0 ≤ i ≤ 4.Using these values and Simpson’s rule with n = 4, we have∫ 10√3x2 + 5 dx ≈ ∆x3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + f(x4)]264 Chapter 2. Techniques of Integration and Computational Methods=0.253[2.23607 + 4(2.27761) + 2(2.39792) + 4(2.58602) + 2.82843]=0.253(29.31486) ≈ 2.4429Aside: The integral∫ 10√3x2 + 5 dx cannot be evaluated by any technique we haveseen so far. Using a technique that we won’t discuss here, one can show that theexact value of this integral to five decimal places is 2.44291.3. [Exercise on page 23](a) Let f(x) =√9− x2. We first need to find the points of intersection betweenthe curve y =√9− x2 and the x-axis. This will give us the interval(s). Set√9− x2 = 0 to find√9− x2 = 09− x2 = 0x2 = 9x = ±√9x = ±3So, the interval is [−3, 3], and the area under the curve y = f(x) and above thex-axis is given by the definite integral∫ 3−3√9− x2 dxWe now approximate this using the trapezoidal rule with n = 6. Firstly, thewidth of each subinterval is∆x =b− an=3− (−3)6=66= 1Using this, we havex0 = −3 x1 = −2 x2 = −1 x3 = 0x4 = 1 x5 = 2 x6 = 32.5. Numerical Integration 265Evaluating f(x) at each xi, we getf(x0) = 0 f(x1) ≈ 2.23607 f(x2) ≈ 2.82843 f(x3) = 3f(x4) ≈ 2.82843 f(x5) ≈ 2.23607 f(x6) = 0By the trapezoidal rule,∫ 3−3√9− x2 dx≈ ∆x[12f(x0) + f(x1) + f(x2) + f(x3) + f(x4) + f(x5) +12f(x6)]= 1[12(0) + 2.23607 + 2.82843 + 3 + 2.82843 + 2.23607 +12(0)]= 13.129(b) We can use the same ∆x and f(xi)’s from the previous part. However, withSimpson’s rule, we have∫ 3−3√9− x2 dx≈ ∆x3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + 4f(x5) + f(x6)]=13[0 + 4(2.23607) + 2(2.82843) + 4(3) + 2(2.82843) + 4(2.23607) + 0]=13(41.20228) ≈ 13.7341(c) Geometrically, the area of the region under the curve y =√9− x2 above thex-axis is half of the area of the circle of radius r = 3. That is,∫ 3−3√9− x2 dx = pir22=pi(3)22=9pi2≈ 14.1372So the exact area, rounded to four decimal places, is 14.1372. Now, we compare theresults found in parts (a) and (b) with the exact area. The distance between the exact area and the result found in part (a) is14.1372− 13.129 = 1.0082266 Chapter 2. Techniques of Integration and Computational Methods The distance between the exact area and the result found in part (b) is14.1372− 13.7341 = 0.4031Since 0.4031 < 1.0082, the result obtained using Simpson’s rule is more accurate.4. [Exercise on page 23]The total growth as t changes from t = 1 to t = 5 is given by the integral∫ 51f(t) dt =∫ 51(4t+ 4+ e−t2)dtWe want to approximate this integral using Simpson’s rule. First, the width of eachsubinterval is∆x =b− an=5− 18=48= 0.5Using this, we havex0 = 1 x1 = 1.5 x2 = 2 x3 = 2.5 x4 = 3x5 = 3.5 x6 = 4 x7 = 4.5 x8 = 5Evaluating f(x) at these points, we getf(x0) = 1.16788 f(x1) ≈ 0.83267 f(x2) ≈ 68498 f(x3) = 0.61731 f(x4) ≈ 0.57155f(x5) ≈ 0.53334 f(x6) ≈ 0.5 f(x7) = 0.47059 f(x8) ≈ 0.44444Using Simpson’s rule, we get∫ 51(4t+ 4+ e−t2)dt≈ ∆x3[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + 4f(x5) + 2f(x6) + 4f(x7) + f(x8)]≈ 2.49017 ftChapter3Differential Equations3.1 Separable Differential Equations1. [Exercise on page 25]Pure-Time Differential Equations A differential equation is an equation relating an unknown functionand its derivatives. A differential equation of the form dydx= g(x) is called pure-time differ-ential equation. A solution of dydx= g(x) is a function y whose derivative is g(x). Forexample, the function y = x2 is a solution of the differential equationdydx= 2xsince the derivative of x2 is 2x. So, a solution of dydx= g(x) is an an-tiderivative of g(x). The family of all antiderivatives of g(x) is called thegeneral solution of the differential equation dydx= g(x). By definitionof the indefinite integral, the general solution of the differential equationdydx= g(x) is given byy =∫g(x) dx = G(x) + Cwhere G(x) is an antiderivative of g(x) and C is an arbitrary constant.267268 Chapter 3. Differential Equations(a) The general solution of the differential equation dydx= x isy =∫x dx =x22+ C(b) The general solution isy =∫(−3x+ 15x2 + 1) dx= −3∫x dx+ 15∫x2 dx+∫1dx= −32x2 +153x3 + x+ C∫xn dx =xn+1n+ 1+ C= −32x2 + 5x3 + x+ C(c) The solution isy =∫ (2x+ x14 − sec2 x)dx= 2∫1xdx+∫x14 dx−∫sec2 x dx= 2 ln |x|+ x5454− tanx+ C∫1xdx = ln |x| + C∫xn dx =xn+1n+ 1+ C∫sec2 x dx = tanx+ C= 2 ln |x|+ 45x54 − tanx+ C abc=acb(d) First we need to put the equation dydx− 7x2 = x3√x into the form dydx = g(x).Adding 7x2 to both sides of the equation, we getdydx=x3√x+ 7x2Integrating this, we gety =∫ (x3√x+ 7x2)dx =∫ (xx13+ 7x2)dx 3√x = x133.1. Separable Differential Equations 269=∫ (x23 + 7x2)dx =∫x23 dx+ 7∫x2 dxxnxm= xn−m=x5353+ 7(x33)+ C =35x53 +73x3 + C∫xn dx =xn+1n+ 1+ C(e)e−3x − 2dydx=x√x2 + 1−2dydx=x√x2 + 1− e−3x Subtract e−3x from both sidesdydx= − x2√x2 + 1+12e−3x Divide both sides by −2y =∫ (− x2√x2 + 1+12e−3x)dxy = −12∫x√x2 + 1dx+12∫e−3x dx∫kf(x) dx = k∫f(x) dxy = −12∫1√u(du2)+12∫e−3x dxUse substitution: u = x2 + 1,du = 2x dx, du2= x dxy = −14∫u−12 du+12∫e−3x dx1√u=1u12= u−12y = −14(u1212)+12(−13e−3x)+ C∫xn dx =xn+1n+ 1+ C,∫ekx dx =1kekx + Cy = −14(21u12)− 16e−3x + Cy = −12(x2 + 1)12 − 16e−3x + C Substitute u by x2 + 1y = −12√x2 + 1− 16e−3x + C(f)1tdydt− sin(4t) = 01tdydt= sin(4t) Add sin(4t) to both sidesdydt= t sin(4t) Multiply both sides by t270 Chapter 3. Differential Equationsy =∫t sin(4t) dty = uv −∫v duIntegrate by parts: Choose u = tand dv = sin(4t). Then du = dtand v = −14cos(4t)y = t(−14cos(4t))−∫ (−14cos(4t))(dt) Substitute u and vy = −14t cos(4t) +14∫cos(4t) dt∫kf(x) dx = k∫f(x) dxy = −14t cos(4t) +14(14sin(4t))+ C∫cos(kx) dx =1ksin(kx) + Cy = −14t cos(4t) +116sin(4t) + C2. [Exercise on page 25]Separable Differential Equations A separable differential equation is a differential equation of the formdydx=p(x)q(y)where p(x) is a function that depends only on x, and q(y) is a functiondepending only on y. To solve a separable differential equation, one can proceed as follows.– Step 1. Separate the variables x and y so that one side of theequation depends only on y and the other side only on x. To do this,we will view dydxas a fraction. Although this is technically incorrect,it turns out that manipulating dx and dy like any other algebraicquantity leads to a correct answer.dydx=p(x)q(y)dy =p(x)q(y)dx Multiply both sides by dxq(y)dy = p(x)dx Multiply both sides by q(y)3.1. Separable Differential Equations 271– Step 2. Take the integral of both sides:∫q(y) dy =∫p(x) dxNote that in the left-hand side the variable is y, while the variableis x in the right-hand side.This method is referred to as separation of variables.(a)dydx=x4y2y2dy = x4dx Multiply both sides by y2dx∫y2 dy =∫x4 dx Take the integral of both sidesy33=x55+ C1 Add only one constanty3 =35x5 + 3C1 Mutiply both sides by 3y3 =35x5 + C Relabel the constant 3C1 as Cy =3√35x5 + C Take the cube root of both sides(b) Multiplying both sides of the equation dydx= x5+3yby ydx, we get ydy = (x5+3)dx.Taking the integral of both sides, we get∫y dy =∫(x5 + 3) dxy22=x66+ 3x+ C1y2 =13x6 + 6x+ 2C1 Multiply both sides by 2y2 =13x6 + 6x+ C Relabel the constant 2C1 as CTaking the square root of both sides, we gety =√13x6 + 6x+ C or y = −√13x6 + 6x+ C272 Chapter 3. Differential EquationsSince we don’t know anything about the sign of y, we don’t know which one tochoose. So, we will leave the general solution in implicit form:y2 =13x6 + 6x+ CNote. In Exercise 2a, we were able to solve explicitly for y because the equationy3 = h(x), where h(x) is a function, has only one solution: y = 3√h(x).(c)dydx=x2 + 1y√xydy =x2 + 1√xdx Multiply both sides by ydx∫y dy =∫x2 + 1√xdx Integrate both sidesy22=∫ (x2x12+1x12)dxa+ bc=ac+bc,√x = x12y22=∫ (x32 + x−12)dxxnxm= xn−m,1xm= x−my22=25x52 + 2x12 + C1∫xn dx =xn+1n+ 1+ Cy2 =45x52 + 4√x+ 2C1 Multiply both sides by 2y2 =45x52 + 4√x+ C Relabel the constant 2C1 as C(d)y4dydx= x3 − x2y4dy = (x3 − x2) dx Multiply both sides by dx∫y4 dy =∫(x3 − x2) dx Integrate both sidesy55=x44− x33+ C∫xn dx =xn+1n+ 1+ Cy5 =54x4 − 53x3 + 5C Multiply both sides by 5y5 =54x4 − 53x3 + C Relabel the constant 5C as C3.1. Separable Differential Equations 273y =5√54x4 − 53x3 + C Take the fifth root of both sides(e)dydx= 5x4ydy = 5x4ydx Multiply both sides by dx1ydy = 5x4dxAssume that y 6= 0 and divideboth sides by y∫1ydy =∫5x4 dx Take the integral of both sidesln |y| = x5 + C∫1ydy = ln |y|+ C|y| = ex5+C Take the natural exponentialof both sides|y| = eCex5 ea+b = eaeby = ±eCex5 |y| = a implies that y = ±ay = Cex5Relabel the constant ±eC as CIn the third line, we assumed y 6= 0 and divided both sides of the equation by y.If y = 0, we cannot divide by y. However, one can easily check that the constantfunction y = 0 is a solution of the differential equation dydx= 5x4y. (Indeed, ify = 0, dydx= 0 and 5x4y = 5x4(0) = 0 = dydx.) So, the solution of the differentialequation isy = Cex5or y = 0Since the solution y = 0 can be obtained from y = Cex5by letting C = 0, thegeneral solution of the differential equation isy = Cex5where C can be any real number.(f)dydx= 7x2y − 5xydydx= (7x2 − 5x)y Factor out y274 Chapter 3. Differential Equations1ydy = (7x2 − 5x)dx Assumed that y 6= 0 and divideboth sides by y∫1ydy =∫(7x2 − 5x) dx Integrate both sidesln |y| = 73x3 − 52x2 + C∫1ydy = ln |y|+ C|y| = e 73x3− 52x2+C Take the natural exponen-tial of both sidesy = ±e 73x3− 52x2eCy = Ce73x3− 52x2 Relabel the constant ±eC as Cy = Ce73x3− 52x2The constant C can be anyreal number including 0(g)(xy3 − xy2)dydx= x3y(xy2 − xy)dydx= x3Assume that y 6= 0 and divideboth sides by y(y2 − y)dydx= x2 Divide both sides by x(y2 − y)dy = x2dx Multiply both sides by dx∫(y2 − y) dy =∫x2 dx Integrate both sides13y3 − 12y2 =13x3 + CThe constant C can be anyreal number including 0Note. We leave the solution in implicit form because the equation is difficult tosolve for y. If y = 0, (xy3 − xy2) dydx= 0 and x3y = 0. So,(xy3 − xy2)dydx= x3y if y = 0Therefore, the constant function y = 0 is a solution of the differential equa-tion. However, it is not possible to derive the that solution from the generalsolution 13y3 − 12y2 = 13x3 + C. In other words, no matter which value of Cwe choose, we won’t be able to get y = 0 from 13y3 − 12y2 = 13x3 + C.3.1. Separable Differential Equations 275Thus, the solution of the differential equation (xy3 − xy2) dydx= x3y is13y3 − 12y2 =13x3 + C or y = 0(h)exdydx= xy2e6xdydx= xy2e5x Divide both sides by ex1y2dydx= xe5xAssume that y 6= 0 and divideboth sides by y21y2dy = xe5xdx Multiply both sides by dx∫1y2dy =∫xe5x dx Integrate both sides−1y=∫xe5x dx An antiderivative of 1y2 is − 1y−1y=15xe5x −∫15e5x dx Integrate by parts−1y=15xe5x − 125e5x + C∫ekx dx =1kekx + C−1y=5xe5x − e5x + 25C25y =−255xe5x − e5x + C Relabel the constant 25C as Cy =−255xe5x − e5x + C or y = 0(i)dydt= 4t√1− y2dy = 4t√1− y2dt Multiply both sides by dt1√1− y2dy = 4tdtAssume that y 6= −1 and y 6= 1,and divide both sides by√1− y2∫1√1− y2 dy =∫4t dt Integrate both sides276 Chapter 3. Differential Equationssin−1(y) = 2t2 + C∫1√1− y2 dy = sin−1 y + Csin(sin−1(y)) = sin(2t2 + C) Take the sine of both sidesy = sin(2t2 + C) sin(sin−1(y)) = y for all y in [−1, 1]y = sin(2t2 + C) or y = −1 or y = 1(j)dydt= ty2 + tdydt= (y2 + 1)t Factor out tdy = (y2 + 1)tdt Multiply both sides by dt1y2 + 1dy = tdt Divide both sides by y2 + 1∫1y2 + 1dy =∫t dt Integrate both sidesarctan(y) =t22+ C∫11 + y2dy = arctan(y) + Ctan(arctan(y)) = tan(t22+ C)Take the tangent of both sidesy = tan(t22+ C)tan(arctan(y)) = y for allreal number y3. [Exercise on page 25]Initial Value ProblemLet (DE) be a differential equation whose unknown is y = y(x). Let x0 and y0be real numbers such that y(x0) = y0. The equation (DE) with the conditiony(x0) = y0 is called an initial value problem. The condition y(x0) = y0 iscalled the initial condition. So,Initial Value Problem ={Differential EquationInitial ConditionTo solve an initial value problem, we first find the general solution of the differential equation.3.1. Separable Differential Equations 277 Then we find the constant C using the initial condition.The solution to an initial value problem is sometimes called particular solu-tion.(a)  The equation dydx= 6x5 − 18x2 implies thaty =∫ (6x5 − 18x2) dx = x6 − 6x3 + C Using the initial condition y(0) = 2, we get(0)6 − 6(0)3 + C = 2C = 2 So the solution is y = x6 − 6x3 + 2 .(b) We first need to find the general solution.x− x2 dydx=√x−x2 dydx= −x+√x Subtract x from both sidesdydx=−x−x2 +√x−x2 Divide both sides by −x2dydx=1x− x− 32 Simplifyy =∫ (1x− x− 32)dx Integrate both sidesy = ln |x| − x− 12−12+ C∫1xdx = ln |x| + C∫xn dx =xn+1n+ 1+ Cy = ln |x|+ 2√x+ CThe initial condition y(1) = 7 is equivalent to ln |1| + 2√1+ C = 7. This latterequation is equivalent to 2 + C = 7 or C = 7 − 2 = 5. (Remember thatln |1| = ln(1) = 0.)278 Chapter 3. Differential EquationsSo, the solution whose initial condition is y(1) = 7 isy = ln |x|+ 2√x+ 5(c) First we find the general solution.dydx=5exeyeydy = 5exdx Multiply both sides by eydx∫ey dy =∫5ex dx Integrate both sidesey = 5ex + C∫ex dx = ex + Cy = ln(5ex + C)Take the natural logarithmof both sidesSince y(0) = 0, it follows thatln(5e0 + C) = 0ln(5 + C) = 0 e0 = 15 + C = e0Take the natural exponen-tial of both sidesC = e0 − 5 = 1− 5 = −4Thus, the solution whose initial condition is y(0) = 0 is y = ln(5ex − 4) .(d)x4dydx− yx 52 = 0x4dydx= yx52 Add yx52 to both sidesdydx= yx52x4Divide both sides by x41ydy = x−32dxAssume that y 6= 0 and divideboth sides by y∫1ydy =∫x−32 dx Integrate both sides3.1. Separable Differential Equations 279ln |y| = x− 12−12+ C∫1ydy = ln |y|+ C,∫xn dx =xn+1n+ 1+ Cln |y| = − 2√x+ Cabc=acb, x−12 =1x12=1√x|y| = eCe− 2√x Take the natural exponentialof both sidesy = ±eCe− 2√x |x| = a implies that x = ±ay = Ce− 2√x Relabel the constant ±eC as Cy = Ce− 2√xThe constant C can be anyreal number including 0We now find the constant C. The initial condition states that y = e−2 whenx = 1. Substituting these into the general solution, we havee−2 = Ce−21 = C Divide both sides by e−2Therefore the solution to the initial value problem is y = e− 2√x .(e) Firstly, we need to find the general solution.dydx=3x+ 5y − 2(y − 2)dy = (3x+ 5)dx Multiply both sides by (y − 2)dx∫(y − 2) dy =∫(3x+ 5) dx Integrate both sides12y2 − 2y = 32x2 + 5x+ C∫xn dx =xn+1n+ 1+ COne can use the quadratic formula −b±√b2−4ac2ato solve this for y. This will givetwo solutions, and since we don’t know anything about y, we won’t know whichone to choose. So we leave the general solution in implicit form.Using the initial condition (y(4) = 12), we have12(12)2 − 2(12) =32(4)2 + 5(4) + CSubstitute x by 4 and y by 12 intothe general solution48 = 44 + C Simplify280 Chapter 3. Differential Equations4 = C Subtract 44 from both sidesSo, the solution whose initial condition is y(4) = 12 is12y2 − 2y = 32x2 + 5x+ 4(f) Finding the general solution, we havedydt= y2 − 1dy = (y2 − 1)dt Multiply both sides by dt1y2 − 1dy = dtAssume y 6= −1 and y 6= 1 anddivide both sides by y2 − 1∫1y2 − 1 dy =∫dt Integrate both sides∫1(y − 1)(y + 1) dy =∫dt Factor∫ ( 12y − 1 −12y + 1)dy =∫dtUse partial fractions to get1(y−1)(y+1) =12y−1 −12y+112∫1y − 1 dy −12∫1y + 1dy =∫dt∫kf(x) dx = k∫f(x) dx12ln |y − 1| − 12ln |y + 1| = t+ CUsing substitution,∫1y−1dy =ln |y − 1| + C and ∫ 1y+1dy =ln |y + 1|+ C12(ln |y − 1| − ln |y + 1|) = t+ C Factor out 1212ln∣∣∣∣y − 1y + 1∣∣∣∣ = t+ C ln a− ln b = ln(ab)ln∣∣∣∣y − 1y + 1∣∣∣∣ = 2t+ 2C Multiply both sides by 2∣∣∣∣y − 1y + 1∣∣∣∣ = e2Ce2t Take the natural exponen-tial of both sidesy − 1y + 1= ±e2Ce2t |x| = a implies x = ±ay − 1y + 1= Ce2tRelabel the constant ±e2Cas Cy − 1 = yCe2t + Ce2t Multiply both sides by y + 13.1. Separable Differential Equations 281y(1− Ce2t) = 1 + Ce2ty =1 + Ce2t1− Ce2t Solve for yTo get this solution, we assumed y 6= −1 and y 6= 1. The solution y = 1 can be obtained from the general solution by lettingC = 0. Not matter which value of C we choose, it is not possible to get y = −1from the general solution.So, the solutions arey =1 + Ce2t1− Ce2t (C ∈ R) and y = −1Now, using the initial condition y(0) = 2, we get2 =1 + Ce01− Ce0Substitute y with 2 and t with 0into the general solution2 =1 + C1− C e0 = 12− 2C = 1 + C Multiply both sides by 1− CC =13Solve for CHence, the solution with y(0) = 2 isy =1 + 13e2t1− 13e2t=3+e2t33−e2t3=3 + e2t3− e2t4. [Exercise on page 26]Since the rate of change of V is 3 + cos (2pi(t+ 1)), it follows thatV =∫[3 + cos (2pi(t+ 1))] dt=∫3 dt+∫cos(2pit+ 2pi) dt=∫3 dt+∫cos(u)(du2pi)Use substitution: u = 2pit + 2pi,du = 2pidt, so that du2pi= dt=∫3 dt+12pi∫cos(u) du∫kf(x) dx = k∫f(x) dx282 Chapter 3. Differential Equations= 3t+12pi(sinu) + C∫cosu du = sinu+ C= 3t+12pisin(2pit+ 2pi) + C Substitute u by 2pit+ 2piUsing the initial condition, we haveV (0) = 83(0) +12pisin(2pi) + C = 80 + 0 + C = 8 sin(2pi) = 0C = 8So, V (t) = 3t+12pisin(2pit+ 2pi) + 85. [Exponential Population Growth] [Exercise on page 26]Rate of ChangeIf y is a quantity depending on time t, the rate of change of y is dydt, thederivative of y with respect to t.Let N be the size of the population at time t. Since the rate of change of N is threetimes the size of the population, it follows thatdNdt= 3NSolving this differential equation, we havedNdt= 3NdN = 3Ndt Multiply both sides by dt1NdN = 3dtAssume that N 6= 0 and divideboth sides by N∫1NdN =∫3 dt Integrate both sidesln |N | = 3t+ C∫1NdN = ln |N |+ C|N | = eCe3t Take the natural exponential of both sides3.1. Separable Differential Equations 283N = ±eCe3t |x| = a implies that x = ±aN = Ce3t Relabel the constant ±eC as CN = Ce3tC is an arbritary nonnegative realnumber including 0Since the initial population is 10, it follows thatN(0) = 10Ce0 = 10C = 10 e0 = 1So, N(t) = 10e3t. The size of the population after 4 years isN(4) = 10e3(4) = 10e12 ≈ 1, 627.547.91 ≈ 1, 627, 548Growth Rate Constant and Per Capita Growth RateIn general, if k is a constant, the general solution of the differential equationdydt= ky isy = CektThe constant k is called the growth rate constant or just the growth con-stant. The growth constant k can be thought of as the difference between thebirth and death rates of the population. If k > 0, the population grows exponentially. If k < 0, the population decays exponentially. If k = 0, the population stays steady.From the equation dydt= ky, we have1ydydt= kThe quantity 1ydydtis referred to as the per capita growth rate of the pop-ulation. In other words, the per capita growth rate is the rate at which thepopulation size changes per individual in the population. For the exponentialgrowth model, the per capita growth rate is a constant.284 Chapter 3. Differential Equations6. [The von Bertalanffy Growth Equation] [Exercise on page 26](a) To find L(t), we need to solve the initial value problemdLdt= k(L∞ − L) and L(0) = 3dLdt= k(L∞ − L)dL = k(L∞ − L)dt Multiply both sides by dt1L∞ − LdL = kdtAssume that L 6= L∞ and divideboth sides by L∞ − L∫1L∞ − LdL =∫kdt Integrate both sides∫1u(−du) =∫kdtUse substitution: u = L∞ − L,du = −dL, so that −du = dL− ln |u| = kt+ C∫1udu = ln |u|+ C− ln |L∞ − L| = kt+ C Substitute u with L∞ − Lln |L∞ − L| = −kt− C Multiply both sides by −1|L∞ − L| = e−Ce−kt Take the natural exponentialof both sidesL∞ − L = ±e−Ce−kt |x| = a implies that x = ±aL∞ − L = Ce−kt Relabel the constant e−C as CL(t) = L∞ − Ce−kt Solve for LThe solution L = L∞ can be obtained from the general solution by letting C = 0.Using the initial condition, we haveL(0) = 3L∞ − Ce0 = 3Substitute t with 0 into thegeneral solutionC = L∞ − 3 e0 = 13.1. Separable Differential Equations 285So, L(t) = L∞ − (L∞ − 3)e−kt .(b) The length of the fish in the long run islimt→∞L(t) = limt→∞(L∞)− (L∞ − 3) limt→∞(e−kt)= L∞ − (L∞ − 3)(0) Since k > 0, limt→∞ e−kt = 0= L∞7. [Restricted Growth] [Exercise on page 26](a) Since the rate of change of y is proportional to the difference between M and y,there exists a constant k such thatdydt= k(M − y)(b) Using the same approach as the one we used to solve the differential equationdLdt= k(L∞ − L) from Exercise 6a, the general solution of the equation dydt =k(M − y) isy = M − Ce−kt(c) Here M = 5000 and the growth constant is k = 25% = 0.25. Soy = 5000− Ce−0.25tSince the initial population is 1500, it follows that y(0) = 1500. Therefore,5000− Ce0 = 15005000− C = 1500 e0 = 1C = 5000− 1500 = 3500Thus, the function describing the population at time t isy = 5000− 3500e−0.25t(d) In 10 years, the population will bey = 5000− 3500e−0.25(10) ≈ 4712.7 ≈ 47138. [Tumor Growth: Gompertz Differential Equation] [Exercise on page 27]286 Chapter 3. Differential Equations(a)dYdt= a (ln b− lnY )YdY = a(ln b− lnY )Y dt Multiply both sides by dt1(ln b− lnY )Y dY = adtAssume that Y 6= b and divideboth sides by (ln b− lnY )Y∫1(ln b− lnY )Y dY =∫a dt Integrate both sides∫1Y ln(bY) dY = ∫ a dt lnx− ln y = ln(xy)To find the integral on the left-hand side, we make the substitution u = ln(bY).Thendu =ddY[ln(bY)]dY =ddY[ bY]bYdYddY[ln(g(Y ))] =ddY[g(Y )]g(Y )du =− bY 2bYdYddY[bY]= bddY[1Y]= b(− 1Y 2)du =(− bY 2)(Yb)dYabcd=ab· dcdu = − 1YdY Simplify−du = 1YdY Multiply both sides by −1So, the equation∫1Y ln( bY )dY =∫a dt becomes∫1ln(bY) ( 1YdY)=∫a dt∫1u(− du) =∫a dt u = ln(bY), −du = 1YdY− ln |u| = at+ C∫1udu = ln |u|+ Cln |u| = −at− C Multiply both sides by −1|u| = e−Ce−at Take the natural exponentialof both sides3.1. Separable Differential Equations 287u = ±e−Ce−at |x| = a implies that x = ±au = Ce−at Relabel the constant e−C as Cln∣∣∣∣ bY∣∣∣∣ = Ce−at Substitute u by ln( bY)∣∣∣∣ bY∣∣∣∣ = eCe−at Take the natural exponentialof both sidesbY= eCe−atb > 0, Y > 0Y (t) =beCe−atThe constant C can be any realnumber including 0Since the initial volume is 2, it follows that Y (0) = 2. Substituting Y with 2and t with 0 into the general solution, we get2 =beCe0eC =b2Multiply both sides by12eCC = ln(b2)Thus,Y (t) =beln(b2)e−at= be− ln(b2)e−at 1ex= e−x= b(e− ln(b2))e−atexy = (ex)y= b(eln(2b ))e−at− ln(x) = ln(1x)= b(2b)e−atelnx = x for all x > 0Y (t) = beln(2b )e−at AB = eB lnA(b) Since a > 0, it follows that limt→∞(e−at) = 0. Therefore,limt→∞Y (t) = beln(2b )(0) = be0 = b(1) = bHence, the asymptotic tumor volume is b .288 Chapter 3. Differential Equations9. [Allometry] [Exercise on page 27]1YdYdt= a1XdXdt1YdY = a1XdX Multiply both sides by dt∫1YdY =∫a1XdX Integrate both sidesln |Y | = a ln |X|+ C∫1udu = ln |u|+ C|Y | = eCea ln |X| Take the natural exponential of both sidesY = ±eCea ln |X| |x| = a implies that x = ±aY = Cea ln |X| Relabel the constant ±eC as CY = C(eln |X|)aexy = (ex)y = (ey)xY = C|X|a elnx = x for all x > 0Y = C(X)aSince the body size, X, is a non-negative number, |X| = X10. [Circadian Cycle] [Exercise on page 27]First, we need to find the general solution.1NdNdt= 3(1 + sin(pit12))1NdN = 3(1 + sin(pit12))dt Multiply both sides by dt∫1NdN =∫3(1 + sin(pit12))dt Integrate both sidesln |N | = 3(t− 12picos(pit12))+ C∫sin(kt) dt = −1kcos(kt) + C|N | = eCe3(t− 12pi cos(pit12)) Take the natural exponentialof both sidesN = ±eCe3(t− 12pi cos(pit12)) |x| = a implies that x = ±aN(t) = Ce3(t−12picos(pit12)) Relabel the constant ±eC as CSince the initial population is 7, it follows that N(0) = 7. Substituting t with 0 and3.1. Separable Differential Equations 289N with 7 into the general solution, we have7 = Ce3(0−12picos(0))7 = Ce3(−12pi ) cos(0) = 17e36pi = C Divide both sides by e−36piSo,N(t) = 7e36pi e3(t−12picos(pit12))The number of cells after 24 hours isN(24) = 7e36pi e3(24−12picos( 24pi12 )) = 7e36pi e72−36pi = 7e36pi e−36pi e72 = 7e7211. [Logistic Equation for Population Growth] [Exercise on page 27](a) First, we will solve the general form of the logistic equation:dydt= ry(1− yK)Once we get the solution, we will substitute r with 0.28 and K with 195.dydt= ry(1− yK)dy = ry(1− yK)dt Multiply both sides by dt1y(1− yK)dy = rdt Assume y 6= 0 and y 6= K anddivide both sides by y(1− yK)∫1y(1− yK)dy = ∫ r dt Integrate both sidesTo find the integral on the left side, we will use partial fractions. Since the denominator y(1− yK)is a product of distinct linear factors, thepartial fraction decomposition has the form1y(1− yK) = Ay+B1− yK290 Chapter 3. Differential Equations Multiplying both sides of the latter equation by y(1− yK), we get1 = A(1− yK)+By1 = A− A yK+By Distribute0y + 1 =(B − AK)y + AThis implies thatB − AK= 0 and A = 1Substituting A with 1 into the equation B − AK= 0, we get B − 1K= 0 orB = 1K. So,1y(1− yK) = Ay+B1− yK=1y+1K1− yKUsing this decomposition, the integral above becomes∫ (1y+1K1− yK)dy =∫r dt∫1ydy +∫ 1K1− yKdy =∫r dt∫1ydy +∫1u(−du) =∫r dtUse substitution: u = 1 − yK,du = − 1Kdy, −du = 1Kdyln |y| − ln |u| = rt+ C∫1ydy = ln |y|+ Cln |y| − ln |1− yK| = rt+ C Substitute u by 1− yKln∣∣∣∣ y1− yK∣∣∣∣ = rt+ C ln a− ln b = ln(ab)∣∣∣∣ y1− yK∣∣∣∣ = ert+C Take the natural exponentialof both sides∣∣∣∣ y1− yK∣∣∣∣ = eCert ea+b = eaeb = ebeay1− yK= ±eCert |x| = a implies that x = ±ay1− yK= Cert Relabel the constant ±eC as C3.1. Separable Differential Equations 291Solving this latter equation for y, we havey = Cert − CyKert Multiply both sides by 1− yKy +CyKert = Cert AddCyKert to both sidesy(1 +CKert)= Cert Factor out yy =Cert1 + CKertDivide both sides by 1 +CKerty =Cert(e−rt)e−rt + CKert(e−rt)Multiply the top and bottomby e−rty =Ce−rt + CKert(e−rt) = e0 = 1y =CCK(KCe−rt + 1) Factor out CKy =C1 + KCe−rt· KCabc=acby =K1 + KCe−rtSimplifyWe assumed y 6= 0 and y 6= K. Clearly, y = 0 satisfies the equation dydt= ry(1− yK). Also, y = K satisfies the equation dydt= ry(1− yK).Since the solutions y = 0 and y = K cannot be obtained from the generalsolution (y = K1+KCe−rt ), the solutions of the logistic equation arey(t) =K1 + KCe−rt, y(t) = 0, and y(t) = KUsing the initial condition y(0) = y0, we haveK1 + KCe0= y0K1 + KC= y0 e0 = 1K = y0 +Ky0CMultiply both sides by 1 +KCK − y0y0=KCSolve forKC292 Chapter 3. Differential EquationsKy0− 1 = KCa− bc=ac− bcSubstituting this into the general solution, we gety(t) =K1 +(Ky0− 1)e−rtSince r = 0.28 and K = 195, it follows thaty(t) =1951 +(195y0− 1)e−0.28tSuppose y0 6= 0. The size of the population in the long run islimt→∞y(t) =1951 +(195y0− 1)limt→∞e−0.28t=1951 +(195y0− 1)(0)limt→∞e−0.28t = 0=1951 + 0= 195This shows that the size of the population in the long run does not depend onthe initial population y0.(b) The graph of y(t) in each case is shown below.0y0 = 40y0 = 125K = 195y0 = 300ty(t)12. [Exercise on page 28]3.2. Equilibria and Stability 293(a) First, we need to find an expression for r. To do this, we will solve the equationN(t) = K1+(KN0−1)e−rtfor r.N(t) =K1 +(KN0− 1)e−rtN(t) +N(t)(K −N0N0)e−rt = KMultiply both sides by1 +(KN0− 1)e−rt(K −N0N0)e−rt =K −N(t)N(t)e−rt =K −N(t)N(t)· N0K −N0−rt = ln(K −N(t)N(t)· N0K −N0)r = −1tln(K −N(t)N(t)· N0K −N0)Substituting K with 180, N0 with 15, t with 2, and N(2) with 100 into this, wefindr = −12ln(180− 100100· 15180− 15)= −12ln(455)≈ 1.3105(b) Here N(0) = 20, N(4) = 38, N(175) ≈ 45, and N(300) ≈ 45. We want toestimate r.Let K be the carrying capacity. From N(175) ≈ 45 and N(300) ≈ 45, it followsthat N(t) is close to 45 for large t. Since the limit of N(t) as t approaches infinityis K, it follows that K ≈ 45. Therefore, by using the formula for r obtained inpart (a), we haver = −1tln(K −N(t)N(t)· N0K −N0)≈ −14ln(45− 3838· 2045− 20)N0 = 20, t = 4, N(4) = 38, K ≈ 45= −14ln(1495)≈ 0.47873.2 Equilibria and Stability1. [Exercise on page 28]294 Chapter 3. Differential EquationsAutonomous Differential Equations and EquilibriaAn autonomous differential equation is a differential equation of the formdydt= g(y)where g(y) depends only on y. For example, dydt= y − 1 is an autonomousdifferential equation, while dydt= t(y − 1) is not.An equilibrium solution (or equilibrium point or fixed point) of theautonomous differential equation dydt= g(y) is a constant function yˆ such thatg(yˆ) = 0. To find the equilibrium points of dydt= g(y), we solve the equationg(y) = 0 for y. For example, consider the differential equation dydt= y − 1.Solving the equation y− 1 = 0 for y, we get y = 1. So, yˆ = 1 is an equilibriumpoint of the differential equation dydt= y − 1.Fact. If yˆ is an equilibrium point of dydt= g(y), then the constant functiony = yˆ is a solution of the differential equation dydt= g(y). Indeed, since y = yˆis a constant function, dydt= 0. On the other hand, since yˆ is an equilibriumpoint, g(yˆ) = 0. So, dydt= 0 = g(yˆ) when y = yˆ.(a) To find the equilibria of the differential equation dydt= 7y, we need to solve theequation 7y = 0 for y.7y = 0y = 0 Divide both sides by 7So, the differential equation dydt= 7y has one equilibrium point: yˆ = 0 .(b)y2 − 9 = 0(y − 3)(y + 3) = 0 a2 − b2 = (a− b)(a+ b)y = 3 or y = −3So, the equilibria are yˆ = −3 and yˆ = 3 .(c)y3 − 2y2 − 8y = 03.2. Equilibria and Stability 295y(y2 − 2y − 8) = 0 Factor out yy(y + 2)(y − 4) = 0 Factory = 0 or y = −2 or y = 4So, the equilibria are yˆ = −2 , yˆ = 0 , and yˆ = 4 .(d)N − 5N + 2= 0N − 5 = 0 Multiply both sides by N + 2N = 5 Add 5 to both sidesSo, Nˆ = 5 .(e)2(N23 − 4)e−3N = 0N23 − 4 = 0 or e−3N = 0N23 = 4 or e−3N = 0N = 432 or e−3N = 0N = 8 or e−3N = 0The equation e−3N = 0 has no solution since ex > 0 for all x. So, Nˆ = 8 .(f) Set x ln(x12)= 0. Thenx = 0 or ln( x12)= 0The solution x = 0 is to be rejected because x > 0. Since ln a = 0 if and only ifa = 1, the second equation implies that x12= 1 or x = 12. So, xˆ = 12 .2. [Exercise on page 28](a)296 Chapter 3. Differential EquationsStability of Equilibrium PointsLet yˆ be an equilibrium point of the differential equationdydt= g(y)We say that yˆ is stable if for any y0 close to yˆ, the solution of the initialvalue problemdydt= g(y), y(0) = y0 (3.2.1)approaches yˆ as t→∞.We say that yˆ is unstable if for any y0 close to yˆ, the solution of theinitial value problem (3.2.1) diverges from yˆ (that is, the solution does notreturn to yˆ).Theorem 3.2.1. Consider the autonomous differential equationdydt= g(y)Let yˆ be an equilibrium point of this equation. If the sign of g(y) changes from positive to negative at yˆ, then theequilibrium point yˆ is stable. This is illustrated by the followingfigure.ySign of g(y)yˆ+ −yˆ is stable If the sign of g(y) changes from negative to positive at yˆ, then theequilibrium point yˆ is unstable. This is illustrated by the followingfigure.ySign of g(y)yˆ− +yˆ is unstable3.2. Equilibria and Stability 297In the plots above, arrows point to the right if g(y) > 0 (that is, if dydt> 0,or if y is increasing). Arrows point to the left if g(y) < 0 (that is, if dydt< 0,or if y is decreasing). These plots are referred to as vector field plots orphase plots.The theorem above says that if the arrows point toward yˆ, then yˆ is stable.If the arrows point away from yˆ, then yˆ is unstable.How to determine the stability of equilibria?Step 1. Find all the equilibrium points of the differential equationdydt= g(y) by solving the equation g(y) = 0 for y.Step 2. Mark the equilibrium points on the real line (or thehorizontal axis). These points determine a number of open intervals.Step 3. Find the sign of g(y) on each interval. To find the sign ofg(y) on an open interval I found in step 2, choose any number, say a, inI. Then substitute y with a into g(y). If g(a) is positive, then so is dydt. This implies that y(t) is an increasingfunction of t. Plot a rightward arrow to show that y(t) is increasing. If g(a) is negative, then so is dydt. This implies that y(t) is a decreasingfunction of t. Plot a leftward arrow to show that y(t) is decreasing.After labeling each interval with direction arrows, we get the phase plotof the differential equation dydt= g(y).Step 4. Use the phase plot to determine whether an equilibriumpoint is stable or unstable. An equilibrium point yˆ is stable if the arrows in the phase plot pointtoward yˆ. An equilibrium point yˆ is unstable if the arrows in the phase plotpoint away from yˆ.For the differential equation dydt= y − 3, g(y) = y − 3. Solving the equation y − 3 = 0, we find y = 3. So, yˆ = 3 is an equilibriumpoint.298 Chapter 3. Differential Equations The equilibrium point yˆ = 3 determines two intervals: (−∞, 3) and (3,∞)as shown in the following figure.−∞ ∞3 y Now, we find the sign of g(y) on each interval.– On (−∞, 3), choose any number and substitute it into g(y). Let’s choosefor example 0. Substituting this into g(y), we get g(0) = 0 − 3 = −3.Since g(0) < 0, it follows that g(y) is negative on the interval (−∞, 3).So, the arrows point to the left as shown in the following figure.ySign of g(y)3−choose 0g(0) = −3 < 0– On (3,∞), choose any number and substitute it into g(y). Let’s choosefor example 4. Substituting this into g(y), we get g(4) = 4 − 3 = 1.Since g(4) > 0, it follows that g(y) is positive on the interval (3,∞). So,the arrows point to the right as shown in the following figure.ySign of g(y)3− +choose 0g(0) = −3 < 0choose 4g(4) = 1 > 0 Since the arrows point away from yˆ = 3, it follows that yˆ = 3 is an unstableequilibrium point.Alternate SolutionAn alternate way to determine the stability of equilibria is to use thefollowing theorem.3.2. Equilibria and Stability 299Stability CriterionTheorem 3.2.2. Let yˆ be an equilibrium point of the autonomousdifferential equationdydt= g(y)Suppose that g is differentiable at yˆ. If g′(yˆ) < 0, then yˆ is stable. If g′(yˆ) > 0, then yˆ is unstable.The test is inconclusive if g′(yˆ) = 0.For the differential equation dydt= y − 3,g(y) = y − 3 Solving the equation g(y) = 0 for y, we find y = 3. So, yˆ = 3 . Taking the derivative of g(y), we getg′(y) =ddy[y − 3] = ddy[y]− ddy[3] = 1− 0 = 1Substituting y with yˆ into g′(y), we get g′(3) = 1. Since g′(3) ispositive, it follows (by Theorem 3.2.2) that yˆ = 3 is an unstableequilibrium.(b) Let g(y) = y2 − 7y + 10. Set g(y) = 0. Then y2 − 7y + 10 = 0. Factoring this,we get (y − 5)(y − 2) = 0. This latter equation implies that y = 5 or y = 2. So,the equilibria areyˆ = 2 and yˆ = 5The phase plot of the differential equation dydt= y2 − 7y + 10 is shown in thefollowing figure.ySign of g(y)2 5+ − +choose 0g(0) = 10 > 0choose 3g(3) = −2 < 0choose 6g(6) = 4 > 0300 Chapter 3. Differential Equations Since the arrows point toward 2, the equilibrium point yˆ = 2 is stable . Since the arrows point away from 5, the equilibrium point yˆ = 5 is unstable .Alternate SolutionThe derivative of g(y) = y2 − 7y + 10 isg′(y) =ddy[y2 − 7y + 10] = 2y − 7 Substituting y with yˆ = 2 into the derivative, we get g′(2) = 2(2)−7 = −3. Since g′(2) < 0, it follows (by Theorem 3.2.2) that yˆ = 2 isa stable equilibrium. Substituting y with yˆ = 5 into the derivative, we get g′(5) = 2(5)−7 = 3. Since g′(5) > 0, it follows (by Theorem 3.2.2) that yˆ = 5 isan unstable equilibrium.(c) Let g(y) = 4y − y3. Solving the equation g(y) = 0, we havey(4− y2) = 0y(2− y)(2 + y) = 0y = 0 or y = 2 or y = −2So, the equilibria are yˆ = −2 , yˆ = 0 , and yˆ = 2 .To determine whether they are stable or unstable, we will use Theorem 3.2.2.First, we need to find derivative of g(y):g′(y) =ddy[4y − y3] = 4− 3y2 Substituting y with yˆ = −2 into the derivative, we getg′(−2) = 4− 3(−2)2 = 4− 3(4) = 4− 12 = −8Since g′(−2) < 0, it follows that yˆ = −2 is a stable equilibrium. If yˆ = 0, g′(0) = 4 − 3(0)2 = 4. Since g′(0) > 0, it follows that yˆ = 0 is anunstable equilibrium. If yˆ = 2, g′(2) = 4− 3(2)2 = −8. Since g′(2) < 0, it follows that yˆ = 2 is astable equilibrium.3.2. Equilibria and Stability 301(d) Let g(x) = 32+x−1. To find the equilibria, we need to solve the equation g(x) = 0for x.32 + x− 1 = 032 + x= 1 Add 1 to both sides3 = 2 + x Multiply both sides by 2 + x1 = x Subtract 2 from both sidesSo, xˆ = 1 is an equilibrium pointThe phase plot of the differential equation dxdt= g(x) is shown below.0 xSign of g(x)1+ −choose 0.5g(0.5) = 0.2 > 0choose 2g(2) = −0.25 < 0Since the arrows point toward 1, the equilibrium point xˆ = 1 is stable .3. [Exercise on page 29]−2 2 yg(y)Geometrically, the solutions of g(y) = 0 are the points of intersection between thegraph of g and the horizontal axis. Here, the graph of g intersects the horizontal axisat −2 and 2. So, the equilibrium points of the differential equation dydt= g(y) areyˆ = −2 and yˆ = 2 .302 Chapter 3. Differential Equations From the graph, g′(−2) > 0 (because g is increasing on an interval containing−2). This implies that yˆ = −2 is an unstable equilibrium (thanks to Theo-rem 3.2.2). Again from the graph, g′(2) < 0 (because g is decreasing on an interval con-taining 2). This implies that yˆ = 2 is a stable equilibrium (thanks to Theo-rem 3.2.2).Alternate SolutionWe will construct the phase plot of the differential equation dydt= g(y). First,the equilibria −2 and 2 determine three intervals: (−∞,−2), (−2, 2), and(2,∞). On the interval (−∞,−2), the graph of g is below the horizontal axis.This means that g(y) < 0 on the interval (−∞,−2). So, the arrows onthat interval point to the left. On the interval (−2, 2), the graph of g is above the horizontal axis. Thismeans that g(y) > 0 on the interval (−2, 2). So, the arrows point to theright. On the interval (2,∞), the graph of g is below the horizontal axis. Thismeans that g(y) < 0 on the interval (2,∞). So, the arrows point to theleft.Using this information, we get the following phase plot.ySign of g(y)−2 2− + −From the phase plot, it follows that yˆ = −2 is unstable, while yˆ = 2 is stable.4. [Exercise on page 29]3.2. Equilibria and Stability 3032 4 yg(y)Here, the graph of g intersects the horizontal axis at 0, 2, and 4. So, the equilibriaare yˆ = 0 , yˆ = 2 , and yˆ = 4 . Since the graph of g is increasing on an interval containing 0, it follows thatg′(0) > 0. This implies that yˆ = 0 is an unstable equilibrium (thanks toTheorem 3.2.2). Since the graph of g is decreasing on an interval containing 2, it follows thatg′(2) < 0. This implies that yˆ = 2 is a stable equilibrium (thanks to Theo-rem 3.2.2). Since the graph of g is increasing on an interval containing 4, it follows thatg′(4) > 0. This implies that yˆ = 4 is an unstable equilibrium (thanks toTheorem 3.2.2).5. [Exercise on page 29](a)Semistable Equilibrium PointsLet yˆ be an equilibrium point of the autonomous differential equationdydt= g(y)If the sign of g(y) does not change at yˆ, we say that yˆ is a semistableequilibrium point. There are two possibilities: The arrows of the phase plots point toward yˆ when y < yˆ, and awayfrom yˆ when y > yˆ. This is illustrated in the following figure.304 Chapter 3. Differential EquationsySign of g(y)yˆ+ +yˆ is semistable The arrows of the phase plots point away from yˆ when y < yˆ, andtoward yˆ when y > yˆ. This is illustrated in the following figure.ySign of g(y)yˆ− −yˆ is semistableFor the differential equation dydt= −y2 + 4y − 4, the function g is g(y) = −y2 +4y−4. Solving the equation g(y) = 0, we get −(y−2)2 = 0 or y = 2. So, yˆ = 2is an equilibrium point. The phase plot is shown below.ySign of g(y)2− −choose 0g(0) = −4 < 0choose 3g(3) = −1 < 0From the phase plot, the arrows to the left of 2 point away from it, and thearrows to the right point toward it. So, yˆ = 2 is a semistable equilibrium point.(b) Let g(y) = (y2 + 2y+ 1)(y− 3) = (y+ 1)2(y− 3). Solving the equation g(y) = 0,we get y = −1 or y = 3. So the equilibrium points are yˆ = −1 and yˆ = 3 .To determine their stability, we make the phase plot. We obtain the followingfigure.ySign of g(y)−1 3− − +choose −2g(−2) = −5 < 0choose 0g(0) = −3 < 0choose 4g(4) = 25 > 03.2. Equilibria and Stability 305From the phase plot, yˆ = −1 is a semistable equilibrium, while yˆ = 3 is anunstable equilibrium.(c)dydt= g(y), where g is the function whose graph is shown below.−2 4 yg(y)The graph of g intersects the horizontal axis at −2 and 4. So, the equilibriumpoints are yˆ = −2 and yˆ = 4 .To determine their stability, we make the phase plot. First, the equilibriumpoints −2 and 4 determine three intervals: (−∞,−2), (−2, 4), and (4,∞). On the interval (−∞,−2), the graph of g is above the horizontal axis. Thismeans that g(y) > 0 on (−∞,−2). So, the arrows on this interval point tothe right. On the interval (−2, 4), the graph of g is below the horizontal axis. Thismeans that g(y) < 0 on (−2, 4). So, the arrows on this interval point to theleft. Likewise, the arrows on the interval (4,∞) point to the left.Using this information, we have the following phase plot.ySign of g(y)−2 4+ − −Form the phase plot, the equilibrium point yˆ = −2 is stable , while the equilib-rium point yˆ = 4 is semistable .6. [Logistic Equation for Population Growth] [Exercise on page 30]306 Chapter 3. Differential Equations(a) It suffices to show that y = 0 is an equilibrium solution of the differential equa-tiondydt= 0.28(1− y195)yLet g(y) = 0.28(1− y195)y. Solving the equation g(y) = 0 for y, we get y = 0 ory = 195. So, the equilibria are yˆ = 0 and yˆ = 195. Since yˆ = 0 is an equilibriumsolution, it follows that the solution of the differential equation whose initialcondition is y0 = 0 is y(t) = 0. In general, we have the following fact.Equilibria and Initial Value ProblemsFact. Let yˆ be an equilibrium point of the autonomous differential equa-tion dydt= g(y). Then, the solution whose initial condition is y0 = yˆ is theconstant functiony(t) = yˆ for all t(b) For the same reason as in part (a), the solution of the differential equation whoseinitial condition is y0 = 195 is N(t) = 195.(c) To find the asymptotic behavior of y(t) without solving the differential equation,we make a phase plot (see the figure below).ySign of g(y)0 195− + −From the phase plot, the arrows point toward 195. This implies that yˆ = 195is a stable equilibrium. And the solution y(t) of the differential equation whoseinitial condition is y0 (y0 > 0) approaches 195 as t → ∞. In other words, ify0 > 0, limt→∞y(t) = 195.7. [The Allee Effect] [Exercise on page 30](a) Let g(N) = rN(N − a) (1− NK). Solving the equation g(N) = 0 for N , weget N = 0 or N = a or N = K. So, the equilibria are Nˆ = 0 , Nˆ = a , andNˆ = K .To determine stability, we construct a phase plot. The equilibria 0, a, and Kdetermine three intervals: (0, a), (a,K), and (K,∞) (remember that N is anonnegative quantity.) Our aim is to find the sign of g(N) on each interval. On the interval (0, a).3.3. Integrating Factors 307– If N < a, then N − a < 0.– If N < a, then N < K since a is less than K. So NK< 1 and therefore,1− NK> 0.Since N − a is negative, and since 1 − NKis positive, it follows that theproduct (N−a) (1− NK)is negative. This implies that the product rN(N−a)(1− NK)is negative (since rN > 0). So, g(N) < 0 on the interval (0, a). If N is in the interval (a,K), then both N − a and 1− NKare positive. Thisimplies that rN(N − a) (1− NK)> 0 (since rN > 0). (The product of threepositive numbers is positive.) So, g(N) > 0 on the interval (a,K). On the interval (K,∞), N is greater thanK and greater than a. This impliesthat N−a is positive and 1− NKis negative. So, the product (N−a) (1− NK)is negative. Since rN is positive, the product rN(N−a) (1− NK)is negative;that is, g(N) < 0 on the interval (K,∞).Using this information, we have the following phase plot.NSign of g(N)0 a K− + −From the phase plot, the equilibrium points Nˆ = 0 and Nˆ = K are stable ,while Nˆ = a is unstable .(b) Let N0 be the initial population.i. If 0 < N0 < a, then according to the phase plot the arrows point toward 0.This means that y(t) approaches 0 as t→∞; that is, limt→∞N(t) = 0 . So, apopulation smaller than a will decline to extinction.ii. If a < N0 < K, then according to the phase plot the arrows point toward K.This means that limt→∞N(t) = K . So, a population whose size is between aand K will increase, converging to the carrying capacity K.iii. IfN0 > K, then according to the phase plot the arrows point towardK. Thismeans that limt→∞N(t) = K . So, a population larger than K will decrease,converging to the carrying capacity K.3.3 Integrating Factors1. [Exercise on page 31]308 Chapter 3. Differential Equations(a)Linear First-Order Differential EquationsA linear first-order differential equation is an equation of the formdydx+ P (x)y = Q(x)where P (x) and Q(x) depend only on x.The integrating factor for the differential equationdydx+ P (x)y = Q(x)is the function I(x) defined byI(x) = e∫P (x)dxTheorem 3.3.1. The general solution of dydx+ P (x)y = Q(x) is given byy =1I(x)(∫I(x)Q(x) dx)+CI(x)(3.3.1)where C is a constant.Proof. First, we haveddx[I(x)y] = I(x)dydx+ I(x)P (x) (3.3.2)Indeed,ddx[I(x)y] = I(x)ddx[y] +ddx[I(x)]y Product Rule= I(x)dydx+ddx[e∫P (x) dx]y I(x) = e∫P (x) dx= I(x)dydx+(e∫P (x) dx) ddx[∫P (x) dx]ddx[eg(x)] = eg(x)ddx[g(x)]= I(x)dydx+(e∫P (x) dx)P (x)ddx[∫P (x) dx]= P (x)= I(x)dydx+ I(x)P (x) e∫P (x) dx = I(x)We now solve the differential equation.dydx+ P (x)y = Q(x)3.3. Integrating Factors 309I(x)dydx+ I(x)P (x) = I(x)Q(x) Multiply both sides by I(x)ddx[I(x)y] = I(x)Q(x) Use (3.3.2)I(x)y =∫I(x)Q(x) dx+ CIntegrate both sideswith respect to xy =1I(x)(∫I(x)Q(x) dx)+CI(x)Divide both sidesby I(x)We now solve dydx+ 7y = 19. Here, P (x) = 7 and Q(x) = 19.To find the integrating factor I(x), we solveI(x) = e∫P (x) dx = e∫7 dx = e7x∫k dx = kx+ CUsing (3.3.1), the general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1e7x(∫e7x(19) dx)+Ce7x=1e7x(19∫e7x dx)+Ce7x∫kf(x) dx = k∫f(x) dx=1e7x(19e7x7)+ Ce−7x∫ekx dx =1kekx + C=197+ Ce−7x SimplifyNote: The constant C is already present in (3.3.1). So, we don’t need to add aconstant again while taking the integrals∫P (x) dx and∫I(x)Q(x) dx.(b) Here, P (x) = −2 and Q(x) = 5. The integrating factor isI(x) = e∫P (x) dx = e∫ −2 dx = e−2x∫k dx = kx+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)310 Chapter 3. Differential Equations=1e−2x(∫e−2x(5) dx)+Ce−2x=1e−2x(5e−2x−2)+ Ce2x∫ekx dx =1kekx + C= −52+ Ce2x Simplify(c) Here, P (x) = −3 and Q(x) = x. The integrating factor isI(x) = e∫P (x) dx = e∫ −3 dx = e−3x∫k dx = kx+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1e−3x(∫e−3x(x) dx)+Ce−3x= e3x(∫xe−3x dx)+ Ce3x1e−a= ea= e3x(uv −∫v du)+ Ce3xIntegrate by parts: let u = xand dv = e−3x. Then du = dxand v = −13e−3x= e3x(−13xe−3x −∫−13e−3x dx)+ Ce3x Substitute u and v= e3x(−13xe−3x +13∫e−3x dx)+ Ce3x∫kf(x) dx = k∫f(x) dx= e3x(−13xe−3x +13(−13e−3x))+ Ce3x∫ekx dx =1kekx + C= e3x(−13xe−3x − 19e−3x)+ Ce3x= e3xe−3x(−13x− 19)+ Ce3x Factor= −13x− 19+ Ce3x eae−a = e0 = 1 for all a(d) First we need to put the equation in the proper form, that is, in the formdydx+ P (x)y = Q(x). Dividing both sides of the equation 2 dydx+ 4xy = 16x by 2,3.3. Integrating Factors 311we getdydx+ 2xy = 8xwhich is in the proper form with P (x) = 2x and Q(x) = 8x.The integrating factor isI(x) = e∫P (x) dx = e∫2x dx = ex2∫xn dx =xn+1n+ 1+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1ex2(∫ex2(8x) dx)+Cex2=1ex2(8∫xex2dx)+Cex2∫kf(x) dx = k∫f(x) dx=1ex2(8∫eu(du2))+ Ce−x2 Use substitution: u = x2, du =2xdx, du2= xdx=1ex2(82eu)+ Ce−x2∫ex dx = ex + C=1ex2(4ex2)+ Ce−x2Substitute u with x2= 4 + Ce−x2Simplify(e) This is already in the proper form with P (x) = 8xandQ(x) = x3. The integratingfactor isI(x) = e∫P (x) dx = e∫8xdx= e8 ln |x| = eln |x|8= eln |x8|∫1xdx = ln |x|+ C= eln(x8) |x8| ≥ 0 for all x= x8 elnx = x for all x > 0The general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)312 Chapter 3. Differential Equations=1x8(∫x8(x3) dx)+Cx8=1x8(∫x11 dx)+Cx8xmxn = xn+m=1x8(x1212)+Cx8∫xn dx =xn+1n+ 1+ C=112x4 +Cx8xnxm= xn−m(f) Dividing both sides by x, we getdydx+4xy = x4 − 1This latter equation is in the proper form with P (x) = 4xand Q(x) = x4 − 1.The integrating factor isI(x) = e∫4xdx = e4 ln |x| = eln |x4| = eln(x4) = x4∫1xdx = ln |x|+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1x4(∫x4(x4 − 1) dx)+Cx4=1x4(∫(x8 − x4) dx)+Cx4=1x4(19x9 − 15x5)+Cx4∫xn dx =xn+1n+ 1+ C=19x5 − 15x+Cx4Distribute and Simplify(g) Subtracting x from both sides, we get 5 dydx+ 10xy = −x. Dividing both sides by5, we getdydx+ 2xy = −15xSo, P (x) = 2x and Q(x) = −15x.3.3. Integrating Factors 313The integrating factor isI(x) = e∫P (x) dx = e∫2x dx = ex2∫xn dx =xn+1n+ 1+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1ex2(∫ex2(−15x)dx)+Cex2=1ex2(−15∫xex2dx)+Cex2∫kf(x) dx = k∫f(x) dx=1ex2(−15(12ex2))+Cex2Use substitution: u = x2,12du = x dx=1ex2(− 110ex2)+Cex2= − 110+ Ce−x2Simplify(h) Subtracting 3xy from both sides, we get 4x3 = x dydx− 3xy. Dividing both sidesby x, we get 4x2 = dydx− 3y. So, the original equation is equivalent todydx− 3y = 4x2which implies that P (x) = −3 and Q(x) = 4x2.The integrating factor isI(x) = e∫P (x) dx = e∫ −3 dx = e−3x∫k dx = kx+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1e−3x(∫e−3x(4x2) dx)+Ce−3x=1e−3x(∫4x2e−3x dx)+ Ce3x1e−3x= e3xTo find∫4x2e−3x dx, we use integration by parts.314 Chapter 3. Differential EquationsChoose u = 4x2 and dv = e−3xdxThen, du = 8xdx and v =∫e−3x dx = −13e−3xWe have∫4x2e−3x dx = uv −∫vdu= (4x2)(−13e−3x)−∫−13e−3x(8xdx) Substitute u and v= −43x2e−3x +83∫xe−3x dx∫kf(x) dx = k∫f(x) dxTo find∫xe−3x dx, we use integration by parts again.Choose u = x and dv = e−3xdxThen, du = dx and v =∫e−3x dx = −13e−3xSo,∫xe−3x dx = x(−13e−3x)−∫−13e−3x dx Substitute again= −13xe−3x +13∫e−3x dx∫kf(x) dx = k∫f(x) dx= −13xe−3x +13(−13e−3x) ∫ekx dx =1kekx + C= −13xe−3x − 19e−3xSubstituting this into the equation∫4x2e−3x dx = −43x2e−3x +83∫xe−3x dxwe get∫4x2e−3x dx = −43x2e−3x +83(−13xe−3x − 19e−3x)= −43x2e−3x − 89xe−3x − 827e−3x= e−3x(−43x2 − 89x− 827)Factor out e−3x3.3. Integrating Factors 315Substituting this this into the general solutiony =1e−3x(∫4x2e−3x dx)+ Ce3xwe arrive at the general solution ofy =1e−3x[e−3x(−43x2 − 89x− 827)]+ Ce3x= −43x2 − 89x− 827+ Ce3x Simplify(i) Dividing both sides of the equation 2tdydt− y = 2t ln t by 2t, we getdydt− 12ty = ln twhich is an equation of the form dydt+ P (t)y = Q(t) with P (t) = − 12tandQ(t) = ln t.The integrating factor isI(t) = e∫P (t) dt = e∫ − 12tdt= e−12ln |t|∫1tdt = ln |t|+ C= e−12ln t |t| = t since t > 0= eln(t−12)= t−12 r ln(a) = ln(ar), eln a = aThe general solution isy =1I(t)(∫I(t)Q(t) dt)+CI(t)=1t−12(∫t−12 ln t dt)+Ct−12= t12(∫t−12 ln t dt)+ Ct121t−m= tmTo find the integral∫t−12 ln t dt, we use integration by parts.Choose u = ln t and dv = t−12dtThen du = 1tdt and v =∫t−12dt = 2t12Therefore,316 Chapter 3. Differential Equations∫t−12 ln t dt = uv −∫vdu= (ln t)(2t12)−∫ (2t12)(1tdt)Substitute u and v= 2√t ln t− 2∫t−12 dt= 2√t ln t− 2(2t12)= 2√t ln t− 4√tSubstituting this into the general solution above, we gety = t12(∫t−12 ln t dt)+ Ct12=√t(2√t ln t− 4√t)+ C√t= 2t ln t− 4t+ C√t2. [Exercise on page 31](a) Here, P (x) = 5 and Q(x) = 12ex. So, the integrating factor isI(x) = e∫P (x) dx = e∫5 dx = e5x∫k dx = kx+ CThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1e5x(∫e5x12ex dx)+Ce5x= e−5x(∫12e6x dx)+ Ce−5x eaeb = ea+b,1ea= e−a= e−5x(126e6x)+ Ce−5x∫ekx dx =1kekx + C= e−5x(2e6x)+ Ce−5x= 2ex + Ce−5x3.3. Integrating Factors 317The initial condition states that y(0) = 11. So,y(0) = 2e0 + Ce0 = 112 + C = 11 e0 = 1C = 9Thus, the solution with initial condition y(0) = 11 isy = 2ex + 9e−5x(b) Dividing both sides by x2 + x, we get dydx+ 2xx2+xy = x+1x2+x. This is equivalent todydx+ 2xx(x+1)y = x+1x(x+1)which, in turn, is equivalent todydx+2x+ 1y =1xSo, P (x) = 2x+1and Q(x) = 1x.The integrating factor isI(x) = e∫P (x) dx = e∫2x+1dx = e2 ln |x+1|∫1x+ 1dx = ln |x+ 1|= eln |(x+1)2| = eln(x+1)2= (x+ 1)2The general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1(x+ 1)2(∫(x+ 1)2(1x)dx)+C(x+ 1)2=1(x+ 1)2(∫x2 + 2x+ 1xdx)+C(x+ 1)2=1(x+ 1)2(∫ (x+ 2 +1x)dx)+C(x+ 1)2a+ bc=ac+bc=1(x+ 1)2(12x2 + 2x+ ln |x|)+C(x+ 1)2∫xn dx =xn+1n+ 1+ C,∫k dx = kx + C,∫1xdx = ln |x|+ C318 Chapter 3. Differential Equations=x2 + 4x+ 2 ln |x|2(x+ 1)2+C(x+ 1)2The initial condition states that y(1) = 0. Buty(1) =1 + 4 + 2 ln |1|2(1 + 1)2+C(1 + 1)2=58+C4So, the equation y(1) = 0 becomes 58+ C4= 0. Subtracting 58from both sides,we get C4= −58. Multiplying both sides by 4, we get C = −52.Thus, the solution with initial condition y(1) = 0 isy =x2 + 4x+ 2 ln |x|2(x+ 1)2− 52(x+ 1)2=x2 + 4x− 5 + 2 ln |x|2(x+ 1)2(c) Adding 3 cos t to both sides of the equation dydt+ (cos t)y − 3 cos t = 0, we getdydt+ (cos t)y = 3 cos tSo, P (t) = cos t and Q(t) = 3 cos t. Finding the integrating factor, we haveI(t) = e∫P (t) dt = e∫cos t dt= esin t∫cos t dt = sin t+ CThe general solution isy(t) =1I(t)(∫I(t)Q(t) dt)+CI(t)=1esin t(∫esin t(3 cos t) dt)+Cesin t= e− sin t(∫3 cos tesin t dt)+ Ce− sin t= e− sin t(∫3eu(du))+ Ce− sin tUse substitution: u = sin t,du = cos tdt= e− sin t (3eu) + Ce− sin t∫ex dx = ex + C= e− sin t(3esin t)+ Ce− sin t Substitute u by sin t3.3. Integrating Factors 319= 3 + Ce− sin t e−aea = e−a+a = e0 = 1Using the initial condition y(pi) = −1, we have3 + Ce− sinpi = −13 + C = −1 sin pi = 0, e0 = 1C = −1− 3 = −4 Subtract 3 from both sidesThus, the solution with initial condition y(pi) = −1 isy(t) = 3− 4e− sin t3. [Exercise on page 31](a) Let y(t) be the amount of insulin present at time t. Then,rate of change of y = (rate in) - (rate out) (3.3.3)whererate in = rate at which insulin is infused into the patient bloodstreamrate out = rate at which insulin is removed from the patient bloodstreamFrom the problem,rate in = 8 and rate out = kywhere k is a positive constant of proportionality. Since the rate of change of yis dydt, the equation (3.3.3) becomesdydt= 8− kyThis gives a differential equation modeling the process.(b) Adding ky to both sides of the equation dydt= 8− ky, we getdydt+ ky = 8which is a differential equation of the form dydt+P (t)y = Q(t) with P (t) = k and320 Chapter 3. Differential EquationsQ(t) = 8. The integrating factor isI(t) = e∫P (t) dt = e∫k dt = ekt∫k dx = kx+ CThe general solution isy =1I(t)(∫I(t)Q(t) dt)+CI(t)=1ekt(∫ekt(8) dt)+Cekt=1ekt(8kekt)+Cekt∫ekx dx =1kekx + Cy(t) =8k+ Ce−kt SimplifyAlternate SolutionThe differential equation dydt= 8− ky can also be solved using the methodof separation of variables:dydt= 8− kydy = (8− ky)dt Multiply both sides by dt18− kydy = dtAssume that y 6= 8kand divideboth sides by 8− ky∫18− ky dy =∫dt Integrate both sides∫1u(du−k)=∫dtUse substitution: u = 8 − ky,du = −kdy, du−k = dy−1kln |u| = t+ C∫1udu = ln |u|+ Cln |u| = −kt− kC Multiply both sides by −k|u| = e−kt+CRelabel the constant −kC asC and take the natural expo-nential of both sidesu = ±eCe−kt |x| = a implies that x = ±a3.3. Integrating Factors 3218− ky = ±eCe−kt Substitute u by m− ky8− ky = Ce−kt Relabel the constant ±eC as Cy(t) =8k− Cke−kt Solve for yy(t) =8k+ Ce−kt Relabel the constant −Ckas CNote. Observe that y = 8ksatisfies the original equation. So, y = 8kis alsoa solution of the differential equation. But this solution can be obtainedfrom the general solution by letting C = 0. So, we don’t need to addy = 8kas a separate solution.(c)  (1) Finding the limit limt→∞y(t) from the solution found in part (b).limt→∞(y) = limt→∞(8k+ Ce−kt)= limt→∞(8k)+ C limt→∞(e−kt)=8k+ C(0) =8kSince k > 0, limt→∞e−kt = e−∞ = 0Hence, the glucose concentration will stabilize at8k. (2) Finding the limit limt→∞y(t) using a phase plot. Consider the differentialequationdydt= 8− kySolving the equation 8−ky = 0 for y, we get y = 8k. So, yˆ = 8kis an equilib-rium point of the differential equation. This equilibrium point determinestwo open intervals: (0, 8k) and ( 8k,∞). (Remember that y is a nonnegativequantity.) To make the phase plot, we need to determine the sign of thefunction g(y) = 8− ky on each of these intervals.– If 0 < y < 8k, then ky < 8. This implies that 0 < 8− ky; that is, g(y) ispositive on (0, 8k).– If y > 8k, then ky > 8. This implies that 0 > 8 − ky; that is, g(y) isnegative on ( 8k,∞).Using this information, we have the following phase plot.322 Chapter 3. Differential Equations0 ySign of g(y)8k+ −Since the arrows are pointing toward 8k, it follows that yˆ = 8kis a stableequilibrium point. This implies that y(t) converges to 8kas t→∞. In otherwords, limt→∞y(t) =8k.4. [Exercise on page 32](a) We haverate of change of y = (rate in) - (rate out) (3.3.4)whererate in = rate at which drug is administered to the patientrate out = rate at which drug is metabolizedFrom the problem,rate in = A(t) and rate out = kywhere k is a positive constant of proportionality. Since the rate of change of yis dydt, the equation (3.3.4) becomesdydt= A(t)− kyThis gives a differential equation modeling the process.(b) We want to solve the differential equation dydt= sin t− y, which is equivalent todydt+ y = sin tThis is a differential equation of the form dydt+ P (t)y = Q(t) with P (t) = 1 andQ(t) = sin t. The integrating factor isI(t) = e∫P (t) dt = e∫1 dt = et3.3. Integrating Factors 323The general solution isy =1I(t)(∫I(t)Q(t) dt)+CI(t)=1et(∫et sin t dt)+Cet= e−t(∫et sin t dt)+ Ce−tTo find the integral∫et sin t dt, we use integration by parts.Choose u = et and dv = sin t dtThen, du = etdt and v =∫sin t dt = − cos tUsing the formula for integration by parts, we get∫et sin t dt = uv −∫v du= −et cos t−∫− cos t et dt= −et cos t+∫et cos t dt∫kf(x) dx = k∫f(x) dx= −et cos t+(wz −∫z dw) Integrate by parts: chosse w =et and dz = cos t. Then dw =etdt and z = sin t= −et cos t+(et sin t−∫et sin t dt)∫et sin t dt = −et cos t+ et sin t−∫et sin t dt2∫et sin t dt = −et cos t+ et sin t Add∫et sin t dt to both sides∫et sin t dt =12et(sin t− cos t) Divide both sides by 2Using this, we havey = e−t(∫et sin t dt)+ Ce−ty = e−t(12et(sin t− cos t))+ Ce−t324 Chapter 3. Differential Equationsy(t) =12(sin t− cos t) + Ce−t ete−t = et−t = e0 = 1(c) Using the initial condition, we havey(0) = 412(sin 0− cos 0) + Ce0 = 4−12+ C = 4 sin 0 = 0 and cos 0 = 1C = 4 +12=92So, y(t) = 12(sin t− cos t) + 92e−t. And the concentration at time t = 2 isy(2) =12(sin 2− cos 2) + 92e−2 ≈ 0.66272 + 0.609 ≈ 1.273.4 Euler’s Method1. [Exercise on page 32]Euler’s MethodSuppose we have a differential equation of the formdydx= g(x, y), with initial condition y(x0) = y0 (3.4.1)For a positive real number h, define a sequence of numbers x0, x1, x2, x3, · · · asx1 = x0 + h x4 = x3 + hx2 = x1 + h...x3 = x2 + h xi+1 = xi + hThe number h is called the step size. Now, define a sequence of numbersy0, y1, y2, y3, · · · asyi+1 = yi + g(xi, yi)h, i ≥ 03.4. Euler’s Method 325That is,y1 = y0 + g(x0, y0)h y4 = y3 + g(x3, y3)hy2 = y1 + g(x1, y1)h...y3 = y2 + g(x2, y2)h yi+1 = yi + g(xi, yi)hLet y = f(x) be the solution of the differential equation dydx= g(x, y) satisfyingthe initial condition f(x0) = y0. Euler’s method says that for every i ≥ 0,the value of f(xi+1) can be approximated by yi+1; that is,f(x1) ≈ y1, f(x2) ≈ y2, · · · , f(xi) ≈ yi, f(xi+1) ≈ yi+1, · · ·Geometrically, Euler’s method consists of using tangent lines to approximatethe solution y = f(x) of the initial value problem (3.4.1). Specifically, considerthe following figure.x0 x1y0y1f(x1)y = f(x)ThaxyFrom the figure, the slope of the tangent line T ism =ahOn the other hand, since dydx= g(x, y) by (3.4.1), the slope of the tangent lineto the graph of y = f(x) at the point (x0, y0) ism =dydx= g(x0, y0)326 Chapter 3. Differential EquationsSo,ah= g(x0, y0)a = g(x0, y0)h Multiply both sides by hy1 − y0 = g(x0, y0)h From the figure, a = y1 − y0y1 = y0 + g(x0, y0)h Add y0 to both sidesFrom the figure, we see that if h is small f(x1) ≈ y1.Similarly, one hasy2 = y1 + g(x1, y1)h, y3 = y2 + g(x2, y2)h, · · · , yi+1 = yi + g(xi, yi)hFor the initial value problemdydx= 6x+ 7y, y(0) = 5we want to use Euler’s method with step size h = 0.1 to approximate the value ofy(0.3). From the initial value problem,x0 = 0, y0 = 5, g(x, y) = 6x+ 7yUsing the formula yi+1 = yi + g(xi, yi)h, we havey1 = y0 + g(x0, y0)h= 5 + g(0, 5)0.1= 5 + [6(0) + 7(5)] 0.1 g(0, 5) = 6(0) + 7(5) = 35= 5 + 35(0.1) = 5 + 3.5 = 8.5y2 = y1 + g(x1, y1)h= 8.5 + g(0.1, 8.5)0.1 x1 = x0 + h = 0 + 0.1 = 0.1= 8.5 + [6(0.1) + 7(8.5)] 0.1 g(0.1, 8.5) = 6(0.1) + 7(8.5) = 60.1= 8.5 + 60.1(0.1) = 8.5 + 6.01 = 14.51y3 = y2 + g(x2, y2)h3.4. Euler’s Method 327= 14.51 + g(0.2, 14.51)0.1 x2 = x1 + h = 0.1 + 0.1 = 0.2= 14.51 + [6(0.2) + 7(14.51)] 0.1= 14.51 + 102.77(0.1) = 14.51 + 10.277 = 24.787Since x3 = x2 + h = 0.2 + 0.1 = 0.3 and since y(x3) ≈ y3, it follows thaty(0.3) ≈ 24.787Using a Tablexi yi yi+1 = yi + g(xi, yi)hx0 = 0 y0 = 5 y1 = 5 + [6(0) + 7(5)]0.1 = 8.5x1 = x0 + h = 0.1 y1 = 8.5 y2 = 8.5 + [6(0.1) + 7(8.5)]0.1 = 14.51x2 = x1 + h = 0.2 y2 = 14.51 y3 = 14.51 + [6(0.2) + 7(14.51)]0.1 = 24.787x3 = x2 + h = 0.3 y3 = 24.7872. [Exercise on page 32](a) Hereg(x, y) = 4xy, x0 = 0, y0 = 2, h = 0.1To estimate y(0.3), we construct the following table.xi yi yi+1 = yi + g(xi, yi)hx0 = 0 y0 = 2 y1 = 2 + [4(0)(2)]0.1 = 2x1 = x0 + h = 0.1 y1 = 2 y2 = 2 + [4(0.1)(2)]0.1 = 2.08x2 = x1 + h = 0.2 y2 = 2.08 y3 = 2.08 + [4(0.2)(2.08)]0.1 = 2.2464x3 = x2 + h = 0.3 2.2464So, y(0.3) ≈ 2.2464(b) Hereg(x, y) = x2 +√y, x0 = 0, y0 = 1, h = 0.1To approximate y(0.5), we construct the following table.328 Chapter 3. Differential Equationsxi yi yi+1 = yi + g(xi, yi)h0 1 y1 = 1 + (02 +√1)0.1 = 1.10.1 1.1 y2 = 1.1 + (0.12 +√1.1)0.1 = 1.2058810.2 1.205881 y3 = 1.205881 + (0.22 +√1.205881)0.1 = 1.3196940.3 1.319694 y4 = 1.319694 + (0.32 +√1.319694)0.1 = 1.4435710.4 1.443571 y5 = 1.443571 + (0.42 +√1.443571)0.1 = 1.57972010.5 1.5797201So, y(0.5) ≈ 1.579720(c) Hereg(x, y) =xexy, x0 = 1, y0 = 5, h = 0.1To approximate y(1.4), we construct the following table.xi yi yi+1 = yi + g(xi, yi)h1 5 y1 = 5 +(1e15)0.1 = 5.054366x1 = x0 + 0.1 = 1.1 y1 = 5.054366 y2 = 5.054366 +(1.1e1.15.054366)0.1 = 5.119746x2 = x1 + h = 1.2 y2 = 5.119746 y3 = 5.119746 +(1.2e1.25.119746)0.1 = 5.197566x3 = x2 + h = 1.3 y3 = 5.197566 y4 = 5.197566 +(1.3e1.35.197566)0.1 = 5.289341x4 = x3 + h = 1.4 5.289341So, y(1.4) ≈ 5.289341(d) First, we need to write the differential equation 2 dydx− 3y = 18 lnx in the formdydx= g(x, y).2dydx− 3y = 18 ln x2dydx= 3y + 18 lnx Add 3y to both sidesdydx=3y + 18 lnx2Divide both sides by 23.4. Euler’s Method 329dydx=32y +18 lnx2a+ bc=ac+bcdydx= 1.5y + 9 lnxSo,g(x, y) = 1.5y + 9 lnx, x0 = 3, y0 = 4, h = 0.2To approximate y(3.6), we construct the following table.xi yi yi+1 = yi + g(xi, yi)h3 4 y1 = 4 + [1.5(4) + 9 ln(3)] 0.2 = 7.1775023.2 7.177502 y2 = 7.177502 + [1.5(7.177502) + 9 ln(3.2)] 0.2 = 11.4244243.4 11.424424 y3 = 11.424424 + [1.5(11.424424) + 9 ln(3.4)] 0.2 = 17.0545473.6 17.054547So,y(3.6) ≈ 17.0545473. [Exercise on page 32]Herex0 = 0, y0 = 5.8, h = 0.1The original differential equation, dydx+ 3x2y = 4x, is equivalent to dydx= 4x − 3x2y.So,g(x, y) = 4x− 3x2yUsing the formula yi+1 = yi + h · g(xi, yi), we havey1 = y0 + h · g(x0, y0)= 5.8 + 0.1 · g(0, 5.8)= 5.8 + 0.1 · [4(0)− 3(0)2(5.8)]= 5.8 + 0.1(0) = 5.8 + 0 = 5.8y2 = y1 + h · g(x1, y1)= 5.8 + 0.1 · g(0.1, 5.8)330 Chapter 3. Differential Equations= 5.8 + 0.1 · [4(0.1)− 3(0.1)2(5.8)] x1 = x0 + h = 0 + 0.1 = 0.1= 5.8 + 0.1(0.226) = 5.8 + 0.0226 = 5.8226y3 = y2 + h · g(x2, y2)= 5.8226 + 0.1 · g(0.2, 5.8226)= 5.8226 + 0.1 · [4(0.2)− 3(0.2)2(5.8226)] x2 = x1 + h = 0.1 + 0.1 = 0.2= 5.8226 + 0.1(0.101288) = 5.832729Proceeding with similar computations, we have the following table.xi yi yi+1 = yi + g(xi, yi)h0 5.8 y1 = 5.80.1 5.8 y2 = 5.82260.2 5.8226 y3 = 5.8327290.3 5.832729 y4 = 5.7952450.4 5.795245 y5 = 5.6770730.5 5.677073 y6 = 5.4512930.6 5.451293 y7 = 5.1025530.7 5.102553 y8 = 4.6324780.8 4.632478 y9 = 4.0630420.9 4.063042 y10 = 3.4357231 3.4357234. [Exercise on page 33](a) Here,g(x, y) = x2y2, x0 = 2, y0 = 1, h = 0.02To approximate y(2.1), we construct the following table.3.4. Euler’s Method 331xi yi yi+1 = yi + g(xi, yi)h2 1 y1 = 1 + [(2)2(1)2] 0.02 = 1.082.02 1.08 y2 = 1.08 + [(2.02)2(1.08)2] 0.02 = 1.1751882.04 1.175188 y3 = 1.175188 + [(2.04)2(1.175188)2] 0.02 = 1.2901362.06 1.290136 y4 = 1.290136 + [(2.06)2(1.290136)2] 0.02 = 1.4314022.08 1.431402 y4 = 1.431402 + [(2.08)2(1.431402)2] 0.02 = 1.608692.1 1.60869So, y(2.1) ≈ 1.60869(b) Solving the differential equation dydx= x2y2, we havedydx= x2y21y2dy = x2dx Multiply both sodes bydxy2∫1y2dy =∫x2 dx Take the integral of both sides∫y−2 dy =∫x2 dx1ym= y−my−1−1 =x33+ C∫xn dx =xn+1n+ 1+ C−1y=x33+ C y−m =1ymy =−1x33+ CSolve for yUsing the initial condition y(2) = 1, we have−1233+ C= 1233+ C = −11C = −1− 83= −33− 83= −113332 Chapter 3. Differential EquationsThus,y(x) =−1x33− 113=−1x3−113=−3x3 − 11Substituting x with 2.1, we gety(2.1) =−3(2.1)3 − 11 =−3−1.739 ≈ 1.725129(c) The results from parts (a) and (b) are close.5. [Exercise on page 33](a) Here,g(x, y) = 2x, x0 = 0, y0 = 1, h = 0.2Approximate values for the solution of the initial value problem dydx= 2x, y(0) =1, with step size h = 0.2 on the interval [0, 1], are given by the following table.xi yi yi+1 = yi + g(xi, yi)h0 1 y1 = 1 + [2(0)] 0.2 = 10.2 1 y2 = 1 + [2(0.2)] 0.2 = 1.080.4 1.08 y3 = 1.08 + [2(0.4)] 0.2 = 1.240.6 1.24 y4 = 1.24 + [2(0.6)] 0.2 = 1.480.8 1.48 y5 = 1.48 + [2(0.8)] 0.2 = 1.81 1.8(b) Solving the differential equation dydx= 2x, we gety(x) =∫2x dx = x2 + CUsing the initial condition y(0) = 1, we find C = 1. So, the exact solution isy = f(x) = x2 + 1 . The graph of y = x2 + 1 and the points (xi, yi) are shownbelow. (The points (xi, yi) are joined by line segments.)3.4. Euler’s Method 3330.2 0.4 0.6 0.8 112y = x2 + 1xy(c) The error made in part (a) in using Euler’s method to approximate the truevalue of f(0.6) isf(0.6)− y3 =[(0.6)2 + 1]− 1.24 = 1.36− 1.24 = 0.126. [Exercise on page 33](a) Here, x0 = 0, y0 = 1, and h = 0.2. Adding x to both sides of the differentialequation −x+ dydx= −y, we get dydx= x− y. So, g(x, y) = x− y.Approximate values for the solution of the initial value problemdydx= x− y, y(0) = 1,with step size h = 0.2 on the interval [0, 1], are given by the following table.xi yi yi+1 = yi + g(xi, yi)h0 1 y1 = 1 + [0− 1] 0.2 = 0.80.2 0.8 y2 = 0.8 + [0.2− 0.8] 0.2 = 0.680.4 0.68 y3 = 0.68 + [0.4− 0.68] 0.2 = 0.6240.6 0.624 y4 = 0.624 + [0.6− 0.624] 0.2 = 0.61920.8 0.6192 y5 = 0.6192 + [0.8− 0.6192] 0.2 = 0.655361 0.65536(b) To solve the differential equation dydx= x− y, we will use the integrating factor334 Chapter 3. Differential Equationsmethod. First, the differential equation dydx= x− y is equivalent todydx+ y = xThis is a differential equation of the form dydx+P (x)y = Q(x) with P (x) = 1 andQ(x) = x. The integrating factor isI(x) = e∫1 dx I(x) = e∫P (x) dx= exThe general solution isy =1I(x)(∫I(x)Q(x) dx)+CI(x)=1ex(∫xex dx)+Cex=1ex(uv −∫vdu)+CexBy parts: Choose u = x anddv = ex. Then du = 1dx andv = ex=1ex(xex −∫exdx)+CexSubstitute u and v=1ex(xex − ex) + Cex∫ex dx = ex + C=1exex(x− 1) + CexFactor out ex= x− 1 + Cex= x− 1 + Ce−x SimplifyUsing the initial condition y(0) = 1, we have0− 1 + Ce0 = 1 Substitute x with 0 into x− 1 + Ce−x−1 + C = 1 e0 = 1C = 1 + 1 = 2 Add 1 to both sidesSo, the exact solution is y = f(x) = x− 1 + 2e−x .The graph of y = x − 1 + 2e−x and the points (xi, yi) are shown below. (Thepoints (xi, yi) are joined by line segments.)3.4. Euler’s Method 3350.2 0.4 0.6 0.8 11y = x− 1 + 2e−xxy(c) The error made in part (a) in using Euler’s method to approximate the truevalue of f(0.6) isf(0.6)− y3 =[0.6− 1 + 2e−0.6]− 0.624 = 0.697623− 0.624 = 0.0736237. [Exercise on page 33]Here,g(t, y) = 0.09y − 0.5y 12 , t0 = 0, y0 = 70, h = 1To approximate y(4), we construct the following table.ti yi yi+1 = yi + g(ti, yi)h0 70 y1 = 70 +[0.09(70)− 0.5(70) 12]= 72.11671 72.1167 y2 = 72.1167 +[0.09(72.1167)− 0.5(72.1167) 12]= 74.3611252 74.361125 y3 = 74.361125 +[0.09(74.361125)− 0.5(74.361125) 12]= 76.7419823 76.741982 y4 = 76.741982 +[0.09(76.741982)− 0.5(76.741982) 12]= 79.2686354 79.268635So, the size of the population at t = 4 is approximately 79 .336 Chapter 3. Differential EquationsChapter4Linear Algebra4.1 Systems of Linear Equations1. [Exercise on page 35]Linear EquationsA linear equation is an equation of the forma1x1 + a2x2 + · · ·+ anxn = bwhere n is a positive integer; a1, a2, · · · , an are real numbers called coefficients; b is a real number called the constant term; x1, x2, · · · , xn are the unknowns (or variables).When n ≤ 3, we use x, y, and z to represent x1, x2, and x3. For example,the equation 4x + 5y + 6z = 7 is a linear equation in n = 3 unknowns. Thecoefficients are a1 = 4, a2 = 5, and a3 = 6, and the constant term is b = 7.(a) The equation x+ y = 0 is of the forma1x+ a2y = bwhere a1 = 1, a2 = 1, and b = 0. So it is a linear equation in n = 2 unknowns.The coefficients are 1 and 1. And the constant term is 0.337338 Chapter 4. Linear Algebra(b) The equation −2x + 4y − 5z = 12is linear . The number of unknowns is 3, thecoefficients are −2, 4, and −5, and the constant term is 12.(c) The equation√x− y = 19 has two unknowns and it is not linear because it isnot of the form a1x+ a2y = b. The term that makes this nonlinear is√x = x12 .(By definition, if an equation is linear, the exponent of each unknown will be 1.)(d) The equation −4x − 20y + cos(z) = −1 is in 3 unknowns and it is not linearbecause it is not of the form a1x + a2y + a3z = b. The term that makes thisnonlinear is cos(z).(e) The equation 5x + y√2 = −47 is linear . The number of unknowns is 2, thecoefficients are 5,√2, and the constant term is −47.(f) The equation −x + 9xy − z = −3 is not linear . The term that makes thisnonlinear is xy as the definition does not allow products of unknowns.(g) The equation 1x+y− z = 24 is in three unknowns and it is not linear because itis not of the form a1x + a2y + a3z = b. The term that makes this nonlinear is1x+y.2. [Exercise on page 35]Solution to a Linear Equation An n−tuple is an ordered list (s1, s2, · · · , sn) of numbers. Here the term“ordered” means that if we switch two numbers of the list, we get adifferent list. For instance, (1, 2) is not the same as (2, 1). One can thinkof this as a point. For example, (1, 2) can be thought of as the point ofthe plane whose x-coordinate is 1 and y-coordinate is 2. A solution to a linear equationa1x1 + a2x2 + · · ·+ anxn = bis an n-tuple (s1, s2, · · · , sn) that satisfies the equation. This means thatif we substitute x1 with s1, x2 with s2, · · · , xn with sn into a1x1 + a2x2 +· · ·+ anxn, we get b.For example, the point (5, 6) is a solution to (or satisfies) the equation3x+ 4y = 39 since 3(5) + 4(6) is 39.4.1. Systems of Linear Equations 339(a)  Substituting x with 3 and y with 3 into 2x+ 3y, we get2(3) + 3(3) = 6 + 9 = 15Since the equation is 2x+ 3y = 14 and since 2(3) + 3(3) is not equal to 14,it follows that (3, 3) is not a solution. Substituting x with 1 and y with 4 into 2x+ 3y, we get2(1) + 3(4) = 2 + 12 = 14So (1, 4) is a solution to the equation 2x+ 3y = 14. Substituting x with 5 and y with 1 into 2x + 3y, we get 2(5) + 3(1) = 13.So (5, 1) is not a solution to the equation 2x+ 3y = 14.(b) Consider the following table.(s1, s2, s3) Equation Substituting(−2, 3, 0) −4x− 5y + 21z = 7 −4(−2)− 5(3) + 21(0) = −7(−2,−3, 0) −4x− 5y + 21z = 7 −4(−2)− 5(−3) + 21(0) = 23(2,−3, 0) −4x− 5y + 21z = 7 −4(2)− 5(−3) + 21(0) = 7From the table, only the point (2,−3, 0) is a solution to the equation −4x−5y + 21z = 7.3. [Exercise on page 35]Systems of Linear EquationsA system of linear equations (or just linear system) is a finite collectionof linear equations. Examples include the following systems.3x − 4y = −15x + 7y = 12x − y + z = 02x − 3y = 2y − z = −3The first has two unknowns and two equations, while the second has threeunknowns and three equations.Note. In a linear system every equation is linear. If one of the equations isnot linear, the system is not a linear system.340 Chapter 4. Linear Algebra(a) Here the system is a linear system since every equation is linear.(b) The system here is a linear system since every equation is linear.(c) The system is not a linear since the third equation, −2x + ez = −2, is notlinear.(d) The system is not a linear since the first equation, 34x − √5y = −4, (as wellas the second equation) is not linear.4. [Exercise on page 36]Solution to a System of Linear EquationsSuppose we have a linear system in n unknowns with m equations. An n-tuple(s1, s2, · · · , sn) is a solution to that system if it satisfies every equation of thesystem. For example, consider the following system.x − y = 1x + y = 3 The point (2, 1) is a solution since it satisfies both equations (2 − 1 = 1and 2 + 1 = 3). The point (4, 3) satisfies the first equation (4 − 3 = 1) but it is nota solution to the system since it does not satisfy the second equation(4 + 3 = 7 6= 3).(a) We need to determine whether (1,−1) is a solution to each equation. Substituting x with 1 and y with −1 into the equation x − 2y = 3, we get1− 2(−1) = 3, that is, 3 = 3. This implies that (1,−1) is a solution to thefirst equation. Substituting x with 1 and y with −1 into the second equation, 3x+ y = 2,we get 3(1)+(−1) = 2, that is, 2 = 2. This implies that (1,−1) is a solutionto the second equation.Since (1,−1) is a solution to every equation of the system, it follows that (1,−1)is a solution to the system .(b)  Substituting x with −2 and y with 3 into the first equation, we get −3(−2)+4(3) = 18, that is, 18 = 18. This shows that (−2, 3) is a solution to the firstequation.4.1. Systems of Linear Equations 341 Substituting x with −2 and y with 3 into the second equation, we get5(−2) − 6(3) = −27, that is, −28 = −27, which is not true. So (−2, 3) isnot a solution to the second equation.Since (−2, 3) does not satisfy all the equations, it follows that (−2, 3) is nota solution to the system .5. [Exercise on page 36](a) First, observe that each equation of the systemx + y = 1−x + y = −1is the equation of a line (a linear equation in one or two unknowns). To sketchthe graph of a line, we need two points. Finding two points on the line x+ y = 1. The idea is to assign a number toone of the unknowns and find the other variable. For example,– if x = 0, then the equation x + y = 1 becomes 0 + y = 1 or y = 1. Sothe point (0, 1) belongs to the line x+ y = 1.– If y = 0, then x+ 0 = 1 or x = 1. So the line x+ y = 1 passes throughthe point (1, 0).Using the points (0, 1) and (1, 0), we get the graph of x + y = 1 as shownbelow. Finding two points on the line −x+ y = −1.– If x = 0, then y = −1. This gives the point (0,−1).– If y = 0, then −x = −1 or x = 1. This gives the point (1, 0).Using these points, we get the graph of −x+ y = −1 as shown below.1 3−11x+ y = 1−x+ y = −1xy342 Chapter 4. Linear AlgebraFrom the figure, the lines intersect at (1, 0) , which is the unique solution tothe system.(b) Sketching the graphs of −x + 2y = 3 and 2x − 4y = −5, we get the followingfigure.−1 1−x+ 2y = 32x− 4y = −5xySince the two lines are parallel, there is no intersection. This means that thesystem has no solution .(c) The figure below shows the graphs of 2x+ 3y = 5 and −3x+ y = 9.−2 2 4310 2x+ 3y = 5−3x+ y = 9xySince the lines intersect at one point, the system has a unique solution Thepoint of intersection, (−2, 3) , is the unique solution to the system.(d) First, observe that if we divide both sides of the second equation, −2x+2y = −4,by −2, we get the first equation. This means that the two equations have thesame graph as shown in the following figure.4.1. Systems of Linear Equations 343−2 2−22 x− y = 2−2x+ 2y = −4xySince the graphs of the two equations coincide, the system has infinitely manysolutions . Every point on the line is a solution to the system.Number of Solutions to a Linear SystemA linear system has no solution, or a unique solution, or infinitely many solutions.We only have these three possibilities.6. [Exercise on page 36]Augmented Matrix of a Linear SystemLet (S) be a linear system. The augmented matrix of (S) is the matrixcontaining all the coefficients and all the constant terms of (S). For example,the augmented matrix of the system5x + 6y = 78x − 9y = 10 is[5 6 78 −9 10]The vertical bar is to separate the coefficients from the constant terms.(a) For the system3x + 4y = 56x − 7y = 8 , the augmented matrix is344 Chapter 4. Linear Algebra[3 4 56 −7 8](b) For the system−5x + 7y = −166x = 11, the augmented matrix is[−5 7 −166 0 11]7. [Exercise on page 36](a) The linear system corresponding to the augmented matrix[1 −1 3−2 4 −8]isx − y = 3−2x + 4y = −8(b) The linear system corresponding to the augmented matrix[1 −2 40 −3 −9]isx − 2y = 4− 3y = −98. [Exercise on page 36]Gauss-Jordan MethodThe Gauss-Jordan method consists of using the elementary row operations totransform the augmented matrix into a matrix from which the solutions can beimmediately read.4.1. Systems of Linear Equations 345Elementary Row OperationsThere are three elementary row operations.(i) Interchange two rows.(ii) Multiply one row by a nonzero number.(iii) Replace one row by the sum of itself and a multiple of another row.Notation. We use the following notation. We write Ri for the ithrow. For example, R1 will represent the first row at each stage, R2 willrepresent the second row at each stage, and so on. Using this notation,the elementary row operations become:(i) Ri ↔ Rj. This means that we interchange Ri and Rj.(ii) aRi → Ri, where a is a nonzero number. This means that wemultiply the row Ri by a and we put the answer in row Ri. Inother words, Ri is replaced by aRi.(iii) Ri + bRj → Ri, where b is a number (not necessarily a nonzeronumber). This means that we replace the row Ri with Ri + bRj.For example, R2 + 3R1 → R2 means that we calculate R2 + 3R1and we put the result in R2.To illustrate the Gauss-Jordan method, let’s solve the following system.x − 3y = −73x + 6y = 9First, we write the augmented matrix:M =[1 −3 −73 6 9]Our goal is to use elementary row operations to transform the augmented ma-trix into a matrix of the form[1 0 α0 1 β]or[1 γ τ0 0 κ]346 Chapter 4. Linear Algebrawhere α, β, γ, τ, and κ are real numbers.We begin by creating a zero below the upper left entry, 1. Performing theoperation R2 − 3R1 → R2 on M , we get[1 −3 −73− 3(1) 6− 3(−3) 9− 3(−7)]=[1 −3 −73− 3 6 + 9 9 + 21]=[1 −3 −70 15 30]= M1Multiplying the second row of M1 by115, we haveM1 =[1 −3 −70 15 30]115R2 → R2[1 −3 −70 1 2]= M2To finish, we perform the operation R1 + 3R2 → R1 on M2 to obtain[1 0 −10 1 2]= M3The linear system corresponding to M3 isx = −1y = 2So, the solution to the original system is (−1,2).(a) Transforming the augmented matrix using the elementary row operations, wehave[1 2 52 −1 −10]R2 − 2R1 → R2[1 2 50 −5 −20]−15R2 → R2[1 2 50 1 4]R1 − 2R2 → R1[1 0 −30 1 4]4.1. Systems of Linear Equations 347The corresponding system isx = −3y = 4So the solution is (−3, 4) .(b) Transforming the augmented matrix using the elementary row operations, wehave[1 1 03 −4 7]R2 − 3R1 → R2[1 1 00 −7 7]−17R2 → R2[1 1 00 1 −1]R1 −R2 → R1[1 0 10 1 −1]The corresponding system isx = 1y = −1So the solution is (1,−1) .(c) The augmented matrix of the system−2x − 10y = −203x + 4y = −3 is[−2 −10 −203 4 −3]Using elementary row operations, we have[−2 −10 −203 4 −3]−12R1 → R1[1 5 103 4 −3]R2 − 3R1 → R2[1 5 100 −11 −33]348 Chapter 4. Linear Algebra− 111R2 → R2[1 5 100 1 3]R1 − 5R2 → R1[1 0 −50 1 3]So the solution is (−5, 3) .(d) Transforming the augmented matrix using the elementary row operations, wehave[2 −3 75 7 3]R2 − 2R1 → R2[2 −3 71 13 −11]R1 ↔ R2[1 13 −112 −3 7]R2 − 2R1 → R2[1 13 −110 −29 29]− 129R2 → R2[1 13 −110 1 −1]R1 − 13R2 → R1[1 0 20 1 −1]So the solution is (2,−1) .9. [Exercise on page 37](a) Transforming the augmented matrix using the elementary row operations, wehave[1 −3 −4−2 6 7]R2 + 2R1 → R2[1 −3 −40 0 −1]The corresponding system isx − 3y = −40 = −14.1. Systems of Linear Equations 349Since 0 = −1 is not true, the system has no solution .(b) Transforming the augmented matrix using the elementary row operations, wehave[−6 9 158 −12 9]−13R1 → R1[2 −3 −58 −12 9]R2 − 4R1 → R2[2 −3 −50 0 29]The second equation is 0 = 29. Since this is not true, the system has no solution .(c) [−5 10 −151 −2 3]R1 ↔ R2[1 −2 3−5 10 −15]R2 + 5R1 → R2[1 −2 30 0 0]The system corresponding to this latter matrix isx − 2y = 30 = 0This reduces to the equation x − 2y = 3. To solve this, we assign one of theunknowns as a parameter. Since the equation is easier to solve for x, we willassign y as a parameter (we will use the letters s, t, · · · for parameters). Lety = s. Then the equation x − 2y = 3 becomes x − 2s = 3. Solving this for x,we get x = 3 + 2s. Thus, the general solution of the system isx = 3 + 2sy = swhere s can be any real number. For each value of s we have a solution. So, thesystem has infinitely many solutions.(d) [7 14 35−4 −8 −20]17R1 → R1−14R2 → R2[1 2 51 2 5]350 Chapter 4. Linear AlgebraR2 −R1 → R2[1 2 50 0 0]The corresponding system is x + 2y = 5. Let y = s. Then x = 5 − 2s and thegeneral solution isx = 5− 2sy = swhere s can be any real number.(e) [4 24 0−3 −18 0]14R1 → R1−13R2 → R2[1 6 01 6 0]R2 −R1 → R2[1 6 00 0 0]The corresponding system is x + 6y = 0. Let y = s. Then x = −6s and thegeneral solution isx = −6sy = swhere s can be any real number.10. [Exercise on page 37]To avoid solving three systems, we will solve the system for arbitrary values of a andb. The we will substitute the values of a and b into the solution in each case.The augmented matrix is [−1 5 a7 −30 b]Using elementary row operations, we have[−1 5 a7 −30 b]−R1 → R1[1 −5 −a7 −30 b]R2 − 7R1 → R2[1 −5 −a0 5 b+ 7a]15R2 → R2[1 −5 −a0 1 b5+ 75a]4.1. Systems of Linear Equations 351R1 + 5R2 → R1[1 0 6a+ b0 1 b5+ 75a]So, the solution isx = 6a+ by =75a+15b(a) If a = 1, b = −2, thenx = 6(1) + (−2) = 6− 2 = 4y =75(1) +15(−2) = 7− 25= 1So, the solution to the system in this case is (4, 1) .(b) If a = −4, b = 18, thenx = 6(−4) + 18 = −24 + 18 = −6y =75(−4) + 15(18) =−28 + 185=−105= −2So, the solution to the system in this case is (−6,−2) .(c) If a = 12, b = −3, thenx = 6(12)+ (−3) = 3− 3 = 0y =75(12)+15(−3) = 710− 35=710− 610=110So, the solution to the system in this case is(0,110).11. [Exercise on page 37][1 −3 b−2 a 0]R2 + 2R1 → R2[1 −3 b0 a− 6 2b]The corresponding system isx − 3y = b(a− 6)y = 2b (4.1.1)352 Chapter 4. Linear AlgebraTo solve the second equation for y, we need to divide both sides by a − 6. But, thedivision by a− 6 makes sense only if a 6= 6. We thus have the following discussion. If a 6= 6, then from the second equation we have y = 2ba−6 . Substituting this intothe first equation, we getx− 3(2ba− 6)= bx− 6ba− 6 = bx = b+6ba− 6x =ab− 6b+ 6ba− 6 a+bc=ac+ bcx =aba− 6So, if a 6= 6 , the system has a unique solution:(aba− 6 ,2ba− 6). Assume a = 6 and b 6= 0. Then the system (4.1.1) becomesx − 3y = b0 = 2bSince 0 = b is not true (because we assumed b 6= 0), the system has no solution.So, if a = 6 and b 6= 0 , the system has no solution . Assume a = 6 and b = 0. Then the system (4.1.1) becomesx − 3y = 00 = 0This reduces to the equation x − 3y = 0. Let y = s. Then x = 3s and thegeneral solution isx = 3sy = swhere s can be any real number. Each value of s gives a solution. So, if a = 6and b = 0 , the system has infinitely many solutions .4.2. Matrix Operations 3534.2 Matrix Operations1. [Exercise on page 37]Size of a MatrixLet A be a matrix. If the number of rows of A is m and the number of columnsis n, we say that A is of size m × n (read m by n). For example, the matrixA =[5 6 40 0 0]is of size 2× 3 because it has two rows and three columns.Matrix AdditionLet A and B be two matrices. Then the sum A + B is defined only if A andB are the same size. In that case, A + B is the matrix in which each entry isthe sum of the corresponding entries on A and B. For example,[1 23 4]+[5 67 8]=[1 + 5 2 + 63 + 7 4 + 8]=[6 810 12]As an immediate consequence of the definition, if A and B have the same size,say m× n, then the sum A+B is of size m× n.(a)A+B =[1 7−3 8]+[−2 65 −14]=[1 + (−2) 7 + 6−3 + 5 8 + (−14)]=[−1 132 −6](b)C +D = 9 −7−3 011 −22+ 17 −9−4 10 19= 9 + 17 −7− 9−3− 4 0 + 111 + 0 −22 + 19 = 26 −16−7 111 −3354 Chapter 4. Linear Algebra(c) The sum A+ C is undefined because A and C do not have the same size.(d) The sum C + D + B is undefined because D and B (as well as C and B) donot have the same size.(e)Scalar MultiplicationLet A be an m × n matrix, and let k be a scalar (by scalar, we mean areal number). Then the scalar multiple kA is the matrix of size m× nobtained by multiplying each entry of A by k. For example,3[4 56 7]=[(3)4 (3)5(3)6 (3)7]=[12 1518 21]For c = 2 and C = 9 −7−3 011 −22, we have2C = 2 9 −7−3 011 −22 = 2(9) 2(−7)2(−3) 2(0)2(11) 2(−22) = 18 −14−6 022 −44(f)3C − 2D = 3 9 −7−3 011 −22− 2 17 −9−4 10 19= 27 −21−9 033 −66+ −34 188 −20 −38= 27− 34 −21 + 18−9 + 8 0− 233 + 0 −66− 38 = −7 −3−1 −233 −1042. [Exercise on page 38]−5A+ 2B − 3I = −5[1 7−3 8]+ 2[−2 65 −14]− 3[1 00 1]4.2. Matrix Operations 355=[−5 −3515 −40]+[−4 1210 −28]+[−3 00 −3]=[−5− 4− 3 −35 + 12 + 015 + 10 + 0 −40− 28− 3]=[−12 −2325 −71]3. [Exercise on page 38]We want to solve the equation X +[5 −31 6]= −6[−2 81 0].Subtracting the matrix[5 −31 6]from both sides, we getX = −6[−2 81 0]−[5 −31 6]=[12− 5 −48 + 3−6− 1 0− 6]=[7 −45−7 −6]4. [Exercise on page 38]Product of Two MatricesLet A be an m × n matrix and let B be an n × p matrix. The product,AB, is the matrix whose entry in the ith row and jth column is obtained bymultiplying the entries in the ith row of A by the corresponding entries of thejth column of B, and by adding these products. Specifically, let aij be the entryin the ith row and jth column of A. Let bij be the entry in the ith row andjth column of B. The entry cij in the ith row and jth column of the productAB is defined bycij = ai1b1j + ai2b2j + · · ·+ ainbnjFor example, if A =[a11 a12a21 a22]and B =[b11 b12b21 b22], thenAB =[a11b11 + a12b21 a11b12 + a12b22a21b11 + a22b21 a21b12 + a22b22]Caution. The product AB is defined only if the number of columns of A isequal to the number of rows of B.356 Chapter 4. Linear Algebra(a)AB =[1 23 4][−3 54 −7]=[1(−3) + 2(4) 1(5) + 2(−7)3(−3) + 4(4) 3(5) + 4(−7)]=[−3 + 8 5− 14−9 + 16 15− 28]=[5 −97 −13](b)BA =[−3 54 −7][1 23 4]=[−3(1) + 5(3) −3(2) + 5(4)4(1)− 7(3) 4(2)− 7(4)]=[−3 + 15 −6 + 204− 21 8− 28]=[12 14−17 −20](c) The product BC is undefined because the number of columns of B is not equalto the number of rows of C.(d)CB = 2 05 −813 −9[ −3 54 −7]= 2(−3) + 0(4) 2(5) + 0(−7)5(−3)− 8(4) 5(5)− 8(−7)13(−3)− 9(4) 13(5)− 9(−7)= −6 + 0 10 + 0−15− 32 25 + 56−39− 36 65 + 63 = −6 10−47 81−75 128(e) The product CD is undefined because the number of columns of C is not equalto the number of rows of D.(f)DC = −2 1 34 8 910 −4 5 2 05 −813 −9= −2(2) + 1(5) + 3(13) −2(0) + 1(−8) + 3(−9)4(2) + 8(5) + 9(13) 4(0) + 8(−8) + 9(−9)10(2)− 4(5) + 5(13) 10(0)− 4(−8) + 5(−9)4.2. Matrix Operations 357= −4 + 5 + 39 0− 8− 278 + 40 + 117 0− 64− 8120− 20 + 65 0 + 32− 45 = 40 −35165 −14565 −13(g)CE = 2 05 −813 −9[ −2 6 93 −7 7]= 2(−2) + 0(3) 2(6) + 0(−7) 2(9) + 0(7)5(−2)− 8(3) 5(6)− 8(−7) 5(9)− 8(7)13(−2)− 9(3) 13(6)− 9(−7) 13(9)− 9(7)= −4 + 0 12 + 0 18 + 0−10− 24 30 + 56 45− 56−26− 27 78 + 63 117− 63 = −4 12 18−34 86 −11−53 141 54(h)EC =[−2 6 93 −7 7] 2 05 −813 −9=[−2(2) + 6(5) + 9(13) −2(0) + 6(−8) + 9(−9)3(2)− 7(5) + 7(13) 3(0)− 7(−8) + 7(−9)]=[−4 + 30 + 117 0− 48− 816− 35 + 91 0 + 56− 63]=[143 −12962 −7](i)DF = −2 1 34 8 910 −4 5 0 9 4−2 −2 −51 −10 −1= −2(0) + 1(−2) + 3(1) −2(9) + 1(−2) + 3(−10) −2(4) + 1(−5) + 3(−1)4(0) + 8(−2) + 9(1) 4(9) + 8(−2) + 9(−10) 4(4) + 8(−5) + 9(−1)10(0)− 4(−2) + 5(1) 10(9)− 4(−2) + 5(−10) 10(4)− 4(−5) + 5(−1)358 Chapter 4. Linear Algebra= 0− 2 + 3 −18− 2− 30 −8− 5− 30− 16 + 9 36− 16− 90 16− 40− 90 + 8 + 5 90 + 8− 50 40 + 20− 5 = 1 −50 −16−7 −70 −3313 48 55(j)A2 = AA =[1 23 4][1 23 4]=[1(1) + 2(3) 1(2) + 2(4)3(1) + 4(3) 3(2) + 4(4)]=[1 + 6 2 + 83 + 12 6 + 16]=[7 1015 22](k) To calculate B3, we first need to find B2.B2 = BB =[−3 54 −7][−3 54 −7]=[−3(−3) + (5)(4) (−3)(5) + 5(−7)4(−3) + (−7)(4) 4(5) + (−7)(−7)]=[9 + 20 −15− 35−12− 28 20 + 49]=[29 −50−40 69]Now,B3 = BBB = BB2 =[−3 54 −7][29 −50−40 69]=[−3(29) + 5(−40) (−3)(−50) + 5(69)4(29) + (−7)(−40) 4(−50) + (−7)(69)]=[−87− 200 150 + 345116 + 280 −200− 483]=[−287 495396 −683](l) The product C2 = CC is undefined because the number of columns of C is notequal to the number of rows.5. [Exercise on page 39]We know that the product AB of a matrix A of size m × n by a matrix B of sizen × p is a matrix of size m × p. Here m = 5, n = 7 and p = 11. So the size of B isn× p = 7× 11 .6. [Exercise on page 39]4.2. Matrix Operations 359(a) If the product BA is of size 17 × 32, then the size of B is of the form 17 × n,and the size of A is of the form n × 32 where n is a positive integer. So B has17 rows .(b) Since the size of A is of the form n× 32, the number of columns of A is 32 .7. [Exercise on page 39]The zero product property does not hold for matrices. Indeed, let A =[1 00 0]andB =[0 00 1]. ThenAB =[0 + 0 0 + 00 + 0 0 + 0]=[0 00 0]Every entry of AB is 0. So AB is the zero matrix. But neither A nor B is the zeromatrix.8. [Exercise on page 39](a)u+ v =[−45]+[3−1]=[−4 + 35 + (−1)]=[−14]−2v = −2[3−1]=[−2(3)−2(−1)]=[−62]6u+ 7v = 6[−45]+ 7[3−1]=[−2430]+[21−7]=[−323](b)Au =[1 23 4][−45]=[1(−4) + 2(5)3(−4) + 4(5)]=[−4 + 10−12 + 20]=[68]Av =[1 23 4][3−1]=[1(3) + 2(−1)3(3) + 4(−1)]=[3− 29− 4]=[15](c)360 Chapter 4. Linear AlgebraMatrix Multiplication PropertiesLet A,B,C be matrices, and let k be a scalar. Then,(1) A(BC) = (AB)C(2) A(B + C) = AB + AC(3) (A+B)C = AC +BC(4) A(kB) = k(AB) = (kA)BUsing property (2), we haveA(3u− 4v) = A(3u) + A(−4v)= 3Au− 4Av Property (4)= 3[68]− 4[15] From part (b), Au = [ 68]and Av =[15]=[1824]+[−4−20]=[144]Calculating A(−4u+ 7v), we haveA(−4u+ 7v) = A(−4u) + A(7v) Property (2)= −4Au+ 7Av Property (4)= −4[68]+ 7[15] From part (b), Au = [ 68]and Av =[15]=[−24−32]+[735]=[−173](d) The graphs of the vector v,Au, and Av are shown in the following figure.4.2. Matrix Operations 361−48uvAuAvxy9. [Exercise on page 40]The Inverse of a 2× 2 MatrixLet A =[a bc d]be a 2× 2 matrix. The determinant of A, denoted detA,is defined bydetA = ad− bcTheorem 4.2.1. (a) The matrix A is invertible if and only if detA 6= 0.(b) If detA 6= 0, the inverse of A is given byA−1 =1ad− bc[d −b−c a](a) For the matrix A =[1 22 6], the determinant isdetA = (1)(6)− 2(2) = 6− 4 = 2Since detA 6= 0, A is invertible andA−1 =12[6 −2−2 1]= 62 −22−2212 = 3 −1−1 12362 Chapter 4. Linear Algebra(b)  The determinant of A =[−6 712 −14]isdetA = (−6)(−14)− (12)(7) = 84− 84 = 0 Since detA = 0, A is not invertible . So A−1 does not exist.(c) The determinant of A =[−2 3−4 8]isdetA = (−2)(8)− (−4)(3) = −16 + 12 = −4Since detA 6= 0, it follows that A is invertible andA−1 =1−4[8 −3−(−4) −2]= 8−4 −3−44−4−2−4 = −2 34−1 12(d) The determinant of A =[15 −18−10 12]isdetA = 15(12)− (−10)(−18) = 180− 180 = 0Since detA = 0, it follows that A is not invertible . So A−1 does not exist.10. [Exercise on page 40]The determinant of A =[k −5−4 k + 1]isdetA = k(k + 1)− (−4)(−5) = k2 + k − 20 = (k − 4)(k + 5)So A is invertible if and only if (k − 4)(k + 5) 6= 0, that is, k 6= 4 and k 6= −5 .11. [Exercise on page 40]Matrix EquationConsider the systemax + by = kcx + dy = l(4.2.1)Associated with this system are the following matrices.4.2. Matrix Operations 363 The coefficient matrix: A =[a bc d]. The vector of unknowns: X =[xy]. The vector of constant terms: B =[kl]The matrix equation of the system (4.2.1) is the equation AX = B.(a) For the system7x − 6y = 13−8x + 9y = 5 , the corresponding matrix equation isAX = B, whereA =[7 −6−8 9], X =[xy], B =[135](b) The matrix equation corresponding to the systemx + y − z = 02x + z = 13y − 4z = 7is AX = B, whereA = 1 1 −12 0 10 3 −4 , X = xyz , B = 01712. [Exercise on page 40]Solving Linear Systems Using Matrix InversesTheorem 4.2.2. Let AX = B be a linear system with n equations in n un-knowns. If the coefficient matrix A is invertible, the system has the uniquesolution X = A−1B.Proof. Suppose that A is invertible. Then we haveAX = B364 Chapter 4. Linear AlgebraA−1AX = A−1B Left-multiply both sides by A−1IX = A−1B Since A−1A = IX = A−1B IX = X because I is the identity matrix(a)  The matrix equation for the system−3x − 6y = 10−x + 3y = 10 is AX = B withA =[−3 −6−1 3], X =[xy], and B =[1010] The determinant of A is detA = (−3)(3)− (−1)(−6) = −9− 6 = −15. Since detA 6= 0, A is invertible andA−1 =1−15[3 61 −3]= 3−15 6−151−15−3−15 = −15 −25− 11515 The solution to the system isX = A−1B =[−15−25− 11515][1010]= −105 − 205−1015+ 105= −105 − 205−1015+ 3015 = [ −3052015]=[−643](b)  Here the matrix equation is AX = B withA =[4 −57 6], X =[xy], and B =[11−3] The determinant of A is detA = (4)(6)− (7)(−5) = 24 + 35 = 59. Since detA 6= 0, A is invertible andA−1 =159[6 5−7 4]= 659 559−7594594.3. Eigenvalues, Eigenvectors, and Matrix Models 365 The solution to the system isX = A−1B = 659 559−759459[ 11−3]= 6659 − 1559−7759− 1259 = 5159−89594.3 Eigenvalues, Eigenvectors, and Matrix Models1. [Exercise on page 41]Eigenvalues and EigenvectorsLet A be a square matrix. Geometrically, a vector x is an eigenvector of A ifAx is parallel to x, that is, Ax is a multiple of x. The factor by which x isstretched (or contracted) is called eigenvalue. To be more precise, we have thefollowing definition.DefinitionLet A be a square matrix. A vector x is called an eigenvector of A ifx 6= 0 and if there exists a scalar λ such thatAx = λxThe scalar λ is called an eigenvalue of A.If Ax = λx, we say that x is an eigenvector corresponding to the eigenvalue λ.We also say that λ is the eigenvalue associated with the eigenvector x.(a)  To determine if u is an eigenvector of A, we need to compute Au, and see ifit is a multiple of u.Au =[3 −3−2 2][−32]=[−9− 66 + 4]=[−1510]= 5[−32]= 5uSince Au = 5u, it follows that u is an eigenvector of A corresponding tothe eigenvalue λ = 5 .366 Chapter 4. Linear Algebra We now compute Av.Av =[3 −3−2 2][1−1]=[3 + 3−2− 2]=[6−4]Clearly, the vector Av =[6−4]is not a multiple of v =[1−1]. Thisimplies that v is not an eigenvector of A .(b)  Computing Au, we getAu = 1 0 −11 −3 04 −13 1 302 = 3 + 0− 23 + 0 + 012 + 0 + 2 = 1314Since Au is not a multiple of u, it follows that u is not an eigenvector of A . Computing Av, we getAv = 1 0 −11 −3 04 −13 1 113 = 1 + 0− 31− 3 + 04− 13 + 3 = −2−2−6 = −2 113Since Av = −2v, the vector v is an eigenvector of A corresponding to theeigenvalue λ = −2 .2. [Exercise on page 41]−4 66uvAuAvxy From the figure, Au is not a multiple of u (because Au does not lie on the sameline as u). So, u is not an eigenvector of A.4.3. Eigenvalues, Eigenvectors, and Matrix Models 367 Since Av lies along the same line as v, it follows that Av is a multiple of v. Fromthe figure, we see that Av = 3v. So, v is an eigenvector of A and the associatedeigenvalue is λ = 3 .3. [Exercise on page 41]8−8uvwAuAvAwxy Since Au lies along the same line as u, it follows that Au is a multiple of u. Thisimplies that u is an eigenvector of A . Moreover, from the figure, we see thatAu = 4u. So, the eigenvalue associated with u is λ = 4 . From the figure, Av = −2v. So, v is an eigenvector of A and the associatedeigenvalue is λ = −2 . From the figure, Aw does not lie along the same line as w. So, w is not aneigenvector of A .4. [Exercise on page 42](a) True by definition.(b) False by definition.(c) True. Indeed,Au =[7 4−3 −1][−23]=[−14 + 126− 3]=[−23]= uSince Au is a multiple of u, it follows that u is an eigenvector of A.368 Chapter 4. Linear Algebra(d) False because Au =[−1114]and the vector[−1114]is not a multiple ofu =[−12].5. [Exercise on page 42]Finding the EigenvaluesThe key result to finding the eigenvalues is the following theorem. Recall thatthe identity matrix of size n× n is denoted by I.Theorem 4.3.1. Let A be a square matrix. Then a scalar λ is an eigenvalueof A if and only if the determinant of A− λI is 0; that is,det(A− λI) = 0It turns out that det(A − λI) is a polynomial in λ of degree n (if A is of sizen× n). This polynomial is called the characteristic polynomial of A. Andthe equation det(A− λI) = 0 is called the characteristic equation of A.So, to find the eigenvalues of A, we first need to find the characteristic polyno-mial. Then we solve the equation det(A− λI) = 0 for λ. The solutions are theeigenvalues of A.(a) Recall the 2× 2 identity matrix I =[1 00 1] First, we form the matrix A− λI.A− λI =[5 −3−10 6]− λ[1 00 1]=[5 −3−10 6]+[−λ 00 −λ]=[5− λ −3 + 0−10 + 0 6− λ]=[5− λ −3−10 6− λ] Now, we compute the determinant of A− λI.det(A− λI) = det[5− λ −3−10 6− λ]4.3. Eigenvalues, Eigenvectors, and Matrix Models 369= (5− λ)(6− λ)− (−3)(−10) det[a bc d]= ad− bc= 30− 5λ− 6λ+ λ2 − 30 Distribute= λ2 − 11λ = λ(λ− 11) Factor Set det(A − λI) = 0. Then λ(λ − 11) = 0. This implies that λ = 0 orλ = 11. So the eigenvalues of A are λ1 = 0 and λ2 = 11 .(b) First, we form the matrix A− λI:A− λI =[8 −116 −9]− λ[1 00 1]=[8− λ −116 −9− λ]The determinant of A− λI isdet(A− λI) = (8− λ)(−9− λ)− (−11)(6) det[a bc d]= ad− bc= −72− 8λ+ 9λ+ λ2 + 66 Distribute= λ2 + λ− 6 = (λ− 2)(λ+ 3) FactorSolving the equation (λ − 2)(λ + 3) = 0, we get λ = 2 or λ = −3. So, theeigenvalues of A are −3 and 2 .(c)det(A− λI) = det([13 4−16 −3]− λ[1 00 1])= det[13− λ 4−16 −3− λ]= (13− λ)(−3− λ)− (4)(−16) det[a bc d]= ad− bc= −39− 13λ+ 3λ+ λ2 + 64 Distribute= λ2 − 10λ+ 25 = (λ− 5)2 FactorSo, A has only one eigenvalue: λ = 5 .370 Chapter 4. Linear Algebra(d)det(A− λI) = det([22 7−1 14]− λ[1 00 1])= det[22− λ 7−1 14− λ]= (22− λ)(14− λ)− (7)(−1) det[a bc d]= ad− bc= 308− 22λ− 14λ+ λ2 + 7 Distribute= λ2 − 36λ+ 315To solve the equation λ2 − 36λ + 315 = 0, we will use the quadratic formula−b±√b2−4ac2a. The coefficients of the polynomial λ2−36λ+315 are a = 1, b = −36,and c = 315. Putting these numbers into the quadratic formula, we getλ =−(−36)±√(−36)2 − 4(1)(315)2(1)=36±√362=36± 62So,λ =36 + 62= 21 or λ =36− 62= 15(e)det(A− λI) = det([0.7 11.21 0.7]− λ[1 00 1])= det[0.7− λ 11.21 0.7− λ]= (0.7− λ)(0.7− λ)− (1)(1.21) det[a bc d]= ad− bc= 0.49− 0.7λ− 0.7λ+ λ2 − 1.21 Distribute= λ2 − 1.4λ− 0.724.3. Eigenvalues, Eigenvectors, and Matrix Models 371Using the quadratic formula, we getλ =−(−1.4)−√(−1.4)2 − 4(1)(−0.72)2(1)=1.4±√4.842=1.4± 2.22So,λ =1.4 + 2.22= 1.8 or λ =1.4− 2.22= −0.4(f) A =[−2.3 30.2 −1.2]det(A− λI) = det([−2.3 30.2 −1.2]− λ[1 00 1])= det[−2.3− λ 30.2 −1.2− λ]= (−2.3− λ)(−1.2− λ)− (3)(0.2) det[a bc d]= ad− bc= 2.76 + 2.3λ+ 1.2λ+ λ2 − 0.6 Distribute= λ2 + 3.5λ+ 2.16Solving the equation λ2 + 3.5λ+ 2.16 = 0 with the quadratic formula, we getλ =−(3.5)±√(3.5)2 − 4(1)(2.16)2(1)=−3.5±√3.612=−3.5± 1.92So,λ =−3.5 + 1.92= −0.8 or λ = −3.5− 1.92= −2.76. [Exercise on page 43]det(A− λI) = det([−1672112−14]− λI)372 Chapter 4. Linear Algebra= det[−16− λ 72112−14− λ]=(−16− λ)(−14− λ)−(72)(112)det[a bc d]= ad− bc=124+16λ+14λ+ λ2 − 724Distribute= λ2 +512λ− 14Multiplying both sides of the equation λ2 + 512λ− 14= 0 by 12, we get12λ2 + 5λ− 3 = 0Solving this latter equation with the quadratic formula, we getλ =−5±√52 − 4(12)(−3)2(12)=−5±√16924=−5± 1324So,λ =−5 + 1324=824=13or λ =−5− 1324=−1824= −347. [Exercise on page 43]Let k be a nonzero number. Let u be an eigenvector of A corresponding to theeigenvalue λ. Then,Au = λu By definitionkAu = kλu Multiply both sides by kA(ku) = λ(ku) Property: kAB = A(kB)The latter equation shows that ku is an eigenvector of A corresponding to λ.8. [Exercise on page 43]Modeling with Leslie MatricesAn age-structured population is a population divided into a certain numberof age classes. Leslie matrices are often used to model the size of age-structured4.3. Eigenvalues, Eigenvectors, and Matrix Models 373populations. If the population is divided into n age classes, the Leslie matrixwill be a square matrix of size n× n.Defining the Leslie Matrix. Suppose a population is divided into n ageclasses:0-year-olds, 1-year-olds, 2-year-olds, · · · , (n− 1)-year-oldsFor 0 ≤ i ≤ n− 1, let bi and si be defined as follows. bi is the average number of offspring produced by an age i individual. si is the fraction of individuals age i at time t that survive to time t+ 1.Then, the Leslie matrix, denoted L, is of the following form. If n = 2, L =[b0 b1s0 0] If n = 3, L = b0 b1 b2s0 0 00 s1 0 If n = 4, L =b0 b1 b2 b3s0 0 0 00 s1 0 00 0 s2 0Note that the entries of the Leslie matrix are 0, except possibly those in thefirst row and in the subdiagonal right below the diagonal.Calculating the future population using the Leslie Matrix. Consider anage-structure population divided into n age classes. Let L be the Leslie matrixof that population. For 0 ≤ i ≤ n − 1, let xi(t) be the number of individualsage i at time t. The population vector at time t is the n × 1 matrix X(t)whose ith entry is xi(t); that is,X(t) =x0(t)x1(t)...xn−1(t)374 Chapter 4. Linear AlgebraIf the population vector at time t is known, then the population vector at timet + 1 is obtained by multiplying the Leslie matrix by the population vector attime t; that is,X(t+ 1) = LX(t)For example, suppose the population vector at time 0 is given. Then thepopulation vectors in year 1, year 2, and year 3 are given byX(1) = LX(0), X(2) = LX(1), and X(3) = LX(2)(a) Let’s interpret the Leslie matrixL = 4 6 1.40.3 0 00 0.5 0Here b0 = 4, b1 = 6, b2 = 1.4, s0 = 0.3, and s1 = 0.5. Since L is a 3×3 matrix, it follows that the population is divided into threeage classes:0-year-olds, 1-year-olds, and 2-year-olds The number s0 = 0.3 means that 30% of the zero-year-olds survive until thenext census. The number s1 = 0.5 means that 50% of the one-year-olds survive until thenext census. The number b0 = 4 means that the average number of offspring producedby an age 0 individual is 4. The number b1 = 6 means that the average number of offspring producedby an age 1 individual is 6. The number b2 = 1.4 means that the average number of offspring producedby an age 2 individual is 1.4.(b) The population vector in year 1 isX(1) = LX(0) = 4 6 1.40.3 0 00 0.5 0 8002405504.3. Eigenvalues, Eigenvectors, and Matrix Models 375= 3200 + 1440 + 770240 + 0 + 00 + 120 + 0 = 5410240120The population vector in year 2 isX(2) = LX(1) = 4 6 1.40.3 0 00 0.5 0 5410240120= 21640 + 1440 + 1681623 + 0 + 00 + 120 + 0 = 2324816231209. [Exercise on page 43] From the sentence “The percentage of individuals age 0 at time t that survive totime t+ 1 is 15%”, we have s0 = 0.15. From the sentence “An individual age 0 produces, on average, 1.8 offspringyearly”, we have b0 = 1.8. From the sentence “An individual age 1 produces, on average, 0.7 offspringyearly”, we have b1 = 0.7.So, the Leslie matrix isL =[b0 b1s0 0]=[1.8 0.70.15 0]Now, we want to find the population size of each age group at time t = 3. From theproblem, X(0) =[2000800]. To find X(3), first we need to find X(1) and X(2).X(1) = LX(0) =[1.8 0.70.15 0][2000800]=[3600 + 560300 + 0]=[4160300]X(2) = LX(1) =[1.8 0.70.15 0][4160300]=[7488 + 210624 + 0]=[7698624]X(3) = LX(2) =[1.8 0.70.15 0][7698624]=[13856.4 + 436.81154.7 + 0]=[14293.21154.7]376 Chapter 4. Linear AlgebraThus, at time 3, the number of individuals age 0 is approximately 14293, and thenumber of individuals age 1 is approximately 1155.4.4 Eigenvalues, Eigenvectors, and Matrix Models (con-tinued)1. [Exercise on page 44]Finding the EigenvectorsLet A be a square matrix, and let λ be an eigenvalue of A. By definition, anonzero vector X is an eigenvector corresponding to λ ifAX = λXSo, to find the eigenvectors corresponding to λ, we need to solve the equationAX = λX for X.If A =[a bc d]is a 2 × 2 matrix, X =[xy], solving the equation AX = λXamounts to solving the linear systemax + by = λxcx + dy = λywhich is equivalent to the system(a− λ)x + by = 0cx + (d− λ)y = 0One can use the Gauss-Jordan method to solve the latter system and get theeigenvectors corresponding to λ.An eigenvalue of the matrix[3 −3−2 2]is λ = 5. We want to find the correspondingeigenvectors. Let X =[xy]be the vector of unknowns.4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 377 First, we form the system AX = λX:3x − 3y = 5x−2x + 2y = 5ySubtracting 5x from both sides of the first equation, and subtracting 5y fromboth sides of the second equation, we get−2x − 3y = 0−2x − 3y = 0 The augmented matrix of the latter system is[−2 −3 0−2 −3 0]Our goal is to use elementary row operations to transform the augmented matrixinto a matrix of the form[1 0 ∗0 1 ∗]or[1 ∗ ∗0 0 ∗][−2 −3 0−2 −3 0]R2 −R1 → R2[−2 −3 00 0 0]−12R1 → R1[1 3200 0 0]The corresponding system is x + 32y = 0. Let y = s. Then x + 32s = 0 or x = −32s.So, the general solution of the system AX = 5X isX =[xy]=[−32ss]= s[−321]where s is a real number. Since an eigenvector is a nonzero vector by definition, theeigenvectors of A corresponding to λ = 5 are of the forms[−321]where s is a nonzero real number378 Chapter 4. Linear Algebra2. Let A =[8 5−12 −9]. [Exercise on page 44]To find the eigenvectors of A =[8 5−12 −9]corresponding to λ = −4, we need tosolve the system8x + 5y = −4x−12x − 9y = −4yAdding 4x to both sides of the first equation, and 4y to both sides of the second, weget12x + 5y = 0−12x − 5y = 0Transforming the augmented matrix, we have[12 5 0−12 −5 0]R2 +R1 → R2[12 5 00 0 0]112R1 → R1[1 51200 0 0]The corresponding system is x + 512y = 0. Let y = s. Then x = − 512s, and theeigenvectors corresponding to λ = −4 are of the form[− 512ss]= s[− 5121]s 6= 03. Find the eigenvalues and eigenvectors of each matrix. [Exercise on page 44](a) Finding the eigenvalues of A.det(A− λI) = det([6 4−6 −5]− λ[1 00 1])I =[1 00 1]= det[6− λ 4−6 −5− λ]= (6− λ)(−5− λ)− (4)(−6) det[a bc d]= ad− bc= −30− 6λ+ 5λ+ λ2 + 24 Distribute= λ2 − λ− 6 = (λ+ 2)(λ− 3) Factor4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 379Solving the equation (λ + 2)(λ − 3) = 0, we find λ = −2 or λ = 3. So, theeigenvalues of A are −2 and 3 .Finding the eigenvectors. Finding the eigenvectors corresponding to λ = −2. We need to solve thesystem AX = −2X; that is, the system6x + 4y = −2x−6x − 5y = −2yAdding 2x to both sides of the first equation, and 2y to both sides of thesecond equation, we get8x + 4y = 0−6x − 3y = 0Transforming the augmented matrix, we have[8 4 0−6 −3 0]14R1 → R1−13R2 → R2[2 1 02 1 0]R2 −R1 → R2[2 1 00 0 0]12R1 → R1[1 1200 0 0]The corresponding system is x + 12y = 0. Let y = s. Then x = −12s, andthe eigenvectors corresponding to λ = −2 are of the form[xy]=[−12ss]= s[−121]s 6= 0 Finding the eigenvectors corresponding to λ = 3. We need to solve thesystem AX = 3X; that is, the system6x + 4y = 3x−6x − 5y = 3ySubtracting 3x from both sides of the first equation, and 3y from both sides380 Chapter 4. Linear Algebraof the second equation, we get3x + 4y = 0−6x − 8y = 0Transforming the augmented matrix, we have[3 4 0−6 −8 0]R2 + 2R1 → R2[3 4 00 0 0]13R1 → R1[1 4300 0 0]The corresponding system is x + 43y = 0. Let y = s. Then x = −43s, andthe eigenvectors corresponding to λ = 3 are of the form[xy]=[−43ss]= s[−431]s 6= 0(b) Finding the eigenvalues.det(A− λI) = det([−12 −323 8]− λ[1 00 1])= det[−12− λ −323 8− λ]= (−12− λ)(8− λ)− (−32)(3) det[a bc d]= ad− bc= −96 + 12λ− 8λ+ λ2 + 96 Distribute= λ2 + 4λ = λ(λ+ 4) Factor out λSolving the equation λ(λ+ 4) = 0, we find λ = 0 or λ = −4. So, the eigenvaluesof A are −4 and 0 .Finding the eigenvectors. Finding the eigenvectors corresponding to λ = −4. We need to solve the4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 381system AX = −4X; that is, the system−12x − 32y = −4x3x + 8y = −4yAdding 4x to both sides of the first equation, and 4y to both sides of thesecond equation, we get−8x − 32y = 03x + 12y = 0Transforming the augmented matrix, we have[−8 −32 03 12 0]−18R1 → R113R2 → R2[1 4 01 4 0]R2 −R1 → R2[1 4 00 0 0]The corresponding system is x + 4y = 0. Let y = s. Then x = −4s, andthe eigenvectors corresponding to λ = −4 are of the form[xy]=[−4ss]= s[−41]s 6= 0 Finding the eigenvectors corresponding to λ = 0. We need to solve thesystem AX = 0X; that is, the system−12x − 32y = 0x = 03x + 8y = 0y = 0Transforming the augmented matrix, we have[−12 −32 03 8 0]−14R1 → R1[3 8 03 8 0]R2 −R1 → R2[3 8 00 0 0]13R1 → R1[1 8300 0 0]The corresponding system is x + 83y = 0. Let y = s. Then x = −83s, and382 Chapter 4. Linear Algebrathe eigenvectors corresponding to λ = 0 are of the form[xy]=[−83ss]= s[−831]s 6= 0(c) Finding the eigenvalues.det(A− λI) = det([3 2−2 7]− λ[1 00 1])= det[3− λ 2−2 7− λ]= (3− λ)(7− λ)− (2)(−2) det[a bc d]= ad− bc= 21− 3λ− 7λ+ λ2 + 4 Distribute= λ2 − 10λ+ 25 = (λ− 5)2 Factor out λSolving the equation (λ− 5)2 = 0, we find λ = 5 .Finding the eigenvectors corresponding to λ = 5. We need to solve the systemAX = 5X; that is, the system3x + 2y = 5x−2x + 7y = 5ySubtracting 5x from both sides of the first equation, and 5y from both sides ofthe second equation, we get−2x + 2y = 0−2x + 2y = 0Transforming the augmented matrix, we have[−2 2 0−2 2 0]−12R1 → R1[1 −1 0−2 2 0]R2 + 2R1 → R2[1 −1 00 0 0]The corresponding system is x − y = 0. Let y = s. Then x = s, and the4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 383eigenvectors corresponding to λ = 5 are of the form[xy]=[ss]= s[11]s 6= 0(d) Finding the eigenvalues.det(A− λI) = det([1.2 5.60.6 −0.4]− λ[1 00 1])= det[1.2− λ 5.60.6 −0.4− λ]= (1.2− λ)(−0.4− λ)− (5.6)(0.6) det[a bc d]= ad− bc= −0.48− 1.2λ+ 0.4λ+ λ2 − 3.36 Distribute= λ2 − 0.8λ− 3.84Using the quadratic formula, we getλ =−(−0.8)±√(−0.8)2 − 4(1)(−3.84)2(1)=0.8±√162=0.8± 42So,λ =0.8 + 42= 2.4 or0.8− 42= −1.6Finding the eigenvectors. Finding the eigenvectors corresponding to λ = 2.4. We need to solve thesystem AX = 2.4X; that is, the system1.2x + 5.6y = 2.4x0.6x − 0.4y = 2.4ySubtracting 2.4x from both sides of the first equation, and 2.4y from bothsides of the second equation, we get−1.2x + 5.6y = 00.6x − 2.8y = 0384 Chapter 4. Linear AlgebraTransforming the augmented matrix, we have[−1.2 5.6 00.6 −2.8 0]−10R1 → R110R2 → R2[12 −56 06 −28 0]12R1 → R1[6 −28 06 −28 0]R2 −R1 → R2[6 −28 00 0 0]16R1 → R1[1 −14300 0 0]The corresponding system is x − 143y = 0. Let y = s. Then x = 143s, andthe eigenvectors corresponding to λ = 2.4 are of the form[xy]=[143ss]= s[1431]s 6= 0 Finding the eigenvectors corresponding to λ = −1.6. We need to solve thesystem AX = −1.6X; that is, the system1.2x + 5.6y = −1.6x0.6x − 0.4y = −1.6yAdding 1.6x to both sides of the first equation, and 1.6y to both sides ofthe second equation, we get2.8x + 5.6y = 00.6x + 1.2y = 0Transforming the augmented matrix, we have[2.8 5.6 00.6 1.2 0]10R1 → R110R2 → R2[28 56 06 12 0]128R1 → R116R2 → R2[1 2 01 2 0]R2 −R1 → R2[1 2 00 0 0]4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 385The corresponding system is x− 2y = 0. Let y = s. Then x = 2s, and theeigenvectors corresponding to λ = −1.6 are of the form[xy]=[2ss]= s[21]s 6= 04. [Exercise on page 44]Since u =[33]is an eigenvector of A corresponding to λ = 2, it follows thatAu = 2u. Let’s compute A2u.(a)A2u = A(Au) = A(2u)= 2Au Property: A(kB) = kAB= 2(2u) Au = 2u= (2)2uComputing A3u, we findA3u = A(A2u) = A(22u) = 22A(u) = 22(2u) = 23uComputing A4, A5, A6, A7, A8, A9, and A10 is the same way, we will getA10 = 210u = 210[33]=[3 · 2103 · 210]=[30723072](b) Pursuing the computations from part (a), we haveAnu = 2nu = 2n[33]=[3 · 2n3 · 2n]In general, we have the following fact.Computing An(u) When u is an Eigenvector of ALet A be a square matrix. Let λ be an eigenvalue of A, and let u be an386 Chapter 4. Linear Algebraeigenvector of A corresponding to λ. Then,Au = λu DefinitionA2u = A(Au) = A(λu) = λAu = λ(λu) = λ2uA3u = A(A2u) = A(λ2u) = λ2Au = λ2(λu) = λ3u...So, we can deduce that for every n ≥ 1,An(u) = λnu5. [Exercise on page 44](a)au1 + bu2 = va[23]+ b[15]=[−65][2a3a]+[b5b]=[−65][2a+ b3a+ 5b]=[−65]This gives the system2a + b = −63a + 5b = 5Transforming the augmented matrix, we have[2 1 −63 5 5]12R1 → R1[1 12−33 5 5]R2 − 3R1 → R2[1 12−30 7214]27R2 → R2[1 12−30 1 4]4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 387R1 − 12R2 → R1[1 0 −50 1 4]So, a = −5 and b = 4 . Thus, −5u1 + 4u2 = v.(b) Since u1 is an eigenvector corresponding to λ1 = −3, Au1 = −3u1. Likewise,Au2 = 4u2. Let n ≥ 1 be an integer. ThenAnv = An(au1 + bu2) v = au1 + bu2= An(au1) + An(bu2) Property: A(B + C) = AB + AC= aAn(u1) + bAn(u2) Property: A(kB) = kAB= a((−3)nu1) + b((4)nu2) An(u1) = (−3)nu1, An(u2) = (4)nu2= (−3)n(−5)[23]+ (4)n(4)[15]Substitute a, b, u1, and u2=[−10(−3)n−15(−3)n]+[4n+15(4)n+1]anam = an+m=[−10(−3)n + 4n+1−15(−3)n + 5(4)n+1]6. [Exercise on page 45](a) Finding the eigenvalues of A.det(A− λI) = det([1.6 0.60.6 0]− λ[1 00 1])= det[1.6− λ 0.60.6 −λ]= (1.6− λ)(−λ)− (0.6)(0.6) det[a bc d]= ad− bc= λ2 − 1.6λ− 0.36 DistributeUsing the quadratic formula, we getλ =−(−1.6)±√(−1.6)2 − 4(1)(−0.36)2(1)388 Chapter 4. Linear Algebra=1.6±√42=1.6± 22So,λ =1.6 + 22= 1.8 or1.6− 22= −0.2Thus, the eigenvalues of A are λ1 = 1.8 and λ2 = −0.2 .Finding the eigenvectors. Finding the eigenvectors corresponding to λ1 = 1.8. We need to solve thesystem AX = 1.8X; that is, the system1.6x + 0.6y = 1.8x0.6x + 0y = 1.8ySubtracting 1.8x from both sides of the first equation, and 1.8y from bothsides of the second equation, we get−0.2x + 0.6y = 00.6x − 1.8y = 0Transforming the augmented matrix, we have[−0.2 0.6 00.6 −1.8 0]−10R1 → R110R2 → R2[2 −6 06 −18 0]12R1 → R1[1 −3 06 −18 0]R2 − 6R1 → R2[1 −3 00 0 0]The corresponding system is x− 3y = 0. Let y = s. Then x = 3s, and theeigenvectors corresponding to λ = 1.8 are of the form[xy]=[3ss]= s[31]s 6= 0 Finding the eigenvectors corresponding to λ2 = −0.2. We need to solve thesystem AX = −0.2X; that is, the system1.6x + 0.6y = −0.2x0.6x + 0y = −0.2y4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 389Adding 0.2x to both sides of the first equation, and 0.2y to both sides ofthe second equation, we get1.8x + 0.6y = 00.6x + 0.2y = 0Transforming the augmented matrix, we have[1.8 0.6 00.6 0.2 0]10R1 → R110R2 → R2[18 6 06 2 0]16R1 → R112R2 → R2[3 1 03 1 0]R2 −R1 → R2[3 1 00 0 0]13R1 → R1[1 1300 0 0]The corresponding system is x + 13y = 0. Let y = s. Then x = −13s, andthe eigenvectors corresponding to λ = −0.2 are of the form[xy]=[−13ss]= s[−131]s 6= 0(b) The population vector after 1 year isX(1) = LX(0) =[1.6 0.60.6 0][1500800]=[2400 + 480900]=[2880900](c) To find a formula for the population vector after n years, one can go throughthe following steps.Step 1. Choose one eigenvector for each eigenvalue. The idea is tochoose values that are easy to work with. Good choices include smallintegers. (Avoid fractions.) From part (a), the eigenvectors corresponding to λ1 = 1.8 are of390 Chapter 4. Linear Algebrathe forms[31], s 6= 0We choose the eigenvector u1 =[31]. From part (a), the eigenvectors corresponding to λ2 = −0.2 areof the forms[−131], s 6= 0The eigenvector[−131]is not a good choice because one of theentries is a fraction. To avoid fractions, we can choose the vectorcorresponding to s = 3; that is, the eigenvector u2 =[−13].Step 2. Find two numbers a and b such that au1 + bu2 = X(0).au1 + bu2 = X(0)a[31]+ b[−13]=[1500800]Substitute u1, u2, and X(0)[3aa]+[−b3b]=[1500800][3a− ba+ 3b]=[1500800]This gives the system3a − b = 1500a + 3b = 800Transforming the augmented matrix, we have[3 −1 15001 3 800]R1 ↔ R2[1 3 8003 −1 1500]R2 − 3R1 → R2[1 3 8000 −10 −900]4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 391− 110R2 → R2[1 3 8000 1 90]R1 − 3R2 → R1[1 0 5300 1 90]So a = 530, b = 90, and 530u1 + 90u2 = X(0).Step 3. Use the result from step 2, as well as the properties ofeigenvalues and eigenvectors, to find a formula for the pop-ulation vector X(n).X(1) = LX(0)X(2) = LX(1) = L [LX(0)] = L2X(0)X(3) = LX(2) = L[L2X(0)]= L3X(0)...X(n) = LnX(0)= Ln (530u1 + 90u2) X(0) = 530u1 + 90u2= Ln(530u1) + Ln(90u2) L(B + C) = LB + LC= 530Ln(u1) + 90Ln(u2) L(kB) = kLB= 530 (λn1u1) + 90 (λn2u2) Using the fact above= 530((1.8)n[31])+ 90((−0.2)n[−13])Since limn→∞(−0.2)n = 0, the first term dominates over the second as n→∞. Sofor large n,X(n) ≈ 530(1.8)n[31]Remarkable Fact. The initial population is not an eigenvector, but the pop-ulation over time approaches a multiple of an eigenvector.7. [Exercise on page 45](a)  From the sentence “The percentage of individuals age 0 at time t that surviveto time t+ 1 is 40%”, we have s0 = 0.4. From the sentence “An individual age 0 produces, on average, 1.1 offspring392 Chapter 4. Linear Algebrayearly”, we have b0 = 1.1. From the sentence “An individual age 1 produces, on average, 2 offspringyearly”, we have b1 = 2.So, the Leslie matrix isL =[b0 b1s0 0]=[1.1 20.4 0]Now, we want to find the number of individuals in each age class at time 3.From the problem, X(0) =[21001260]. To find X(3), first we need to find X(1)and X(2).X(1) = LX(0) =[1.1 20.4 0][21001260]=[2310 + 2520840 + 0]=[4830840]X(2) = LX(1) =[1.1 20.4 0][4830840]=[5313 + 16801932 + 0]=[69931932]X(3) = LX(2) =[1.1 20.4 0][69931932]=[7692.3 + 38642797.2 + 0]=[11556.32797.2]Thus, at time 2, the number of individuals age 0 is approximately 11556 ; The number of individuals age 1 is approximately 2797 .(b) Finding the eigenvalues of L.det(L− λI) = det([1.1 20.4 0]− λ[1 00 1])I =[1 00 1]= det[1.1− λ 20.4 0− λ]4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 393= (1.1− λ)(−λ)− 0.8 det[a bc d]= ad− bc= λ2 − 1.1λ− 0.8 DistributeUsing the quadratic formula−b±√b2 − 4ac2a, we getλ =−(−1.1)±√(−1.1)2 − 4(1)(−0.8)2(1)=1.1±√4.412=1.1± 2.12So,λ =1.1 + 2.12= 1.6 or1.1− 2.12= −0.5Thus, the eigenvalues of A are λ1 = 1.6 and λ2 = −0.5 .Finding the eigenvectors of L. Finding the eigenvectors corresponding to λ1 = 1.6. We need to solve thesystem AX = 1.6X; that is, the system1.1x + 2y = 1.6x0.4x + 0y = 1.6ySubtracting 1.6x from both sides of the first equation, and 1.6y from bothsides of the second equation, we get−0.5x + 2y = 00.4x − 1.6y = 0Transforming the augmented matrix, we have[−0.5 2 00.4 −1.6 0]−10R1 → R110R2 → R2[5 −20 04 −16 0]15R1 → R1[1 −4 04 −16 0]R2 − 4R1 → R2[1 −4 00 0 0]The corresponding system is x− 4y = 0. Let y = s. Then x = 4s, and the394 Chapter 4. Linear Algebraeigenvectors corresponding to λ = 1.6 are of the form[xy]=[4ss]= s[41]s 6= 0 Finding the eigenvectors corresponding to λ2 = −0.5. We need to solve thesystem AX = −0.5X; that is, the system1.1x + 2y = −0.5x0.4x + 0y = −0.5yAdding 0.5x to both sides of the first equation, and 0.5y to both sides ofthe second equation, we get1.6x + 2y = 00.4x + 0.5y = 0Transforming the augmented matrix, we have[1.6 2 00.4 0.5 0]10R1 → R110R2 → R2[16 20 04 5 0]14R1 → R1[4 5 04 5 0]R2 −R1 → R2[4 5 00 0 0]14R1 → R1[1 5400 0 0]The corresponding system is x + 54y = 0. Let y = s. Then x = −54s, andthe eigenvectors corresponding to λ = −0.5 are of the form[xy]=[−54ss]= s[−541]s 6= 0(c) To find a formula for the population vector after n years, one can go throughthe following three steps.Step 1. Choose one eigenvector for each eigenvalue. The idea is to choosevalues that are easy to work with. Good choices include small integers. (Avoid4.4. Eigenvalues, Eigenvectors, and Matrix Models (continued) 395fractions.) From part (b), the eigenvectors corresponding to λ1 = 1.6 are of the forms[41], s 6= 0. We choose the eigenvector u1 =[41]. From part (b), the eigenvectors corresponding to λ2 = −0.5 are of the forms[−541], s 6= 0. The eigenvector[−541]is not a good choice because oneof the entries is a fraction. To avoid fractions, we can choose the vectorcorresponding to s = 4; that is, the eigenvector u2 =[−54].Step 2. Find two numbers a and b such that au1 + bu2 = X(0).au1 + bu2 = X(0)a[41]+ b[−54]=[21001260]Substitute u1, u2, and X(0)[4aa]+[−5b4b]=[21001260][4a− 5ba+ 4b]=[21001260]This gives the system4a − 5b = 2100a + 4b = 1260Transforming the augmented matrix, we have[4 −5 21001 4 1260]R1 ↔ R2[1 4 12604 −5 2100]R2 − 4R1 → R2[1 4 12600 −21 −2940]− 121R2 → R2[1 4 12600 1 140]R1 − 4R2 → R1[1 0 7000 1 140]So a = 700, b = 140, and 700u1 + 140u2 = X(0).Step 3. Use the result from step 2, as well as the properties of eigen-396 Chapter 4. Linear Algebravalues and eigenvectors, to find a formula for the population vectorX(n).X(1) = LX(0)X(2) = LX(1) = L [LX(0)] = L2X(0)X(3) = LX(2) = L[L2X(0)]= L3X(0)...X(n) = LnX(0)= Ln (700u1 + 140u2) X(0) = 700u1 + 140u2= Ln(700u1) + Ln(140u2) L(B + C) = LB + LC= 700Ln(u1) + 140Ln(u2) L(kB) = kLB= 700 (λn1u1) + 140 (λn2u2)If λ is an eigenvalue of A, and uis an eigenvector correpondingto λ, Anu = λnu for all n ≥ 1= 700((1.6)n[41])+ 140((−0.5)n[−54])Since limn→∞(−0.5)n = 0, the first term dominates over the second as n→∞. Sofor large n,X(n) ≈ 700(1.6)n[41](d) From part (c), we haveX(15) ≈ 700(1.6)15[41]≈ 807045.053[41]≈[3, 228, 180807, 045](e) From part (b), the eigenvalues are λ1 = 1.6 and λ2 = −0.5. The largest eigen-value is λ1 = 1.6 and the corresponding eigenvector is u1 =[xy](from part (c))with x = 4 and y = 1. So, the stable distribution isxx+ y=44 + 1=45andyx+ y=14 + 1=15Chapter5Infinite Series5.1 Geometric Series1. [Exercise on page 47]Geometric Series A sequence is an ordered list of numbersa1, a2, a3, · · · , an, · · ·The summation of the terms of a sequence,a1 + a2 + a3 + · · ·+ an + · · ·is called an infinite series (or just a series), and is denoted by∑∞n=1 an.or just∑an. So,∞∑n=1an = a1 + a2 + a3 + a4 + · · ·For example,∞∑n=11n=11+12+13+14+ · · · If∑∞k=1 ak is a series, the sum of the first n terms, denoted Sn, is givenbySn = a1 + a2 + a3 + · · ·+ an =n∑k=1akThe sum Sn is called nth partial sum. The series∑∞k=1 ak is said to397398 Chapter 5. Infinite Seriesbe convergent if limn→∞Sn is a real number, and divergent otherwise. Iflimn→∞Sn = s, where s is a real number, we say that s is the sum of theseries, and we write∞∑n=1an = s A geometric series is a series of the form∞∑n=1arn−1 = a+ ar + ar2 + ar3 + · · ·+ arn−1 + · · ·where a 6= 0 and r are real numbers. The number r is referred to as thecommon ratio of the series. Note that in a geometric series each termdivided by the previous term is r. The following theorem gives a conditionthat guarantees the convergence of the geometric series∑∞n=1 arn−1.Theorem 5.1.1. Let∑∞k=1 ark−1 be a geometric series.(i) The nth partial sum of the series isSn =a(1− rn)1− r(ii) If −1 < r < 1 (or |r| < 1), then the geometric series converges toa1−r . That is,If |r| < 1, then∞∑n=1arn−1 =a1− r(iii) If |r| ≥ 1, the geometric series is divergent.Note that the number a is the first term of the geometric series∑∞n=1 arn−1.(a) For the series∞∑n=1(13)n−1, we have a = 1 and r = 13. Since r belongs to the openinterval −1 < r < 1, it follows from Theorem 5.1.1 that the series is convergent,and its sum is ∞∑n=1(13)n−1=a1− r =11− 13=123=325.1. Geometric Series 399(b) The series∞∑n=15(−34)n−1is a geometric series with a = 5 and r = −34= −0.75.Since −1 < r < 1, the series is convergent, and its sum is∞∑n=15(−34)n−1=a1− r =51− (−34) ==51 + 34=574= 5(47)abc=acb=207(c) The series∞∑n=14(32)n−1is a geometric series with a = 4 and r = 32= 1.5. Since|r| ≥ 1, the series is divergent .(d) The common ration of the geometric series∞∑n=19(67)nis r = 67. The first termisa = 9(67)1 Substitute n with 1 into thegeneral term an = 9(67)n=547Since |r| < 1, the series is convergent, and its sum is∞∑n=19(67)n=a1− r =5471− 67=547171− ab=b− ab=547(71) abcd=adbc= 54(e) First, we need to write the series∞∑n=13(5n)4n−1in the form∑∞n=1 arn−1.∞∑n=13(5n)4n−1=∞∑n=13(5)(5)n−14n−1bk = bbk−1=∞∑n=115(54)n−1bkck=(bc)k400 Chapter 5. Infinite SeriesSo, this is a geometric series with a = 15 and r = 54. Since |r| ≥ 1, the series isdivergent .(f) First, we need to write the series∞∑n=172n81−n in the form∑∞n=1 arn−1.∞∑n=172n81−n =∞∑n=1(72)n(8)(8)−n =∞∑n=1(49)n(8)18n=∞∑n=18(498)n=∞∑n=18(498)(498)n−1=∞∑n=149(498)n−1So, this is a geometric series with a = 49 and r = 498. Since |r| ≥ 1, it followsthat the series is divergent .(g) Rewriting the series∞∑n=011n+2(−10)n+1 , we have∞∑n=011n+2(−10)n+1 =∞∑n=0(11)n(11)2(−10)n(−10) bk+m = bkbm=∞∑n=0−12110(−1110)nbkck=(bc)kThis is a geometric series with common ratio r = −1110. Since |r| ≥ 1, the seriesis divergent .(h) Rewriting the series∞∑n=0(−4)n+352n+1, we have∞∑n=0(−4)n+352n+1=∞∑n=0(−4)n(−4)3(52n)(5)bk+m = bkbm=∞∑n=0(−4)n(−4)3[(52)n] (5)bkm =(bk)m=∞∑n=0(−4)n(−64)(25)n(5)=∞∑n=0−645(− 425)nbkck=(bc)kThis is a geometric series with r = − 425. The first term, a, is obtained bysubstituting n with 0 (we substitute n with 0 because the series starts at n = 0);that is,a = −645(− 425)0= −645(1) = −6455.1. Geometric Series 401Since |r| < 1, the series is convergent, and its sum is∞∑n=0(−4)n+352n+1=a1− r =−6451− (− 425) = −6451 + 425=−6452925= −645(2529)= −32029(i)Sum of Two SeriesLet∑an and∑bn be convergent series, and let c be a constant. Thenthe series∑can,∑(an+ bn), and∑(an− bn) are convergent, and we have∑can = c∑an∑(an + bn) =∑an +∑bn∑(an − bn) =∑an −∑bn• The series ∑( 34n)=∑3(14)n=∑34(14)n−1is a geometric series witha = 34and r = 14. Since r belongs to the interval −1 < r < 1, the series isconvergent, and∞∑n=1=a1− r =341− 14=3434= 1• Regarding the series ∑ 67(2n+3), we have∞∑n=167 (2n+3)=∞∑n=167(12)4(12)n−1=∞∑n=167(116)(12)n−1=∞∑n=1356(12)n−1=3561− 12=35612=328So, the given series is convergent and we have∞∑n=1(34n+67 (2n+3))=∞∑n=1(34n)+∞∑n=167 (2n+3)= 1 +328=3128(j)∞∑n=13 + 4n9n=∞∑n=1(39n+4n9n)=∞∑n=139n+∑ 4n9n402 Chapter 5. Infinite Series=∞∑n=139(19)n−1+∞∑n=149(49)n−1=∞∑n=113(19)n−1+∞∑n=149(49)n−1=131− 19+491− 49=1389+4959If |r| < 1,∞∑n=1arn−1 =a1− r=(13)(98)+(49)(95)=38+45=4740So, the series∑∞n=13+4n9nconverges to4740.2. [Exercise on page 47](a)23+49+827+1681+ · · · = 23+(23)2+(23)3+(23)4+ · · ·=231− 23If |r| < 1,∞∑n=1arn−1 =a1− r=2313=23(31)= 2So, the series converges to 2.(b)− 52+54− 58+516− 532+ · · ·= 5[(−12)1+(−12)2+(−12)3+(−12)4+(−12)5+ · · ·]= 5[−121− (−12)] If |r| < 1, ∞∑n=1arn−1 =a1− r= 5[ −121 + 12]= 5[−1232]= 5(−13)= −53So, the series converges to −53.5.2. Taylor Series 4033. [Exercise on page 47]1.999999 . . . = 1 + 0.999999 . . .= 1 + 0.9 + 0.09 + 0.009 + 0.0009 + · · ·= 1 +910+9100+91000+910000+ · · ·= 1 +910+9102+9103+9104+ · · ·= 1 + 9[(110)1+(110)2+(110)3+(110)4+ · · ·]= 1 + 9[ 1101− 110]If |r| < 1,∞∑n=1arn−1 =a1− r= 1 + 9( 110910)= 1 + 9(19)= 1 + 1 = 2So, the number 2 can be written 1.999999 . . .5.2 Taylor Series1. [Exercise on page 48]Taylor SeriesLet y = f(x) be a function, and let a be a number. Let n ≥ 0 be an integer. We write f (n)(a) for the nth derivative of f(x)evaluated at a.– If n = 0, f (0)(a) = f(a). This is just f(x) evaluated at a.– If n = 1, f (1)(a) = f ′(a). This is the first derivative of f(x) at a.– If n = 2, f (2)(a) = f ′′(a). This is the second derivative of f(x)evaluated at a.– Etc. Let n ≥ 1 be an integer. The factorial of n, denoted n!, is defined byn! = n(n− 1)(n− 2) · · · (3)(2)(1)404 Chapter 5. Infinite SeriesBy convention, 0! = 1. For example, 4! = 4(3)(2)(1) = 24. If f(x) has the derivative of all orders at a, the Taylor series of f at a(or about a or centered at a) is the series∞∑n=0f (n)(a)n!(x−a)n = f(a)+f ′(a)(x−a)+f′′(a)2!(x−a)2+f′′′(a)3!(x−a)3+· · ·If the Taylor series converges to f(x) for every x in an open interval(a−R, a+R), we writef(x) =∞∑n=0f (n)(a)n!(x− a)n for |x− a| < R– If the Taylor series is convergent for |x − a| < R, and divergent for|x− a| > R, the radius of convergence is defined to be R.– If the Taylor series is convergent for every real number x, the radiusof convergence is defined to be R =∞.(a) To find the first four nonzero terms of the Taylor series of f(x) = e−4x at a = 0,we first need to find the successive derivatives of f(x). Recall the chain rule forthe natural exponential function.ddx[eg(x)]= eg(x)ddx[g(x)]f(x) = e−4x f(0) = 1f ′(x) = −4e−4x f ′(0) = −4f ′′(x) = 16e−4x f ′′(0) = 16f ′′′(x) = −64e−4x f ′′′(0) = −64So, the first four nonzero terms of the Taylor series of f(x) at a = 0 aref(0) + f ′(0)(x− 0) + f′′(0)2!(x− 0)2 + f′′′(0)3!(x− 0)3= 1− 4(x− 0) + 162!(x− 0)2 + −643!(x− 0)3= 1− 4x+ 162(1)(x)2 +−643(2)(1)(x)35.2. Taylor Series 405= 1− 4x+ 8x2 − 323x3(b) Recall the following differentiation rules:ddx[lnx] =1xandddx[xn] = nxn−1We calculate the successive derivatives of f(x) = sinx at a = pi, and we willstop when we get four nonzero terms.f(x) = ln x f(1) = ln 1 = 0f ′(x) =ddx[lnx] =1x= x−1 f ′(1) = 1f ′′(x) =ddx[x−1]= (−1)x−2 f ′′(1) = −1f ′′′(x) =ddx[(−1)x−2] = (−1)(−2)x−3 f ′′′(1) = 2f (4)(x) =ddx[(−1)(−2)x−3] = (−1)(−2)(−3)x−4 f (4)(1) = −6So, the first four nonzero terms of the Taylor series of f(x) = ln x at a = 1 aref(1) + f ′(1)(x− 1) + f′′(1)2!(x− 1)2 + f′′′(1)3!(x− 1)3 + f(4)(1)4!(x− 1)4= 0 + 1(x− 1) + −12!(x− 1)2 + 23!(x− 1)3 + −64!(x− 1)4= (x− 1) + −12(1)(x− 1)2 + 23(2)(1)(x− 1)3 + −64(3)(2)(1)(x− 1)4= (x− 1)− 12(x− 1)2 + 13(x− 1)3 − 14(x− 1)4(c) Recall the following differentiation rules:ddx[sinx] = cos x andddx[cosx] = − sinxWe calculate the successive derivatives of f(x) = sinx at a = pi, and we willstop when we get four nonzero terms.f(x) = sin x f(pi) = sin(pi) = 0f ′(x) = cos x f ′(pi) = cos(pi) = −1406 Chapter 5. Infinite Seriesf ′′(x) = − sinx f ′′(pi) = − sin(pi) = 0f ′′′(x) = − cosx f ′′′(pi) = − cos(pi) = −(−1) = 1f (4)(x) = −(− sinx) = sinx f (4)(pi) = sin(pi) = 0f (5)(x) = cos x f (5)(pi) = cos(pi) = −1f (6)(x) = − sinx f (6)(pi) = − sin(pi) = 0f (7)(x) = − cosx f (7)(pi) = − cos(pi) = −(−1) = 1So, the first four nonzero terms of the Taylor series of f(x) = sin x at a = pi aregiven byf(pi) + f ′(pi)(x− pi) + f′′(pi)2!(x− pi)2 + f′′′(pi)3!(x− pi)3 + f(4)(pi)4!(x− pi)4+f (5)(pi)5!(x− pi)5 + f(6)(pi)6!(x− pi)6 + f(7)(pi)7!(x− pi)7= 0− (x− pi) + 0 + 13!(x− pi)3 + 0− 15!(x− pi)5 + 0 + 17!(x− pi)7= −(x− pi) + 16(x− pi)3 − 1120(x− pi)5 + 15040(x− pi)7(d) Recall the following differentiation rules:ddx[arcsin(g(x))] =ddx[g(x)]√1− [g(x)]2 andddx[[g(x)]n] = n[g(x)]n−1ddx[g(x)]Calculating the 0th and 1st derivatives of f(x) at a = 14, we havef(x) = arcsin(2x) f(14)= arcsin(12)=pi6f ′(x) =ddx[2x]√1− (2x)2 =2√1− 4x2 = 2(1− 4x2)−12 f ′(14)=2√1− 14=4√3Finding the second derivative.f ′′(x) =ddx[2(1− 4x2)− 12]= 2(1− 4x2)− 12−1 ddx[1− 4x2]= 2(−12)(1− 4x2)− 32 (−8x) = 8x(1− 4x2)− 325.2. Taylor Series 407Evaluating this at 14, we getf ′′(14)= 8(14)(1− 4(14)2)− 32= 2(1− 14)− 32= 2(34)− 32= 21(34) 32= 21(3)32(4)32= 213(3)124(4)12= 24(4)123(3)12= 283√3=163√3Finding the third derivative.f ′′′(x) =ddx[8x(1− 4x2)− 32]=ddx[8x](1− 4x2)− 32 + 8x ddx[(1− 4x2)− 32]= 8(1− 4x2)− 32 + 8x(−32)(1− 4x2)− 32−1(−8x)= 8(1− 4x2)− 32 + 96x2(1− 4x2)− 52Evaluating this at 14, we getf ′′′(14)= 8(34)− 32+ 6(34)− 52= 883√3+ 6329√3=643√3+643√3=1283√3The first four nonzero terms of the Taylor series of f(x) = arcsin(2x) at 14isf(14)+ f ′(14)(x− 14)+f ′′(14)2!(x− 14)2+f ′′′(14)3!(x− 14)3=pi6+4√3(x− 14)+163√32!(x− 14)2+1283√33!(x− 14)3=pi6+4√3(x− 14)+83√3(x− 14)2+649√3(x− 14)32. [Exercise on page 48]Maclaurin SeriesThe Maclaurin series of a function f(x) is the Taylor of f at a = 0; that is,408 Chapter 5. Infinite Seriesthe series∞∑n=0f (n)(0)n!xn = f(0) + f ′(0)x+f ′′(0)2!x2 +f ′′′(0)3!x3 + · · ·Some Important Maclaurin SeriesThe following are some important Maclaurin series and their radii of conver-gence R, and intervals of convergence I.11− x =∞∑n=0xn = 1 + x+ x2 + x3 + x4 · · · , R = 1, I = (−1, 1)ex =∞∑n=0xnn!= 1 +x1!+x22!+x33!+x44!+ · · · , R =∞, I = (−∞,∞)sinx =∞∑n=0(−1)n x2n+1(2n+ 1)!= x− x33!+x55!− x77!+ · · · , R =∞, I = (−∞,∞)cosx =∞∑n=0(−1)n x2n(2n)!= 1− x22!+x44!− x66!+ · · · , R =∞, I = (−∞,∞)arctanx =∞∑n=0(−1)n x2n+12n+ 1= x− x33+x55− x77+ · · · , R = 1, I = [−1, 1]ln(1 + x) =∞∑n=1(−1)n−1xnn= x− x22+x33− x44+ · · · , R = 1, I = (−1, 1](1+x)k = 1+kx+k(k − 1)2!x2+k(k − 1)(k − 2)3!x3+· · · , R = 1, I = (−1, 1)The latter series is referred to as the binomial series.5.2. Taylor Series 409(a) From the table above,ex =∞∑n=0xnn!Substituting x with 3x into this, we get the Maclaurin series of e3x:e3x =∞∑n=0(3x)nn!=∞∑n=03nn!xn(b) From the table, we have11− x =∞∑n=0xnSubstituting x with x2 into this, we get the Maclaurin series for 11−x2 :11− x2 =∞∑n=0(x2)n=∞∑n=0x2n(c) From the table, we havecosx =∞∑n=0(−1)n x2n(2n)!Substituting x with x3 into this, we get the Maclaurin series for cos(x3):cos(x3) =∞∑n=0(−1)n (x3)2n(2n)!=∞∑n=0(−1)n x6n(2n)!(d) From the table, we havearctan(x) =∞∑n=0(−1)n x2n+12n+ 1Substituting x with 4x into this, we get the Maclaurin series for arctan(4x):arctan(4x) =∞∑n=0(−1)n (4x)2n+12n+ 1=∞∑n=0(−1)n42n+1x2n+12n+ 1(e) Using the table, we getsin(2x) =∞∑n=0(−1)n (2x)2n+1(2n+ 1)!=∞∑n=0(−1)n22n+1 x2n+1(2n+ 1)!410 Chapter 5. Infinite SeriesMultiplying this by x, we get the Maclaurin series for x sin(2x):x sin(2x) = x∞∑n=0(−1)n22n+1 x2n+1(2n+ 1)!=∞∑n=0(−1)n22n+1 x2n+2(2n+ 1)!(f) First, we need to find the Maclaurin series for ln(1 + x3). From the table abovethe Maclaurin series of ln(1 + x) isln(1 + x) =∞∑n=1(−1)n−1xnnSubstituting x with x3, we getln(1 +x3)=∞∑n=1(−1)n−1(x3)nn=∞∑n=1(−1)n−1xn3nn=∞∑n=1(−1)n−1 xnn3nMultiplying both sides of this by x2, we getx2 ln(1 +x3)= x2∞∑n=1(−1)n−1 xnn3n=∞∑n=1(−1)n−1x2xnn3n=∞∑n=1(−1)n−1xn+2n3nxmxk = xm+k(g) From the seriesex =∞∑n=0xnn!,we havexe−2x = x∞∑n=0(−2x)nn!= x∞∑n=0(−1)n2nxnn!=∞∑n=0(−1)n2nxn+1n!(h)ln(1 + 5x)x=1xln(1 + 5x)=1x∞∑n=1(−1)n−1 (5x)nnSubstitute x with 5x intoln(1 + x) =∑∞n=1(−1)n−1 xnn=1x∞∑n=1(−1)n−15nxnn(ab)n = anbn5.2. Taylor Series 411=∞∑n=1(−1)n−15n 1xxnnMove1xinside the sum=∞∑n=1(−1)n−15nxn−1nxnxm= xn−m(i)5x = ex ln 5 BA = eA lnB=∞∑n=0(x ln 5)nn!Substitute x with x ln 5 intoex =∑∞n=0xnn!=∞∑n=0(ln 5)nxnn!(ab)n = anbn(j) First,√1 + x3 = (1 + x3)12 . Substituting x with x3, and k with 12into thebinomial series(1 + x)k = 1 + kx+k(k − 1)2!x2 +k(k − 1)(k − 2)3!x3 + · · ·we get√1 + x3 = 1 +12(x3) +12(12− 1)2!(x3)2 +12(12− 1) (12− 2)3!(x3)3 + · · ·= 1 +12x3 +12(12− 1)2!x6 +12(12− 1) (12− 2)3!x9 + · · ·3. [Exercise on page 48](a) Consider the Taylor series of sinx about a = 0.sinx = x− x33!+x55!− x77!+ · · ·This tells us that the if x is close to 0, sin x can be approximated by x.Note. sinx can also be approximated by x − x33!. Actually, the more terms weadd, the better the approximation will be.(b) The approximation cos x ≈ 1− x22makes sense because the Taylor series of cosxabout 0 iscosx = 1− x22!+x44!− x66!+ · · ·412 Chapter 5. Infinite Series(c) The approximation ex ≈ 1+x makes sense because the Taylor series of ex about0 isex = 1 +x1!+x22!+x33!+x44!+ · · ·(d) First, observe that (1 − x)n = (1 + (−x))n. Substituting x with −x into thebinomial series(1 + x)n = 1 + nx+n(n− 1)2!x2 +n(n− 1)(n− 2)3!x3 + · · ·we get(1− x)n = 1 + n(−x) + n(n− 1)2!(−x)2 + n(n− 1)(n− 2)3!(−x)3 + · · ·= 1− nx+ n(n− 1)2!x2 − n(n− 1)(n− 2)3!x3 + · · ·The first two terms of this series are 1 and −nx. So, it makes sense to approxi-mate (1− x)n by 1− nx for small values of x.4. [Exercise on page 48](a) We need to find the first terms of the Maclaurin series off(x) = 5√(1− 3x)2= [(1− 3x)2] 15 p√x = x 1p= (1− 3x) 25 (xn)m = xnm= [1 + (−3x)] 25Substituting x with −3x and k with 25into the binomial series(1 + x)k = 1 + kx+k(k − 1)2!x2 +k(k − 1)(k − 2)3!x3 + · · ·we get[1 + (−3x)] 25 = 1 + 25(−3x) +25(25− 1)2!(−3x)2 + · · ·= 1− 65x+25(25− 1)2!(−3x)2 + · · ·Using the first two terms, we get f(x) ≈ 1− 65x .5.2. Taylor Series 413(b) First, we need to find a “small” number x such that f(x) = 5√(0.88)2.5√(1− 3x)2 = 5√(0.08)2(1− 3x) 25 = (0.88) 25 p√xq = x qp1− 3x = 0.88 Raise both sides to the power of 52−3x = −0.12 Subtract 1 from both sidesx = 0.04 Divide both sides by −3Substituting x with 0.04 into the approximation from part (a), we get5√(0.88)2 = f(0.04) ≈ 1− 65(0.04) ≈ 0.9525. [Exercise on page 48](a)ex2=∞∑n=0(x2)nn!Substitute x with x2 into∑∞n=0xnn!=∞∑n=0x2nn!= 1 +x21!+x42!+x63!+ · · ·(b) Taking the integral term by term, we get∫ex2dx = C +∫1 dx+∫x21!dx+∫x42!dx+ · · · C is the integra-tion constant= C +∞∑n=0∫x2nn!dx= C +∞∑n=0x2n+1(2n+ 1)n!∫xn dx =xn+1n+ 1+ C(c) ∫ 10ex2dx =∞∑n=0∫ 10x2nn!dx414 Chapter 5. Infinite Series=∞∑n=0x2n+1(2n+ 1)n!∣∣∣∣10=∞∑n=0(12n+1(2n+ 1)n!− 02n+1(2n+ 1)n!)Substitute the bounds∞∑n=01(2n+ 1)n!(d) From part (c), ∫ 10ex2dx =∞∑n=01(2n+ 1)n! If n = 0,1(2n+ 1)n!=1(2(0) + 1)0!= 1 0! = 1 If n = 1,1(2n+ 1)n!=1(2(1) + 1)1!=13 If n = 2,1(2n+ 1)n!=1(2(2) + 1)2!=15(2)=110 If n = 3,1(2n+ 1)n!=1(2(3) + 1)3!=17(6)=142So, by considering the first four nonzero terms of the series, we get∫ 10ex2dx ≈ 1 + 13+110+142≈ 1.45714Chapter6Continuous Probability Distributionsand Linear Regression6.1 Probability Density Functions and Cumulative Dis-tribution Functions1. [Exercise on page 51](a)Random Variables The sample space of an experiment (or random trial) is defined asthe the set of all possible outcomes or results of that experiment.The sample space is usually denoted Ω. A random variable X is a function from the sample space to R,the set of real numbers. For example, if we toss a coin two times,then the sample space is Ω = {HH,HT, TH, TT}, where H standsfor Head and T for Tail. Define X to be the number of heads ineach result.– For the result HH, X = 2, for the result HT , X = 1, etc.– So, X takes on values 0, 1, or 2. If the values taken by a random variable X form a discrete set (forexample a subset of the set of integers), we say that X is a discreterandom variable. In the example above, X is discrete. If the values taken by a random variable X form a continuous set(for example an interval), we say that X is a continuous random415416 Chapter 6. Continuous Probability Distributions and Linear Regressionvariable. For example, the clotting time of blood, T , of a personchosen at random is a continuous random variable because the valuesof T range over an interval of real numbers.Probability Density Function (PDF)Let X be a continuous random variable that takes on values in the interval[a, b], and let f be a function. The function f is called a probabilitydensity function (or just PDF) of X in the interval [a, b] if the followingtwo conditions are satisfied.(1) f(x) ≥ 0 for all x in [a, b].(2) The integral of f(x) from x = a to x = b is equal to 1; that is,∫ baf(x) dx = 1The first condition says that the probability of an event is always greaterthan or equal to 0. And the second condition says that the “sum” of allprobabilities over the interval [a, b] is 1.We come back to the function f(x) = 2x defined on [0, 1]. To determine whetherf is a PDF, we need to check conditions (1) and (2).(1) Clearly, 2x ≥ 0 for all x in [0, 1]. So, Condition (1) holds.(2) For the second condition, we need to find the integral∫ 10f(x) dx.∫ 10f(x) dx =∫ 102x dx= x2∣∣∣∣10∫xn dx =xn+1n+ 1+ C= (1)2 − (0)1 = 1 F (x)∣∣∣∣ba= F (b)− F (a)So, Condition (2) is also satisfied.Since both conditions are satisfied, it follows that f(x) is a PDF .(b)Condition (1) holds because x293≥ 0 for all x in [4, 7]. For the second condition,6.1. Probability Density Functions and Cumulative Distribution Functions 417we have ∫ 74x293dx =193x33∣∣∣∣74∫xn dx =xn+1n+ 1+ C=1279x3∣∣∣∣74=1279(73 − 43) F (x)∣∣∣∣ba= F (b)− F (a)=1279(343− 64)=1279(279) = 1Since both conditions hold, f(x) is a PDF .(c)The function f is greater than or equal to 0 if and only if x3 − 4 ≥ 0, that is,x3 ≥ 4 or x ≥ 3√4 ≈ 1.5874. So, if x is less than 1.5874, f(x) is negative. Takefor example x = 1.2 in the interval [1, 3]. Evaluating f(x) at 1.2, we getf(1.2) =1.23 − 412=1.728− 412≈ −0.1893Since f(x) is negative for some values of x in the interval [1, 3], it follows thatCondition (1) is not satisfied. Therefore, f is not a PDF .(d)(1) Clearly, 27x4 ≥ 0 for all x in the interval [0, 2]. So, Condition (1) holds.(2) For Condition (2), we have∫ 2027x4 dx =27x55∣∣∣∣20=235x5∣∣∣∣20=235(25 − 05) = 6435.Since the integral is not equal to 1, it follows that Condition (2) is notsatisfied.Since only one of the two conditions is satisfied, f is not a PDF .(e)(1) This is not obvious! We need to find the sign of f(x).Set f(x) equal to 0 and solve for x.6305(−x2 + 3x+ 10) = 0418 Chapter 6. Continuous Probability Distributions and Linear Regression−x2 + 3x+ 10 = 0 Multiply both sides by 3056(x+ 2)(−x+ 5) = 0 Factorx = −2 or x = 5Marking these numbers on the real line, we get the intervals (−∞,−2),(−2, 5), and (5,∞).To get the sign of f(x) in an interval, choose any number in that interval ansubstitute it into f(x). If the result is positive, then the function is positive.If the result is negative, then the function is negative. The sign of f(x) isshown in the following figure.−∞ ∞−2 5− + −choose −3f(−3) = −0.157 < 0choose 0f(0) > 0choose 6f(6) = −0.157 < 0From the figure, it follows that f(x) is positive for all x in the interval(−2, 5). In particular, f(x) is positive for all x in the interval [−1, 4]. So,Condition (1) is satisfied.(2) ∫ 4−16305(−x2 + 3x+ 10) dx=6305∫ 4−1(−x2 + 3x+ 10) dx∫kf(x) dx = k∫f(x) dx=6305(−x33+32x2 + 10x)∣∣∣∣4−1∫xn dx =xn+1n+ 1+ C=6305[(−433+32(42) + 10(4))−(−(−1)33+32(−1)2 + 10(−1))]F (x)∣∣∣∣ba= F (b)− F (a)=6305[(−643+482+ 40)−(13+32− 10)]=6305(−643+482+ 40− 13− 32+ 10)6.1. Probability Density Functions and Cumulative Distribution Functions 419=6305(−1286+1446+2406− 26− 96+606)=6305(−128 + 144 + 240− 2− 9 + 606)=6305(3056)= 1So, Condition (2) holds.Since both conditions are satisfied, f(x) is a PDF .(f) Condition (1). Since sinx is always between −1 and 1, it follows that 3 + sin xis positive for every real number x. This implies that 13pi(3 + sinx) ≥ 0 for all xin[−pi2, pi2]. So, condition (1) is satisfied.Condition (2).∫ pi2−pi2f(x) dx =∫ pi2−pi213pi(3 + sin x) dx=13pi(3x− cosx)∣∣∣∣pi2−pi2∫sinx dx = − cosx+ C=13pi(3pi2− cos pi2)− 13pi(−3pi2− cos(−pi2))=13pi(3pi2− 0)− 13pi(−3pi2− 0)cos(pi2)= 0 = cos(−pi2)=12+12= 1So, condition (2) is also satisfied. Since conditions (1) and (2) are both met, thefunction f(x) is a PDF .2. [Exercise on page 51](a)∫ 62kx2 dx = k∫ 62x2 dx∫kf(x) dx = k∫f(x) dx= kx33∣∣∣∣62∫xn dx =xn+1n+ 1+ C420 Chapter 6. Continuous Probability Distributions and Linear Regression=k3(63 − 23) F (x)∣∣∣∣ba= F (b)− F (a)=k3(216− 8)=208k3For the function f(x) to be a PDF, the integral must be equal to 1.208k3= 1k =3208Multiply both sides by3208Clearly, f(x) = 3208x2 is greater than or equal to 0 for all x in [2, 6].(b)∫ 94f(x) dx =∫ 94k(x)12 dx= kx3232∣∣∣∣∣94∫xn dx =xn+1n+ 1+ C=2k3x32∣∣∣∣94abc=acb=2k3(932 − 4 32)F (x)∣∣∣∣ba= F (b)− F (a)=2k3(27− 8)=38k3For the function to be a PDF, the integral must be equal to 1. This holds when38k3= 1. Solving this latter equation for k, we get k =338. For this value of k,one can easily check that f(x) ≥ 0 for all x in [4, 9].(c)∫ 10(3√x2 + kx)dx =∫ 10(x23 + kx)dx p√xq = xqp6.1. Probability Density Functions and Cumulative Distribution Functions 421=(x5353+ kx22)∣∣∣∣∣10∫xn dx =xn+1n+ 1+ C=(35x53 +k2x2)∣∣∣∣10abc=acb=(35(1)53 +k2(1)2)−(35(0)53 +k2(0)2)F (x)∣∣∣∣ba= F (b)− F (a)=(35+k2)− 0 = 35+k2For the function to be a PDF, the integral must be equal to 1.35+k2= 1k2= 1− 35k2=25k =25(2) =45Clearly,3√x2 + 45x ≥ 0 for all x in [0, 1].3. [Exercise on page 52](a)Condition (1) holds since 3125x2 ≥ 0 for all x in [0, 5]. Moreover,∫ 503125x2 dx =3125x33∣∣∣∣50∫xn dx =xn+1n+ 1+ C=1125x3∣∣∣∣50=1125(53 − 03) F (x)∣∣∣∣ba= F (b)− F (a)=1125(125) = 1So, f(x) is a PDF .(b)422 Chapter 6. Continuous Probability Distributions and Linear RegressionProbability that c ≤X ≤ dLet X be a continuous random variable, and let f be the probabilitydensity function of X on some interval. Let c and d be numbers. Then,the probability that X is between c and d, denoted P (c ≤ X ≤ d),is given byP (c ≤ X ≤ d) =∫ dcf(x) dxUsing this, we haveP (1 ≤ X ≤ 2) =∫ 21f(x) dx=∫ 213125x2 dx=1125x3∣∣∣∣21∫xn dx =xn+1n+ 1+ C=1125(23 − 13) F (x)∣∣∣∣ba= F (b)− F (a)=1125(8− 1) = 7125(c)P (3 ≤ X ≤ 5) =∫ 53f(x) dx=∫ 533125x2 dx=1125x3∣∣∣∣53∫xn dx =xn+1n+ 1+ C=1125(53 − 33) F (x)∣∣∣∣ba= F (b)− F (a)=1125(125− 27) = 981254. [Exercise on page 52]6.1. Probability Density Functions and Cumulative Distribution Functions 423(a)For Review: Improper IntegralsLet a be a real number, and let f be a function. The improper integral∫∞af(x)dx is defined to be the limit of∫ taf(x)dx as t approaches∞. Thatis, ∫ ∞af(x)dx = limt→∞∫ taf(x)dxClearly, f(x) ≥ 0 for all x in [0,∞). So, Condition (1) holds. For the secondcondition, since the function is defined on the interval [0,∞), we need to findthe integral∫∞0f(x) dx. By definition of the improper integral, we have∫ ∞0f(x) dx = limt→∞∫ t023(1 + 2x)43dxTo find the integral∫ t023(1 + 2x)43dx, let u = 1 + 2x. Then, du = 2dx.If x = 0, u = 1. If x = t, u = 1 + 2tThe integral becomes∫ t023(1 + 2x)43dx =∫ 1+2t113u43du Substitute, change bounds=13∫ 1+2t1u−43 du∫kf(x) dx = k∫f(x) dx=13u−13−13∣∣∣∣∣1+2t1∫xn dx =xn+1n+ 1+ C=13(−311u13)∣∣∣∣1+2t1abc=acb= − 1u13∣∣∣∣1+2t1= − 1(1 + 2t)13−(− 1113)F (x)∣∣∣∣ba= F (b)− F (a)= − 1(1 + 2t)13+ 1424 Chapter 6. Continuous Probability Distributions and Linear RegressionTaking the limit of this as t approaches ∞, we getlimt→∞(− 1(1 + 2t)13+ 1)= limt→∞(− 1(1 + 2t)13)+ limt→∞(1)= 0 + 1 = 1So,∫∞0f(x) dx = 1. This shows that Condition (2) is satisfied. Since bothconditions hold, f(x) is a PDF .(b)P (0 ≤ X ≤ 2) =∫ 20f(x) dx=∫ 2023(1 + 2x)43dx=∫ 5113u43du Substitute u = 1 + 2x=13∫ 51u−43 du∫kf(x) dx = k∫f(x) dx= − 1u13∣∣∣∣51∫xn dx =xn+1n+ 1+ C= − 1513−(− 1113)F (x)∣∣∣∣ba= F (b)− F (a)= − 13√5+ 1 (≈ 0.4152)(c)Probability that X ≥ aLet X be a continuous random variable, and let f be the probabilitydensity function of X on [a,∞). Let c be a number greater than or equalto a. Then,P (X ≥ c) = P (X > c) =∫ ∞cf(x) dx6.1. Probability Density Functions and Cumulative Distribution Functions 425Using this, we haveP (X ≥ 13) =∫ ∞13f(x) dx=∫ ∞1323(1 + 2x)43dx= limt→∞∫ t1323(1 + 2x)43dx= limt→∞− 1u13∣∣∣∣1+2t27Substitute u = 1 + 2x,∫xn dx =xn+1n+ 1+ C= limt→∞(− 1(1 + 2t)13+1(27)13)F (x)∣∣∣∣ba= F (b)− F (a)= limt→∞(− 1(1 + 2t)13)+ limt→∞(13)= 0 +13=135. [Exercise on page 52](a)Since 3x2 ≥ 0 and e−x3 > 0 for all x in [0,∞), it follows that 3x2e−x3 ≥ 0 for allx in [0,∞). So, Condition (1) is satisfied. For Condition (2), we have∫ ∞03x2e−x3dx = limt→∞∫ t03x2e−x3dx= limt→∞∫ −t30eu(−du) Substitute u = −x3, change bounds= limt→∞− eu∣∣∣∣−t30∫ex dx = ex + C= limt→∞(−e−t3 − (−e0))F (x)∣∣∣∣ba= F (b)− F (a)= limt→∞(−e−t3 + 1)= limt→∞(−e−t3) + 1= 0 + 1 = 1426 Chapter 6. Continuous Probability Distributions and Linear RegressionSo, f(x) is a PDF .(b)P (0 ≤ X ≤ 1) =∫ 103x2e−x3dx=∫ −10eu(−du) Substitute u = −x3, change bounds= −eu∣∣∣∣−10∫ex dx = ex + C= −e−1 + 1 (≈ 0.6321) F (x)∣∣∣∣ba= F (b)− F (a)(c)P (X > 1) = P (X ≥ 1)=∫ ∞1f(x) dx=∫ ∞13x2e−x3dx= limt→∞∫ t13x2e−x3dx= limt→∞∫ −t3−1eu(−du) Substitute u = −t3,change bounds= limt→∞− eu∣∣∣∣−t3−1= limt→∞(−e−t3 + e−1)= limt→∞(−e−t3)+ limt→∞(e−1)= 0 + e−1 = e−1 (≈ 0.3679)Alternate SolutionWe have P (X > 1) =∫∞1f(x) dx. Using the property∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx,we have∫∞0f(x) dx =∫ 10f(x) dx +∫∞1f(x) dx. Solving this latter equa-6.1. Probability Density Functions and Cumulative Distribution Functions 427tion for∫∞1f(x) dx, we get∫ ∞1f(x) dx =∫ ∞0f(x) dx−∫ 10f(x) dx= 1− (−e−1 + 1) From (a) and (b)= 1 + e−1 − 1= e−1 ≈ (0.3679)(d)Since X takes on values on the interval [0,∞), the inequality X ≤ 2 is equivalentto 0 ≤ X ≤ 2. So,P (X ≤ 2) = P (0 ≤ X ≤ 2)=∫ 203x2e−x3dx=∫ −80eu(−du) Substitute u = −x3= −eu∣∣∣∣−80= −e−8 + 1 (≈ 0.9997)6. [Exercise on page 52](a)(1) First, note that f is defined on the interval [0,∞). For all x in the interval [0, 1], f(x) = 45x3 is greater than or equal to 0. Clearly, f(x) = 45x2is greater than 0 for all x in [1,∞).So, f(x) ≥ 0 for all x in [0,∞).(2) ∫ ∞0f(x) dx =∫ 10f(x) dx+∫ ∞1f(x) dx=∫ 1045x3 dx+∫ ∞145x2dx By definition of f(x)428 Chapter 6. Continuous Probability Distributions and Linear Regression=45x44∣∣∣∣10+ limt→∞∫ t145x2dx∫xn dx =xn+1n+ 1+ C=15x4∣∣∣∣10+ limt→∞45∫ t11x2dx∫kf(x) dx = k∫f(x) dx=15x4∣∣∣∣10+ limt→∞45(−1x)∣∣∣∣t1∫xn dx =xn+1n+ 1+ C=15(14 − 04) + limt→∞(− 45t+45)F (x)∣∣∣∣ba= F (b)− F (a)=15(1)− 45limt→∞(1t)+ limt→∞(45)=15− 45(0) +45=15+45=55= 1So, Condition (2) holds.Since both conditions are satisfied, f(x) is a PDF .(b)P (0 ≤ X ≤ 1) =∫ 10f(x) dx=∫ 1045x3 dx=15x4∣∣∣∣10∫xn dx =xn+1n+ 1+ C=15(1− 0) F (x)∣∣∣∣ba= F (b)− F (a)=15(c)P (0 ≤ X ≤ 3) =∫ 30f(x) dx=∫ 10f(x) dx+∫ 31f(x) dx6.1. Probability Density Functions and Cumulative Distribution Functions 429=∫ 1045x3 dx+∫ 3145x2dx=15x4∣∣∣∣10+(− 45t)∣∣∣∣31∫xn dx =xn+1n+ 1+ C=15(1− 0) +[(− 45(3))−(− 45(1))]F (x)∣∣∣∣ba= F (b)− F (a)=15− 415+45=1115(d)Probability that X ≤ aLet X be a continuous random variable, and let f be the probabilitydensity function of X on [a,∞). Let c be a number in [a,∞). Then,P (X < c) = P (X ≤ c) = P (a ≤ X ≤ c) =∫ caf(x) dx.Note: These equalities remain unchanged if the interval is (a,∞). If a = −∞ and f(x) is defined on (−∞,∞), thenP (X < c) = P (X ≤ c) =∫ c−∞f(x) dx.Here, f(x) is defined on [0,∞). So,P (X < 5) = P (X ≤ 5)= P (0 ≤ X ≤ 5)=∫ 50f(x) dx=∫ 1045x3 dx+∫ 5145x2dx=15x4∣∣∣∣10+(− 45x)∣∣∣∣51∫xn dx =xn+1n+ 1+ C=15(1− 0)− 425+45F (x)∣∣∣∣ba= F (b)− F (a)430 Chapter 6. Continuous Probability Distributions and Linear Regression=21257. [Exercise on page 52]Cumulative Distribution Function (CDF)Let X be a continuous random variable, and let f(x) be the probability densityfunction of X on the interval [a, b]. The cumulative distribution functionof X is defined asF (x) = P (X ≤ x) = P (a ≤ X ≤ x) =∫ xaf(t) dtfor x ≥ a. If x < a, F (x) is defined to be 0. This definition remains unchanged if f(x) is defined on [a,∞) or (a,∞). If f(t) is defined on an interval of the form (−∞, b), or (−∞, b], or(−∞,∞), F (x) is defined asF (x) =∫ x−∞f(t) dtNote. The derivative of F (x) is f(x). Indeed, by the Fundamental Theoremof Calculus (part I),ddx[∫ xaf(t) dt]= f(x)In other words, the probability density function is the derivative of the cu-mulative distribution function. Or, the cumulative distribution function is anantiderivative of the probability density function.(a) The probability distribution function f(x) is defined on the interval (−∞,−1].The cumulative distribution function, F (x), is defined on the same interval, andwe haveF (x) = P (X ≤ x) =∫ x−∞f(t) dt=∫ x−∞1t2dt= limz→−∞∫ xz1t2dt∫ a−∞f(x) dx = limz→−∞∫ azf(x) dx6.1. Probability Density Functions and Cumulative Distribution Functions 431= limz→−∞− 1t∣∣∣∣xz∫1x2dx = −1x+ C= limz→−∞(−1x+1z)= limz→−∞(−1x)+ limz→−∞(1z)= −1x+ 0 = −1x(b)P (X < −5) = P (X ≤ −5)= F (−5) P (X ≤ x) = F (x) by definition= − 1(−5) F (x) = −1xfrom part (a)=158. [Exercise on page 53](a) By the Fundamental Theorem of Calculus (part I), we haveF (x) =∫ xaf(t) dt implies F ′(x) = f(x)So, the probability density function is the derivative of the cumulative distribu-tion function. If x < 0, f(x) = ddx[F (x)] = ddx[0] = 0. If 0 < x < 3,f(x) =ddx[F (x)] =ddx[19x2]=19ddx[x2] =19(2x) =29x If x > 3, f(x) = ddx[F (x)] = ddx[1] = 0.Thus,f(x) =29x if 0 < x < 30 if x < 0 or x > 3432 Chapter 6. Continuous Probability Distributions and Linear Regression(b)P (X ≤ 2) = F (2) P (X ≤ x) = F (x)=19(2)2 =49(c)P (1 ≤ X < 2) =∫ 21f(x) dx P (a ≤ X ≤ b) =∫ baf(x) dx= F (x)∣∣∣∣21F (x) is an antiderivative of f(x)= F (2)− F (1) F (x)∣∣∣∣ba= F (b)− F (a)=19(2)2 − 19(1)2=49− 19=39=139. [Exercise on page 53](a) The probability that T is between 2 and 7 seconds isP (2 ≤ T ≤ 7) =∫ 72f(t) dt =∫ 721(ln 16)tdt=1ln 16∫ 721tdt∫kf(x) dx = k∫f(x) dx=1ln(16)ln t∣∣∣∣72∫1xdx = ln |x|+ C=1ln(16)(ln 7− ln 2) F (x)∣∣∣∣ba= F (b)− F (a)=≈ 0.4518(b) The probability that T is greater than 8 seconds isP (T > 8) =∫ 168f(t) dtIf a PDF f(x) is defined on[a, b], P (X > c) =∫ bcf(x) dxfor c in [a, b]6.2. Expected Value and Variance of Continuous Random Variables 433=∫ 1681(ln 16)tdt=1ln(16)ln t∣∣∣∣168∫1xdx = ln |x|+ C=1ln(16)(ln(16)− ln(8)) F (x)∣∣∣∣ba= F (b)− F (a)= 0.25(c) The probability that T is at most 12 seconds isP (T ≤ 12) =∫ 121f(t) dtIf a PDF f(x) is defined on[a, b], P (X ≤ c) = ∫ caf(x) dxfor c in [a, b]=∫ 1211(ln 20)tdt=1ln(16)ln t∣∣∣∣121∫1xdx = ln |x|+ C=1ln(16)(ln(12)− ln(1)) F (x)∣∣∣∣ba= F (b)− F (a)=ln(12)ln(16)≈ 0.8962 ln 1 = 06.2 Expected Value and Variance of Continuous Ran-dom Variables1. [Exercise on page 53](a)Expected Value, Variance, and Standard DeviationLet X be a continuous random variable, and let f(x) be the probabilitydensity function of X on the interval [a, b]. The expected value of X, denoted E(x), is given byE(X) =∫ baxf(x) dx434 Chapter 6. Continuous Probability Distributions and Linear Regression The variance of X, denoted Var(X), isVar(X) =∫ ba(x− µ)2f(x) dxwhere µ = E(X). The standard deviation of X, denoted σ, is the square root of thevariance. That is,σ =√Var(X)Alternative Formula for VarianceLet X be a continuous random variable, and let f(x) be the probabilitydensity function of X on the interval [a, b]. One can prove thatVar(X) =(∫ bax2f(x) dx)− (µ)2This formula can be rewritten as follows. If n is a real number, define the expected value of Xn asE(Xn) =∫ baxnf(x) dx.For n = 2, E(X2) =∫ bax2f(x) dx Using this, an alternative formula for variance is given byVar(X) = E(X2)− [E(X)]2This latter formula is commonly used.Let’s find E(X),Var(X), and σ for the probability density function f(x) = 14defined on [2, 6].Finding E(X).E(X) =∫ 62xf(x) dx6.2. Expected Value and Variance of Continuous Random Variables 435=∫ 62x(14)dx=∫ 6214x dx=14x22∣∣∣∣62∫xn dx =xn+1n+ 1+ C=18x2∣∣∣∣62=18(62 − 22) F (x)∣∣∣∣ba= F (b)− F (a)=18(36− 4) = 4Finding Var(X). We will use the alternative formula. First, we need to findE(X2).E(X2) =∫ 62x2f(x) dx=∫ 6214x2 dx=14x33∣∣∣∣62∫xn dx =xn+1n+ 1+ C=112x3∣∣∣∣62=112(63 − 23) F (x)∣∣∣∣ba= F (b)− F (a)=112(216− 8) = 20812=523Now, find the variance:Var(X) = E(X2)− [E(X)]2 = 523− [4]2 = 523− 16 = 523− 483=43The standard deviation isσ =√Var(X) =√43=2√3=2√33(≈ 1.1547)436 Chapter 6. Continuous Probability Distributions and Linear Regression(b)The expected value of X isE(X) =∫ 20xf(x) dx=∫ 20x(−13x+56)dx=∫ 20(−13x2 +56x)dx=(−13x33+56x22)∣∣∣∣20∫xn dx =xn+1n+ 1+ C=(−19x3 +512x2)∣∣∣∣20=(−19(2)3 +512(2)2)− (0) F (x)∣∣∣∣ba= F (b)− F (a)= −89+53= −89+159=79Finding the variance. First,E(X2) =∫ 20x2f(x) dx=∫ 20x2(−13x+56)=∫ 20(−13x3 +56x2)dx=(−13x44+56x33)∣∣∣∣20∫xn dx =xn+1n+ 1+ C=(− 112x4 +518x3)∣∣∣∣20=(− 112(2)4 +518(2)3)− (0) F (x)∣∣∣∣ba= F (b)− F (a)= −43+209= −129+209=896.2. Expected Value and Variance of Continuous Random Variables 437The variance isVar(X) = E(X2)− [E(X)]2 = 89−[79]2=89− 4981=7281− 4981=2381The standard deviation isσ =√Var(X) =√2381=√239(≈ 0.5329)(c)The expected value isE(X) =∫ 91xf(x) dx=∫ 91x(116(1 +2√x))dx=∫ 91(116x+x8(x)12)dx=∫ 91(116x+18x12)dx=(116x22+18x3232)∣∣∣∣∣91∫xn dx =xn+1n+ 1+ C=(132x2 +112x32)∣∣∣∣91abc=acb=(132(9)2 +112(9)32)−(132(1)2 +112(1)32)F (x)∣∣∣∣ba= F (b)− F (a)=143Finding the variance.E(X2) =∫ 91x2f(x) dx=∫ 91x2(116(1 +2√x))dx=∫ 91(116x2 +x28(x)12)dx438 Chapter 6. Continuous Probability Distributions and Linear Regression=∫ 91(116x2 +18x32)dx=(116x33+18x5252)∣∣∣∣∣91∫xn dx =xn+1n+ 1+ C=(148x2 +120x52)∣∣∣∣91abc=acb=(148(9)3 +120(9)52)−(148(1)3 +120(1)52)F (x)∣∣∣∣ba= F (b)− F (a)=72948+24320− 148− 120=40915The variance isVar(X) = E(X2)− [E(X)]2 = 40915−[143]2=40915− 1969=24745The standard deviation isσ =√24745(≈ 2.3428)(d)The expected value isE(X) =∫ ∞1xf(x) dx=∫ ∞1x(4x5)dx=∫ ∞14x4dx= limt→∞∫ t14x−4 dx= limt→∞4x−3−3∣∣∣∣t1∫xn dx =xn+1n+ 1+ C= limt→∞(− 43x3∣∣∣∣t1)= limt→∞(− 43t3−(− 43(1)3))F (x)∣∣∣∣ba= F (b)− F (a)6.2. Expected Value and Variance of Continuous Random Variables 439= limt→∞(− 43t3)+ limt→∞(43)= 0 +43=43Finding the variance.E(X2) =∫ ∞1x2f(x) dx=∫ ∞1x2(4x5)dx=∫ ∞14x3dx= limt→∞∫ t14x−3 dx= limt→∞4x−2−2∣∣∣∣t1∫xn dx =xn+1n+ 1+ C= limt→∞(− 2x2∣∣∣∣t1)= limt→∞(− 2t2−(− 2(1)2))F (x)∣∣∣∣ba= F (b)− F (a)= limt→∞(− 2t2)+ limt→∞(2)= 0 + 2 = 2The variance isVar(X) = E(X2)− [E(X)]2 = 2−[43]2= 2− 169=29The standard deviation isσ =√Var(X) =√29=√23(≈ 0.4714)2. [Exercise on page 54](a)440 Chapter 6. Continuous Probability Distributions and Linear RegressionThe expected value isE(X) =∫ 161xf(x) dx=∫ 161x(328x−14)dx=∫ 161328x34 dx=328x7474∣∣∣∣∣161∫xn dx =xn+1n+ 1+ C=328(47x74)∣∣∣∣161abc=acb=349x74∣∣∣∣161=349(1674 − 1 74)F (x)∣∣∣∣ba= F (b)− F (a)=349(128− 1)=38149(≈ 7.78)(b)To find Var(X), we first need to find E(X2).E(X2) =∫ 161x2f(x) dx=∫ 161x2(328x−14)dx=∫ 161328x74 dx=328x114114∣∣∣∣∣161∫xn dx =xn+1n+ 1+ C=328(411x74)∣∣∣∣161abc=acb=377x114∣∣∣∣1616.2. Expected Value and Variance of Continuous Random Variables 441=377(16114 − 1 114)F (x)∣∣∣∣ba= F (b)− F (a)=377(2048− 1)=614177Now, the variance isVar(X) = E(X2)− [E(X)]2 = 614177−[38149]2≈ 19.29The standard deviation isσ =√Var(X) =√19.29 ≈ 4.39(c)Median of a Continuous Random VariableLet X be a continuous random variable, and let f(x) be the probabilitydensity function of X on the interval [a, b]. The median of X is the realnumber m such that ∫ maf(x) dx =12 In other words, m is the median of X if P (a ≤ X ≤ m) = 12= 50%. Since the total probability is 1 (P (a ≤ X ≤ b) = ∫ baf(x) dx = 1), itfollows that the probability that X is between m and b is 12. Thatis, P (m ≤ X ≤ b) = 12= 50%.Let’s find the median m of X for the probability density function f(x) = 328x−14defined on the interval [1, 16].First, we find the integral∫ m1f(x) dx in terms of m.∫ m1f(x) dx =∫ m1328x−14 dx=328x3434∣∣∣∣∣m1∫xn dx =xn+1n+ 1+ C442 Chapter 6. Continuous Probability Distributions and Linear Regression=328(43x34)∣∣∣∣m1abc=acb=17x34∣∣∣∣m1=17m34 − 17(1)34 F (x)∣∣∣∣ba= F (b)− F (a)=17m34 − 17We now solve the equation∫ m1f(x) dx = 12for m.17m34 − 17=1217m34 =12+17Add17to both sides17m34 =914m34 =92Multiply both sides by 7m = (4.5)43 Raise both sides to the power of43m ≈ 7.43(d)In part (a), we found the mean: µ = E(X) ≈ 7.78. The probability that X isgreater than the mean is P (X > 7.78) = P (X ≥ 7.78). Since X takes on valuesin the interval [1, 16], it follows that X ≤ 16. So, the inequality X ≥ 7.78 isequivalent to 7.78 ≤ X ≤ 16. Therefore,P (X > 7.48) = P (X ≥ 7.48)= P (7.48 ≤ X ≤ 16)=∫ 167.48f(x) dx=∫ 167.48328x−14 dx=17x34∣∣∣∣167.78∫xn dx =xn+1n+ 1+ C6.2. Expected Value and Variance of Continuous Random Variables 443=17(1634 − (7.78) 34)F (x)∣∣∣∣ba= F (b)− F (a)≈ 0.48(e)Probability that X is Within n Standard Deviations of the MeanLet X be a continuous random variable, and let f(x) be the probabilitydensity function of X on the interval [a, b]. The probability that X is within n standard deviations of the mean,µ = E(X), is the probability that X is between µ− nσ and µ+ nσ. So, the probability that X is within 1 standard deviation of the meanisP (µ− σ ≤ X ≤ µ+ σ) =∫ µ+σµ−σf(x) dxWe need to find the probability that X is between µ− σ and µ+ σ. From Part(a), µ = E(X) ≈ 7.78. From part (b), σ ≈ 4.39. So,µ− σ = 7.78− 4.39 = 3.39 and µ+ σ = 12.17And the probability that X is between µ− σ and µ+ σ isP (3.39 ≤ X ≤ 12.17) =∫ 12.173.39f(x) dx=∫ 12.173.39328x−14 dx=17x34∣∣∣∣12.173.39∫xn dx =xn+1n+ 1+ C=17((12.17)34 − (3.39) 34)F (x)∣∣∣∣ba= F (b)− F (a)≈ 0.573. [Exercise on page 54]First, note that f is defined on [0,∞).444 Chapter 6. Continuous Probability Distributions and Linear RegressionFinding the expected value.E(X) =∫ ∞0xf(x) dx=∫ 10xf(x) dx+∫ ∞1xf(x) dx=∫ 10x(127x3)dx+∫ ∞1x(127x4)dx=∫ 10127x4 dx+ limt→∞∫ t1127x−3 dx=127x55∣∣∣∣10+ limt→∞127x−2−2∣∣∣∣t1∫xn dx =xn+1n+ 1+ C=1235x5∣∣∣∣10+ limt→∞(− 67x2∣∣∣∣t1)=1235(15 − 05) + limt→∞(− 67t2+67(1)2)F (x)∣∣∣∣ba= F (b)− F (a)=1235+ 0 +67=1235+3035=4235=65Finding the variance.E(X2) =∫ ∞0x2f(x) dx=∫ 10x2f(x) dx+∫ ∞1x2f(x) dx=∫ 10x2(127x3)dx+∫ ∞1x2(127x4)dx=∫ 10127x5 dx+ limt→∞∫ t1127x−2 dx=127x66∣∣∣∣10+ limt→∞127x−1−1∣∣∣∣t1∫xn dx =xn+1n+ 1+ C=27x6∣∣∣∣10+ limt→∞(−127x∣∣∣∣t1)=27(16 − 06) + limt→∞(−127t+127(1))F (x)∣∣∣∣ba= F (b)− F (a)6.2. Expected Value and Variance of Continuous Random Variables 445=27+ 0 +127=147= 2So, the variance isVar(X) = E(X2)− [E(X)]2 = 2−[65]2= 2− 3625=1425The standard deviation isσ =√Var(X) =√1425=√145≈ 0.74834. [Exercise on page 54](a) The expected value of T isE(T ) =∫ 161tf(t)dt=∫ 161t1(ln 16)tdt =∫ 1611ln 16dt Simplify=1ln 16t∣∣∣∣161=1ln 16(16− 1)=15ln 16≈ 5.41 s(b) Finding the variance. First, we need to find E(T 2).E(T 2) =∫ 161t2f(t) dt=∫ 161t21(ln 16)tdt =∫ 161tln 16dt Simplify=1ln 16(t22)∣∣∣∣161=12 ln 16(162 − 12)=2552 ln 16446 Chapter 6. Continuous Probability Distributions and Linear RegressionThe variance isVar(T ) = E(T 2)− [E(T )]2=2552 ln 16−(15ln 16)2=2552 ln 16− 225(ln 16)2=255 ln 162(ln 16)2− 4502(ln 16)2=−450 + 255 ln 162(ln 16)2≈ 16.7166 ≈ 16.72 sFinding the standard deviation.σ =√Var(X) =√16.7166 ≈ 4.0886 ≈ 4.09 s5. [Exercise on page 54](a) Finding the mean.E(X) =∫ 80xh(x) dx=∫ 80x4√x+ 1dx=∫ 91u− 14√uduUse substitution: Let u = x + 1.Then du = dx. Changing thebounds: if x = 0, u=1. If x = 8,u = 9. From u = x + 1, we havex = u− 1=14∫ 91(u− 1)u− 12 du 1√u=1u12= u−12=14∫ 91(uu−12 − u− 12)du Distribute=14∫ 91(u12 − u− 12)du unum = un+m=14(u3232− u1212)∣∣∣∣91∫un du =un+1n+ 1+ C6.2. Expected Value and Variance of Continuous Random Variables 447=14(23u32 − 2u 12)∣∣∣∣91abc=acb=14(23(9)32 − 2(9) 12)− 14(23(1)32 − 2(1) 12)=14(18− 6)− 14(23− 2)=14(12)− 14(−43)= 3 +13=93+13=103Finding the variance.E(X2) =∫ 80x2h(x) dx=∫ 80x24√x+ 1dx=∫ 91(u− 1)24√uduUse substitution: Let u = x + 1.Then du = dx. Changing thebounds: if x = 0, u=1. If x = 8,u = 9. From u = x + 1, we havex = u− 1=14∫ 91(u− 1)2u− 12 du 1√u=1u12= u−12=14∫ 91(u2 − 2u+ 1)u− 12 du (a− b)2 = a2 − 2ab+ b2=14∫ 91(u32 − 2u 12 + u− 12)du Distribute=14(u5252− 2u3232+u1212)∣∣∣∣91∫un du =un+1n+ 1+ C=14(25u52 − 43u32 + 2u12)∣∣∣∣91abc=acb=14(25(9)52 − 43(9)32 + 2(9)12)− 14(25(1)52 − 43(1)32 + 2(1)12)=14(4865− 36 + 6)− 14(25− 43+ 2)=14(4865− 1805+305)− 14(615− 2015+3015)448 Chapter 6. Continuous Probability Distributions and Linear Regression=14(3365)− 14(1615)=845− 415=25215− 415=24815The variance of X isVar(X) = E(X2)− [E(X)]2 = 24815−(103)2=24815− 1009=74445− 50045=24445Finding the standard deviation.σ =√24445(≈ 2.3286)(b) The probability that the height of a tree chosen randomly is greater than themean height isP (X > E(X)) = P(X >103)E(X) =103=∫ 8103f(x) dx =∫ 810314√x+ 1dx=∫ 913314√uduUse substitution: Let u = x + 1.Then du = dx. Changing thebounds: if x = 103, u = 133. Ifx = 8, u = 9=14∫ 9133u−12 du1√u=1u12= u−12=14u1212∣∣∣∣9133∫un du =un+1n+ 1+ C=12u12∣∣∣∣9133abc=acb=12(912 −(133) 12)=12(3−√133)(≈ 0.4592)6.2. Expected Value and Variance of Continuous Random Variables 4496. [Exercise on page 55](a) Let T be the life span of a cable. The expected value of T isE(T ) =∫ ∞0tf(t) dt =∫ ∞0125te−t25 dt = limb→∞∫ b0125te−125t dtTo find the integral∫ b0125te−125t dt, we use integration by parts.Choose u = 125t and dv = e−125tThen, du = 125dt and v =∫e−125tdt = 1− 15e−125t = −25e− 125 tThe indefinite integral∫125te−125t dt becomes∫125te−125t dt = uv −∫vdu=(125t)(−25e− 125 t)−∫ (−25e− 125 t)( 125dt)Substitute u and v= −te− 125 t +∫e−125t dt= −te− 125 t +(−25e− 125 t)+ C∫ekx dx =1kekx + C= (−t− 25)e− 125 t + CEvaluating this from t = 0 to t = b, we get∫ b0125te−125t dt = (−t− 25)e− 125 t∣∣∣∣b0= (−b− 25)e− 125 b − (−0− 25)e0 F (x)∣∣∣∣ba= F (b)− F (a)= (−b− 25)e− 125 b + 25Taking now the limit as b approaches ∞, we getE(T ) =∫ ∞0tf(t) dt=∫ ∞0125te−t25 dt= limb→∞∫ b0125te−125t dt450 Chapter 6. Continuous Probability Distributions and Linear Regression= limb→∞((−b− 25)e− 125 b + 25)From above= limb→∞(−b− 25)e− 125 b + limb→∞(25)= − limb→∞(be−125b)− 25 limb→∞(e−125b)+ 25= −(0)− 25(0) + 25 = 25 limb→∞e−kb = 0For Review - Limits - Exponential FunctionsLet g(x) be a function such that limx→∞g(x) = −∞. Then, for every nonnegative real number n,limx→∞xneg(x) = 0 n ≥ 0So, if P (x) is a polynomial and limx→∞g(x) = −∞, thenlimx→∞P (x)eg(x) = 0(b)To find the standard deviation, we first need to find the variance.Finding the variance.E(T 2) =∫ ∞0t2f(t) dt =∫ ∞0125t2e−125t dt = limb→∞∫ b0125t2e−125t dtTo find the integral∫ b0125t2e−125t dt, we use integration by parts. We will firstfind the indefinite integral∫125t2e−125t dt.Choose u = 125t2 and dv = e−125tThen, du = 225tdt and v =∫e−125tdt = 1− 125e−125t = −25e− 125 tThe integral∫125t2e−125t dt becomes∫125t2e−125t dt =(125t2)(−25e− 125 t)−∫ (−25e− 125 t)( 225tdt)Substitute u and v= −t2e− 125 t + 2∫te−125t dtIn part (a), we found that∫125te−125t dt = (−t−25)e− 125 t. Multiplying both sides6.2. Expected Value and Variance of Continuous Random Variables 451of this latter equation by 25, we get∫te−125t dt = (−25t− 625)e− 125 t + 25CSo, the integral∫125t2e−15t dt becomes∫125t2e−125t dt =(125t2)(−25e− 125 t)−∫ (−25e− 125 t)( 225tdt)= −t2e− 15 t + 2∫te−15t dt= −t2e− 125 t + 2((−25t− 625)e− 15 t + 25C)From above= −t2e− 125 t + (−50t− 1250)e− 125 t +K K = 50C= (−t2 − 50t− 1250)e− 125 t +KEvaluating this from t = 0 to t = b, we get∫ b0125t2e−125t dt= (−t2 − 50t− 1250)e− 125 t∣∣∣∣b0From above= (−b2 − 50b− 1250)e− 125 b − (−0− 0− 1250)e0 F (x)∣∣∣∣ba= F (b)− F (a)= (−b2 − 50b− 1250)e− 125 b + 1250Taking now the limit of this as b→∞, we getE(T 2) = limb→∞∫ b0125t2e−125t dt= limb→∞[(−b2 − 50b− 1250)e− 125 b]+ limb→∞(1250) From above= 0 + 1250 = 1250Now, the variance of T isVar(T ) = E(T 2)− [E(T )]2 = 1250− [25]2 = 1250− 625 = 625452 Chapter 6. Continuous Probability Distributions and Linear RegressionAnd the standard deviation isσ =√Var(T ) =√625 = 25(c)From part (a), the mean is µ = E(T ) = 25. So, the probability that T is lessthan the mean number of months is P (T < 25), which is equal to P (T ≤ 25).Since the domain of the probability density function is [0,∞), it follows thatT ≥ 0. So, the inequality T ≤ 25 is equivalent to 0 ≤ T ≤ 25. Therefore,P (T < 25) = P (T ≤ 25)= P (0 ≤ T ≤ 25)=∫ 250f(t) dt=∫ 250125e−125t dt=1251− 125e−125t∣∣∣∣250∫ekx dx =1kekx + C= −e− 125 t∣∣∣∣250abc=acb= −e−1 + e0 F (x)∣∣∣∣ba= F (b)− F (a)= −e−1 + 1 (≈ 0.6321)(d)Let m be the median life of the tower. Then, by definition, we have∫ m0f(t) dt =12.Finding the integral∫ m0f(t) dt.∫ m0f(t) dt =∫ m0125e−125t dt= −e− 125 t∣∣∣∣m0∫ekx dx =1kekx + C= −e− 125m + e0 F (x)∣∣∣∣ba= F (b)− F (a)= −e− 125m + 1 e0 = 16.3. Uniform, Exponential, and Normal Distributions 453Set∫ m0f(t) dt equal to 12and solve for m.−e− 125m + 1 = 12−e− 125m = −12Subtract 1 from both sidese−125m = 0.5 Multiply both sides by −1− 125m = ln(0.5) Take the natural logarithm of both sidesm = −25 ln(0.5) ≈ 17.33 Multiply both sides by −256.3 Uniform, Exponential, and Normal Distributions1. [Exercise on page 55]Uniform DistributionA continuous random variable X is said to be uniformly distributed if itsprobability density function (PDF) is a constant function. Suppose X takes onvalues on the interval [a, b]. Then X has the normal distribution if and only ifits PDF isf(x) =1b− a, a ≤ x ≤ bThe graph of y = f(x) is shown in the following figure.a b1b−ay = f(x)xyFinding the expected value of X.E(X) =∫ baxf(x) dx =∫ bax(1b− a)dx=1b− a∫ bax dx∫kf(x) dx = k∫f(x) dx454 Chapter 6. Continuous Probability Distributions and Linear Regression=1b− ax22∣∣∣∣ba∫xn dx =xn+1n+ 1+ C=12(b− a)(b2 − a2) F (x)∣∣∣∣ba= F (b)− F (a)=12(b− a)(b− a)(b+ a) b2 − a2 = (b− a)(b+ a)E(X) =b+ a2SimplifyFinding the variance of X. First, we need to find the expected value of X2.E(X2) =∫ bax2f(x) dx =∫ bax2(1b− a)dx=1b− a∫ bax2 dx∫kf(x) dx = k∫f(x) dx=1b− ax33∣∣∣∣ba∫xn dx =xn+1n+ 1+ C=13(b− a)(b3 − a3) F (x)∣∣∣∣ba= F (b)− F (a)=13(b− a)(b− a)(b2 + ba+ a2) b3 − a3 = (b− a)(b2 + ba+ a2)=b2 + ba+ a23SimplifyThe variance of X isVar(X) = E(X2)− [E(X)]2 Var(X) = E(X2)− [E(X)]2=b2 + ba+ a23−(b+ a2)2=b2 + ba+ a23− b2 + 2ba+ a24(b+ a)2 = b2 + 2ba+ a2=4(b2 + ba+ a2)12− 3(b2 + 2ba+ a2)12=4b2 + 4ba+ 4a2 − 3b2 − 6ba− 3a212Distribute=b2 − 2ba+ b2126.3. Uniform, Exponential, and Normal Distributions 455Var(X) =(b− a)212b2 − 2ba+ a2 = (b− a)2Finding the standard deviation of X.σ =√Var(X) =√(b− a)212=b− a√12Thus,E(X) =12(b+a), Var(X) =112(b−a)2, and σ = 1√12(b−a)(a) Here,f(x) =132 ≤ x ≤ 5, a = 2, b = 5=15− 2 =1b− aSo, X has the uniform distribution. Using the formulas above, we haveE(X) =12(b+ a) E(X) =12(b+ a)=12(5 + 2) = 3.5 a = 2 and b = 5Var(X) =112(b− a)2 Var(X) = 112(b− a)2=112(5− 2)2 = 912= 0.75 a = 2 and b = 5σ =√Var(X) =√0.75 ≈ 0.866(b) For the function f(x) = 29,−0.5 ≤ x ≤ 4, a = −0.5 and b = 4. Since1b− a =14− (−0.5) =14 + 0.5=14.5=192=29abc=acb= f(x)456 Chapter 6. Continuous Probability Distributions and Linear Regressionit follows that X is uniformly distributed. The expected value, variance, andstandard deviation of X areE(X) =12(b+ a) E(X) =12(b+ a)=12[4 + (−0.5)] = 12(3.5) = 1.75 a = −0.5 and b = 4Var(X) =112(b− a)2 Var(X) = 112(b− a)2=112[4− (−0.5)]2 = 112[4.5]2 = 1.6875 a = −0.5 and b = 4σ =√Var(X) =√1.6875 ≈ 1.2992. [Exercise on page 55](a) Here, f(x) = 0.02 is defined on the interval [450, 500]. So, a = 450 and b = 500.Observe that1b− a =1500− 450 =150= 0.02 = f(x)Since f(x) = 1b−a , a ≤ x ≤ b, it follows that X has the uniform distribution. Theexpected value of X isE(X) =12(b+ a) E(X) =12(b+ a)=12[500 + 450] =12(950) = 475 a = 450 and b = 500The standard deviation of X isσ =1√12(b− a) σ = 1√12(b− a)=1√12(500− 450) ≈ 14.434 a = 450, b = 500(b) The probability that X is less than $486 isP (X < 486) =∫ 486450f(x) dxIf a ≤ c ≤ b, P (X < c) =∫ caf(x) dx=∫ 4864500.02 dx f(x) = 0.026.3. Uniform, Exponential, and Normal Distributions 457= 0.02x∣∣∣∣486450∫k dx = kx+ C= 0.02(486− 450) = 0.72(c) The probability that X is between the mean (µ) and one standard deviation (σ)above the mean isP (µ ≤ X ≤ µ+ σ) =∫ µ+σµf(x) dx P (x1 ≤ X ≤ x2) =∫ x2x1f(x) dx=∫ µ+σµ0.02 dx f(x) = 0.02= 0.02x∣∣∣∣µ+σµ∫k dx = kx+ C= 0.02(µ+ σ − µ) = 0.02σ F (x)∣∣∣∣ba= F (b)− F (a)= 0.02(14.434) ≈ 0.2887 From part (a), σ ≈ 14.4343. [Exercise on page 55](a) Let X be the life span of the mosquito. Since the random variable X is uniformlydistributed over the interval [a, b], with a = 2 and b = 7, its probability densityfunction isf(x) =1b− a =17− 2 =15, 2 ≤ x ≤ 7The expected life span isE(X) =12(b+ a) E(X) =12(b+ a)=12(7 + 2) =92(b) The probability that X is greater than 5 months isP (X > 5) =∫ 75f(x) dxIf a ≤ c ≤ b, P (X > c) =∫ bcf(x) dx=∫ 7515dx f(x) =15=15x∣∣∣∣75∫k dx = kx+ C458 Chapter 6. Continuous Probability Distributions and Linear Regression=15(7− 5) = 25F (x)∣∣∣∣ba= F (b)− F (a)4. [Exercise on page 56]Exponential DistributionA continuous random variable X is said to be exponentially distributed ifits probability density function is of the formf(x) = λe−λx, for x in [0,∞),where λ is a positive constant. The graph of y = f(x) is shown in the followingfigure.λf(x) = λe−λxxyUsing integration by parts, one can show that the expected value, variance,and standard deviation of X are given byE(X) =1λ, Var(X) =1λ2, and σ =1λ(a) The probability density function f(x) = 5e−5x, x ≥ 0, is of the form f(x) =λe−λx with λ = 5. So, X is exponentially distributed. The expected value,variance, and standard deviation of X are given byE(X) =1λ=15= 0.2Var(X) =1λ2=152= 0.04σ =1λ=15= 0.26.3. Uniform, Exponential, and Normal Distributions 459(b) Observe that f(x) is of the form f(x) = λe−λx, x ≥ 0, with λ = 23. This meansthat X has the exponential distribution.E(X) =1λ=123E(X) =1λ=32= 1.5abc=acbVar(X) =1λ2=1(23)2 Var(X) = 1λ2=149=94= 2.25abc=acbσ =1λ=123σ =1λ=32= 1.5abc=acb5. [Exercise on page 56](a) Let X be the lifetime of the organism. From the problem, X is exponentiallydistributed. This implies that the probability density function of X is of theformf(x) = λe−λx, for x in [0,∞)Since the mean of X is µ = 48 and since the mean is given by the formula µ = 1λ,it follows that 1λ= 48. Solving this latter equation for λ, we find λ = 148. So,the probability density function of X isf(x) = 148e−148x, for x in [0,∞)(b) The probability that an organism lives between 45 and 50 years isP (45 ≤ X ≤ 50) =∫ 5045f(x) dx =∫ 5045148e−148x dx=148∫ 5045e−148x dx∫kf(x) dx = k∫f(x) dx=148(1− 148e−148x)∣∣∣∣5045∫ekx dx =1kekx + C460 Chapter 6. Continuous Probability Distributions and Linear Regression=148(−481e−148x)abc=acb= −e− 148x∣∣∣∣5045Simplify= −e− 148 (50) −(−e− 148 (45))F (x)∣∣∣∣ba= F (b)− F (a)= −e− 5048 + e− 4548≈ −0.352866 + 0.3916056 ≈ 0.0387(c) The probability that the number of people is greater than 52 isP (X > 52) =∫ ∞52f(x) dx If c is in [0,∞), P (X > c) =∫ ∞cf(x) dx=∫ ∞52148e−148x dx f(x) =148e−148x= limt→∞∫ t52148e−148x dx∫ ∞cf(x) dx = limt→∞∫ tcf(x) dx= limt→∞(−e− 148x∣∣∣∣t52)From part (b), an antiderivative off(x) = 148e−148x is F (x) = −e− 148x= limt→∞(−e− 148 t + e− 148 (52))F (x)∣∣∣∣ba= F (b)− F (a)= limt→∞(−e− t48)+ limt→∞(e−148(52))= 0 + e−148(52) For k > 0, limt→∞e−kt = 0= e−5248 ≈ 0.33856. [Exercise on page 56]• We first need to find the probability density function for the price. Let X bethe price. From the problem, X has the exponential distribution. This meansthat the probability density function of X is of the formf(x) = λe−λx, for x in [0,∞)Since the mean of X is µ = 2.5 and since the mean is given by the formula µ = 1λ,it follows that 1λ= 2.5. Solving this latter equation for λ, we find λ = 12.5= 0.4.6.3. Uniform, Exponential, and Normal Distributions 461So, the probability density function of X isf(x) = 0.4e−0.4x, for x in [0,∞)• The probability that the price is less than $7000 (7 thousand dollars) isP (X < 7) =∫ 70f(x) dx If c is in [0,∞), P (X < c) =∫ c0f(x) dx=∫ 700.4e−0.4x dx f(x) = 0.4e−0.4x= 0.4(1−0.4e−0.4x)∣∣∣∣70∫ekx dx =1kekx + C= −e−0.4x∣∣∣∣70Simplify= −e−0.4(7) + e−0.4(0) F (x)∣∣∣∣ba= F (b)− F (a)= −e−2.8 + e0≈ −0.0608 + 1 ≈ 0.93927. [Exercise on page 56]Normal DistributionLet µ and σ be real numbers with σ > 0. A random variable X has the normaldistribution with parameters µ and σ if its probability density function (PDF)isf(x) =1σ√2pie−(x−µ)22σ2 , for x in (−∞,∞)The graph of y = f(x) is shown in the following figure.µ− σ µ µ+ σ1σ√2pif(x) = 1σ√2pie−(x−µ)22σ2Inflection Pointsxy462 Chapter 6. Continuous Probability Distributions and Linear RegressionIf X has the normal distribution with parameters µ and σ, we writeX ∼ N(µ, σ2)One can prove that the expected value of X is µ and the standard deviation isσ; that is,E(X) = µ and Var(X) = σ2If µ = 0 and σ = 1, we say that X has the standard normal distribution.The PDF for the standard normal distribution is thenf(x) =1√2pie−x22 , for x in (−∞,∞)The following fact enables us to convert probabilities for the normal distributionto probabilities for the standard normal distribution.Converting N(µ,σ2) into N(0,1)Theorem 6.3.1. Let µ and σ be real numbers with σ > 0. Let X be arandom variable. Let Z be a random variable defined byZ =X − µσIf X has the normal distribution with parameters µ and σ, then Z hasthe standard normal distribution. In other words,if X ∼ N(µ, σ2), then Z ∼ N(0, 1)Suppose Z ∼ N(0, 1). To find probabilities for Z, we need to find the integral∫ ba1√2pie−x22 dxSince it is impossible to find the exact value of this integral, one can use nu-merical methods to approximate it. (Numerical methods include the midpointrule, the trapezoidal rule, and Simpson’s rule.)6.3. Uniform, Exponential, and Normal Distributions 463Using the Table for the Standard Normal Distribution to Find ProbabilitiesThe table at the end of this book contains approximate values of the integralP (Z ≤ z) =∫ z−∞f(x) dx, where f(x) =1√2pie−x22 ,for different values of z. For example,P (Z ≤ 0.45) = 0.6736, P (Z < 2.1) = 0.9821, P (Z ≤ −1.04) = 0.1492Suppose X ∼ N(µ, σ2).Finding the probability P (X ≤ a).Step 1. Use Theorem 6.3.1 to transform P (X ≤ a) into a probability of theform P (Z ≤ z):P (X ≤ a) = P(X − µσ≤ a− µσ)= P(Z ≤ a− µσ)= P (Z ≤ z) Where z = a− µσStep 2. Use the table to find P (Z ≤ z). Geometrically, P (Z ≤ z) is the areaof the shaded region in the following figure.zShaded area= P (Z ≤ z)xyFinding the probability P (X ≥ b).Step 1. Use Theorem 6.3.1 to transform P (X ≥ b) into a probability of theform P (Z ≤ z):P (X ≥ b)464 Chapter 6. Continuous Probability Distributions and Linear Regression= P(X − µσ≥ b− µσ)= P(Z ≥ b− µσ)= P (Z ≥ z) =∫ ∞zf(x) dx Where z =b− µσ=∫ −∞zf(x) dx+∫ ∞−∞f(x) dx∫ baf(x) dx =∫ caf(x) dx+∫ bcf(x) dx=∫ −∞zf(x) dx+ 1∫∞−∞ f(x) dx = 1 since f(x) isa PDF on (−∞,∞)= −∫ z−∞f(x) dx+ 1∫ baf(x) dx = −∫ abf(x) dx= 1− P (Z ≤ z)Step 2. Use the table to get P (Z ≤ z). Geometrically, P (Z ≥ z) is the areaof the shaded region in the following figure.zShaded area= P (Z ≥ z)xyFinding the probability P (a ≤ X ≤ b). Using the same approach as before, wehaveP (a ≤ X ≤ b) = P(a− µσ≤ X − µσ≤ b− µσ)= P (z1 ≤ Z ≤ z2) =∫ z2z1f(x)dx Where z1 =a− µσ, z2 =b− µσ=∫ −∞z1f(x)dx+∫ z2−∞f(x) dx=∫ z2−∞f(x) dx−∫ z1−∞f(x) dx6.3. Uniform, Exponential, and Normal Distributions 465= P (Z ≤ z2)− P (Z ≤ z1)These two latter probabilities can be obtained from the table. Geometrically,P (z1 ≤ Z ≤ z2) is the area of the shaded region in the following figure.z1 z2Shaded area= P (z1 ≤ Z ≤ z2)xy(a) Here X ∼ N(µ, σ2) with µ = 55 and σ = 20.P (X ≤ 60) = P(X − µσ≤ 60− µσ)= P(X − 5520≤ 60− 5520)= P (Z ≤ 0.25) Z = X − 5520= 0.5987 From the table, P (Z ≤ 0.25) = 0.5987(b)P (X > 25) = P(X − µσ>25− µσ)= P(X − 5520>25− 5520)= P (Z > −1.5) Z = X − 5520= 1− P (Z ≤ −1.5) P (Z > a) = 1− P (Z ≤ a)= 1− 0.0668 From the table, P (Z ≤ −1.5) = 0.0668466 Chapter 6. Continuous Probability Distributions and Linear Regression= 0.9332(c)P (45 < X ≤ 80)= P(45− µσ<X − µσ≤ 80− µσ)= P(45− 5520<X − 5520≤ 80− 5520)Substitute µ and σ= P (−0.5 < Z ≤ 1.25) Z = X − µσ= P (Z ≤ 1.25)− P (Z < −0.5) P (z1 ≤ Z ≤ z2) = P (Z ≤ z2)− P (Z ≤ z1)= 0.8944− 0.3085 = 0.5859 From the table, P (Z ≤ 1.25) = 0.8944and P (Z ≤ −0.5) = 0.30858. [Exercise on page 56](a) Let X be the weight of an individual in the population. From the problem,X has the normal distribution with mean µ = 4700 and standard deviationσ = 645. Finding the probability that a randomly selected individual has aweight greater than 6000 g amounts to finding the probability that X is greaterthan 6000; that is, P (X > 6000).P (X > 6000) = P(X − 4700645>6000− 4700645)= P (Z > 2.02)= 1− P (Z ≤ 2.02) P (Z > a) = 1− P (Z ≤ a)= 1− 0.9783 From the table,P (Z ≤ 2.02) = 0.9783= 0.0217Thus, 2.17% of the population has a weight that exceeds 6000 g.(b) Finding the probability that a randomly selected individual has a weight between3500 g and 5200 g amounts to finding the probability that X is between 3500and 5200.P (3500 ≤ X ≤ 5200)= P(3500− 4700645≤ X − 4700645≤ 5200− 4700645)6.3. Uniform, Exponential, and Normal Distributions 467= P (−1.86 ≤ Z ≤ 0.78) Z = X − 4700645= P (Z ≤ 0.78)− P (Z ≤ −1.86) P (z1 ≤ Z ≤ z2) = P (Z ≤ z2)− P (Z ≤ z1)= 0.7823− 0.0314 From the table, P (Z ≤ 0.78) = 0.7823and P (Z ≤ −1.86) = 0.0314= 0.7509Thus, 75.09% of the population has a weight between 3500 g and 5200 g.9. [Exercise on page 57]Let X be the height of the individual. From the problem, X is normally distributed,with a mean of µ = 1.15 and a standard deviation of σ = 0.3. We need to find x1and x2 such thatP (z1 ≤ Z ≤ z2) = 0.5, z1 = x1 − µσ, z2 =x2 − µσ, and z2 = −z1where Z = X−µσ. Using the table, we can find z1 and z2 as follows. Consider thefollowing figure in which y = f(x) represents the probability density function of thestandard normal distribution.z1 z2Shaded area = 0.5y = f(x)xySince the graph is symmetric about the y-axis, it follows that the area to the left ofz1 is equal to the area to the right of z2. Since the area under the curve over theinterval (−∞,∞) is 1 (because y = f(x) is a probability density function), we haveP (Z ≤ z1) + P (z1 ≤ Z ≤ z2) + P (Z ≥ z2) = 12P (Z ≤ z1) + 0.5 = 1P (Z ≤ z1) = P (Z ≥ z2),P (z1 ≤ Z ≤ z2) = 0.5468 Chapter 6. Continuous Probability Distributions and Linear RegressionP (Z ≤ z1) = 1− 0.52= 0.25From the table, the value of z1 for which P (Z ≤ z1) is closest to 0.25 is −0.67. Thisimplies that z1 = −0.67. And z2 = −z1 = 0.67.Solving the equation z1 =x1−µσfor x1, we havex1 − µσ= z1x1 − µ = z1σ Multiply both sides by σx1 = µ+ z1σ Add µ to both sides= 1.15 + (−0.67)(0.3) µ = 1.15, σ = 0.3, z1 = −0.67= 1.15− 0.201 = 0.949Likewise,x2 = µ+ z2σ = 1.15 + (0.67)(0.3) = 1.351Thus, the interval centered at the mean such that the height of an individual belongsto the interval is 50% is [0.949, 1.351] .6.4 Linear Regression1. [Exercise on page 57]Finding the Line of Best Fit Using the Method of Least SquaresConsider the following datax x1 x2 · · · xny y1 y2 · · · ynSuppose we want to fit a straight line, y = ax+ b, to this data. For 1 ≤ i ≤ n,define ei asei = yi − (axi + b)The most common approach to find a and b is called the method of least squaresor linear regression. The method of least squares consists of finding theline of best fit. Specifically, it consists of finding the values of a and b that6.4. Linear Regression 469minimize the sumn∑i=1e2i = e21 + e22 + · · ·+ e2nUsing advanced techniques from Calculus, one can show that the values of aand b that minimize the sum above are given bya =∑ni=1(xi − x)(yi − y)∑ni=1(xi − x)2and b = y − ax (6.4.1)where x is the mean of x and y is the mean of y.x =1nn∑i=1xi and y =1nn∑i=1yiHere we have the following data.x 1 2 3 4 5y 5 8 10 9 14To find the least-square line (y = ax+ b), we need to find a and b using (6.4.1).Step 1. Find x, the mean of x, and y, the mean of y.x =1nn∑i=1xi =155∑i=1xi =15(1 + 2 + 3 + 4 + 5) =15(15) = 3y =1nn∑i=1yi =155∑i=1yi =15(5 + 8 + 10 + 9 + 14) =15(46) = 9.2Step 2. Start off with the original data and add the column titled xi − x.x y xi − x1 5 1− 3 = −22 8 2− 3 = −13 10 3− 3 = 04 9 4− 3 = 15 14 5− 3 = 2470 Chapter 6. Continuous Probability Distributions and Linear RegressionStep 3. Add the column yi − y.x y xi − x yi − y1 5 −2 5− 9.2 = −4.22 8 −1 8− 9.2 = −1.23 10 0 10− 9.2 = 0.84 9 1 9− 9.2 = −0.25 14 2 14− 9.2 = 4.8Step 4. Add the column (xi − x)(yi − y).x y xi − x yi − y (xi − x)(yi − y)1 5 −2 −4.2 (−2)(−4.2) = 8.42 8 −1 −1.2 (−1)(−1.2) = 1.23 10 0 0.8 (0)(0.8) = 04 9 1 −0.2 (1)(−0.2) = −0.25 14 2 4.8 (2)(4.8) = 9.6Step 5. Add the column (xi − x)2.x y xi − x yi − y (xi − x)(yi − y) (xi − x)21 5 −2 −4.2 8.4 (−2)2 = 42 8 −1 −1.2 1.2 (−1)2 = 13 10 0 0.8 0 (0)2 = 04 9 1 −0.2 −0.2 (1)2 = 15 14 2 4.8 9.6 (2)2 = 4Step 6. Add up the numbers from the column titled (xi − x)(yi − y). Also, add upthe numbers from the column titled (xi − x)2.6.4. Linear Regression 471x y xi− x yi− y (xi− x)(yi− y) (xi− x)21 5 −2 −4.2 8.4 42 8 −1 −1.2 1.2 13 10 0 0.8 0 04 9 1 −0.2 −0.2 15 14 2 4.8 9.6 4Sum 19 10Step 7. Find a and b.a =∑5i=1(xi − x)(yi − y)∑ni=1(xi − x)2=1910= 1.9b = y − ax = 9.2− 1.9(3) = 9.2− 5.7 = 3.5Thus, the least-squares line is y = 1.9x+ 3.5 . Plotting this line along with the data,we get the following figure.2 4 651015xy2. [Exercise on page 57]x 3 9 13 16 24y 11 10 9.5 6 5.5First, we need to find the mean of x, x, and the mean of y, y.472 Chapter 6. Continuous Probability Distributions and Linear Regressionx =1nn∑i=1xi =155∑i=1xi =15(3 + 9 + 13 + 16 + 24) =15(65) = 13y =1nn∑i=1yi =155∑i=1yi =15(11 + 10 + 9.5 + 6 + 5.5) =15(42) = 8.4We now find a and b using (6.4.1). To ease the computation, we build the followingtable.x y xi− x yi− y (xi− x)(yi− y) (xi− x)23 11 −10 2.6 −26 1009 10 −4 1.6 −6.4 1613 9.5 0 1.1 0 016 6 3 −2.4 −7.2 924 5.5 11 −2.9 −31.9 121Sum −71.5 246So,a =∑5i=1(xi − x)(yi − y)∑5i=1(xi − x)2=−71.5246≈ −0.29065b = y − ax = 8.4− (−0.29065)(13) = 8.4 + 3.77845 = 12.17845Thus, the least-squares line is y = −0.29065x+ 12.17845 . Plotting this line alongwith the data, we get the following figure.5 10 15 20 25510xy6.4. Linear Regression 4733. [Exercise on page 57]Here we have the following data.x −10.7 −5.2 −3 −1 2.5 7 10 14y −3.4 −1.6 −2.8 2 1 4 7.3 9First, we need to find x, the mean of x, and y, the mean of y.x =1nn∑i=1xi =188∑i=1xi=18(−10.7− 5.2− 3− 1 + 2.5 + 7 + 10 + 14) = 18(13.6) = 1.7y =188∑i=1yi =188∑i=1yi=18(−3.4− 1.6− 6− 2.8 + 2 + 1 + 4 + 7.3 + 9) = 18(15.5) = 1.9375We now find a and b using (6.4.1). To ease the computation, we construct thefollowing table.x y xi− x yi− y (xi− x)(yi− y) (xi− x)2−10.7 −3.4 −12.4 −5.3375 66.185 153.76−5.2 −1.6 −6.9 −3.5375 24.40875 47.61−3 −2.8 −4.7 −4.7375 22.26625 22.09−1 2 −2.7 0.0625 −0.16875 7.292.5 1 0.8 −0.9375 −0.75 0.647 4 5.3 2.0625 10.93125 28.0910 7.3 8.3 5.3625 44.50875 68.8914 9 12.3 7.0625 86.86875 151.29Sum 254.25 479.66So,a =∑8i=1(xi − x)(yi − y)∑8i=1(xi − x)2=254.25479.66≈ 0.53474 Chapter 6. Continuous Probability Distributions and Linear Regressionb = y − ax = 1.9375− (0.53)(1.7) = 1.0364Thus, the linear regression line is y = 0.53x+ 1.0364 . Plotting this line along withthe data, we get the following figure.−10 −5 5 10 15−5510xy4. [Exercise on page 57]Length of a chromosome (x) 39 40 41 43 46 49 50 52Life span (y) 74 73 70 68 63 59 55 53(a) Plotting the data, we get the following figure.40 42 44 46 48 50 52 54 56506070xyFrom the figure, the relationship between the length of the chromosome and thelife span is approximately linear .6.4. Linear Regression 475(b) First, we need to find x, the mean of x, and y, the mean of y.x =1nn∑i=1xi =188∑i=1xi=18(39 + 40 + 41 + 43 + 46 + 49 + 50 + 52) =18(360) = 45y =188∑i=1yi =188∑i=1yi=18(74 + 73 + 70 + 68 + 63 + 59 + 55 + 53) =18(515) = 64.375We now find a and b using (6.4.1). To ease the computation, we construct thefollowing table.x y xi− x yi− y (xi− x)(yi− y) (xi− x)239 74 −6 9.625 −57.75 3640 73 −5 8.625 −43.125 2541 70 −4 5.625 −22.5 1643 68 −2 3.625 −7.25 446 63 1 −1.375 −1.375 149 59 4 −5.375 −21.5 1650 55 5 −9.375 −46.875 2552 53 7 −11.375 −79.625 49Sum −280 172So,a =∑8i=1(xi − x)(yi − y)∑8i=1(xi − x)2=−280172≈ −1.6279b = y − ax = 64.375− (−1.6279)(45) ≈ 137.6305Thus, the linear regression line is y = −1.6279x+ 137.6305 .(c) To predict the life span of an individual whose length of the chromosome is 54,476 Chapter 6. Continuous Probability Distributions and Linear Regressionwe substitute x with 54 into the linear regression equation from part (b):y = −1.6279(54) + 137.6305 = 49.7239 ≈ 505. [Exercise on page 58]Leaf width (in cm) 2.3 2.9 3.5 3.8 4.2 4.8 5 5.2 5.5 6Leaf length (in cm) 3.4 3.8 4.2 5 5.3 5.2 6 6.3 7 7.2(a) Let x be the leaf width and y be the leaf length. Plotting the data, we get thefollowing figure.1 2 3 4 5 6246xyFrom the figure, the relationship between the length of the chromosome and thelife span is approximately linear .(b) First, we need to find x, the mean of x, and y, the mean of y.x =1nn∑i=1xi =11010∑i=1xi=110(2.3 + 2.9 + 3.5 + 3.8 + 4.2 + 4.8 + 5 + 5.2 + 5.5 + 6)=110(43.2) = 4.32y =11010∑i=1yi =11010∑i=1yi=110(3.4 + 3.8 + 4.2 + 5 + 5.3 + 5.2 + 6 + 6.3 + 7 + 7.2) =110(53.4) = 5.34We now find a and b using (6.4.1). To ease the computation, we construct thefollowing table.6.4. Linear Regression 477x y xi− x yi− y (xi− x)(yi− y) (xi− x)22.3 3.4 −2.02 −1.94 3.9188 4.08042.9 3.8 −1.42 −1.54 2.1868 2.01643.5 4.2 −0.82 −1.14 0.9348 0.67243.8 5 −0.52 −0.34 0.1768 0.27044.2 5.3 −0.12 −0.04 0.0048 0.01444.8 5.2 0.48 −0.14 −0.0672 0.23045 6 0.68 0.66 0.4488 0.46245.2 6.3 0.88 0.96 0.8448 0.77445.5 7 1.18 1.66 1.9588 1.39246 7.2 1.68 1.86 3.1248 2.8224Sum 13.532 12.736So,a =∑10i=1(xi − x)(yi − y)∑10i=1(xi − x)2=13.53212.736= 1.0625b = y − ax = 5.34− (1.0625)(4.32) = 0.75Thus, the linear regression line is y = 1.0625x+ 0.75 .Plotting the graph of y = 1.0625x+0.75 along with the data, we get the followingfigure.1 2 3 4 5 6246xy478 Chapter 6. Continuous Probability Distributions and Linear Regression(c) If the width of a leaf is 6.7 cm, its length is approximatelyy = 1.0625(6.7) + 0.75 Substitute x with 6.7 into the regression line= 7.86875 ≈ 7.9 cm479Table of the Normal Distributionz xyz 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09−3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010−2.9 0.0019 0.0018 0.0017 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014−2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019−2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026−2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036−2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048−2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064−2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084−2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110−2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143−2.0 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183−1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233−1.8 0.0359 0.0352 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294−1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367−1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455−1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559−1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0722 0.0708 0.0694 0.0681−1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823−1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985−1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170−1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379−0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611−0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867−0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148−0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451−0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776−0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121−0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483−0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859−0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247−0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641480 Table of the Normal DistributionTable of the Normal Distribution (continued)z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.83891.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.93191.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.97672.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.99362.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.99863.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990The table above gives the area of the region under the standard normal curve y =1√2pie−x22 over the interval (−∞, z]. In other words, the table gives the values of the proba-bility P (Z ≤ z) for different values of z.
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