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Calculus Volume 1

4.10 Antiderivatives

Calculus Volume 14.10 Antiderivatives
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 4.10.1. Find the general antiderivative of a given function.
  • 4.10.2. Explain the terms and notation used for an indefinite integral.
  • 4.10.3. State the power rule for integrals.
  • 4.10.4. Use antidifferentiation to solve simple initial-value problems.

At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their applications. We now ask a question that turns this process around: Given a function f,f, how do we find a function with the derivative ff and why would we be interested in such a function?

We answer the first part of this question by defining antiderivatives. The antiderivative of a function ff is a function with a derivative f.f. Why are we interested in antiderivatives? The need for antiderivatives arises in many situations, and we look at various examples throughout the remainder of the text. Here we examine one specific example that involves rectilinear motion. In our examination in Derivatives of rectilinear motion, we showed that given a position function s(t)s(t) of an object, then its velocity function v(t)v(t) is the derivative of s(t)s(t)—that is, v(t)=s(t).v(t)=s(t). Furthermore, the acceleration a(t)a(t) is the derivative of the velocity v(t)v(t)—that is, a(t)=v(t)=s(t).a(t)=v(t)=s(t). Now suppose we are given an acceleration function a,a, but not the velocity function vv or the position function s.s. Since a(t)=v(t),a(t)=v(t), determining the velocity function requires us to find an antiderivative of the acceleration function. Then, since v(t)=s(t),v(t)=s(t), determining the position function requires us to find an antiderivative of the velocity function. Rectilinear motion is just one case in which the need for antiderivatives arises. We will see many more examples throughout the remainder of the text. For now, let’s look at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. We examine various techniques for finding antiderivatives of more complicated functions later in the text (Introduction to Techniques of Integration).

The Reverse of Differentiation

At this point, we know how to find derivatives of various functions. We now ask the opposite question. Given a function f,f, how can we find a function with derivative f?f? If we can find a function FF derivative f,f, we call FF an antiderivative of f.f.

Definition

A function FF is an antiderivative of the function ff if

F(x)=f(x)F(x)=f(x)

for all xx in the domain of f.f.

Consider the function f(x)=2x.f(x)=2x. Knowing the power rule of differentiation, we conclude that F(x)=x2F(x)=x2 is an antiderivative of ff since F(x)=2x.F(x)=2x. Are there any other antiderivatives of f?f? Yes; since the derivative of any constant CC is zero, x2+Cx2+C is also an antiderivative of 2x.2x. Therefore, x2+5x2+5 and x22x22 are also antiderivatives. Are there any others that are not of the form x2+Cx2+C for some constant C?C? The answer is no. From Corollary 22 of the Mean Value Theorem, we know that if FF and GG are differentiable functions such that F(x)=G(x),F(x)=G(x), then F(x)G(x)=CF(x)G(x)=C for some constant C.C. This fact leads to the following important theorem.

Theorem 4.14

General Form of an Antiderivative

Let FF be an antiderivative of ff over an interval I.I. Then,

  1. for each constant C,C, the function F(x)+CF(x)+C is also an antiderivative of ff over I;I;
  2. if GG is an antiderivative of ff over I,I, there is a constant CC for which G(x)=F(x)+CG(x)=F(x)+C over I.I.

In other words, the most general form of the antiderivative of ff over II is F(x)+C.F(x)+C.

We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.

Example 4.50

Finding Antiderivatives

For each of the following functions, find all antiderivatives.

  1. f(x)=3x2f(x)=3x2
  2. f(x)=1xf(x)=1x
  3. f(x)=cosxf(x)=cosx
  4. f(x)=exf(x)=ex

Solution

  1. Because
    ddx(x3)=3x2ddx(x3)=3x2

    then F(x)=x3F(x)=x3 is an antiderivative of 3x2.3x2. Therefore, every antiderivative of 3x23x2 is of the form x3+Cx3+C for some constant C,C, and every function of the form x3+Cx3+C is an antiderivative of 3x2.3x2.
  2. Let f(x)=ln|x|.f(x)=ln|x|. For x>0,f(x)=ln(x)x>0,f(x)=ln(x) and
    ddx(lnx)=1x.ddx(lnx)=1x.

    For x<0,f(x)=ln(x)x<0,f(x)=ln(x) and
    ddx(ln(x))=1x=1x.ddx(ln(x))=1x=1x.

    Therefore,
    ddx(ln|x|)=1x.ddx(ln|x|)=1x.

    Thus, F(x)=ln|x|F(x)=ln|x| is an antiderivative of 1x.1x. Therefore, every antiderivative of 1x1x is of the form ln|x|+Cln|x|+C for some constant CC and every function of the form ln|x|+Cln|x|+C is an antiderivative of 1x.1x.
  3. We have
    ddx(sinx)=cosx,ddx(sinx)=cosx,

    so F(x)=sinxF(x)=sinx is an antiderivative of cosx.cosx. Therefore, every antiderivative of cosxcosx is of the form sinx+Csinx+C for some constant CC and every function of the form sinx+Csinx+C is an antiderivative of cosx.cosx.
  4. Since
    ddx(ex)=ex,ddx(ex)=ex,

    then F(x)=exF(x)=ex is an antiderivative of ex.ex. Therefore, every antiderivative of exex is of the form ex+Cex+C for some constant CC and every function of the form ex+Cex+C is an antiderivative of ex.ex.
Checkpoint 4.49

Find all antiderivatives of f(x)=sinx.f(x)=sinx.

Indefinite Integrals

We now look at the formal notation used to represent antiderivatives and examine some of their properties. These properties allow us to find antiderivatives of more complicated functions. Given a function f,f, we use the notation f(x)f(x) or dfdxdfdx to denote the derivative of f.f. Here we introduce notation for antiderivatives. If FF is an antiderivative of f,f, we say that F(x)+CF(x)+C is the most general antiderivative of ff and write

f(x)dx=F(x)+C.f(x)dx=F(x)+C.

The symbol is called an integral sign, and f(x)dxf(x)dx is called the indefinite integral of f.f.

Definition

Given a function f,f, the indefinite integral of f,f, denoted

f(x)dx,f(x)dx,

is the most general antiderivative of f.f. If FF is an antiderivative of f,f, then

f(x)dx=F(x)+C.f(x)dx=F(x)+C.

The expression f(x)f(x) is called the integrand and the variable xx is the variable of integration.

Given the terminology introduced in this definition, the act of finding the antiderivatives of a function ff is usually referred to as integrating f.f.

For a function ff and an antiderivative F,F, the functions F(x)+C,F(x)+C, where CC is any real number, is often referred to as the family of antiderivatives of f.f. For example, since x2x2 is an antiderivative of 2x2x and any antiderivative of 2x2x is of the form x2+C,x2+C, we write

2xdx=x2+C.2xdx=x2+C.

The collection of all functions of the form x2+C,x2+C, where CC is any real number, is known as the family of antiderivatives of 2x.2x. Figure 4.85 shows a graph of this family of antiderivatives.

The graphs for y = x2 + 2, y = x2 + 1, y = x2, y = x2 − 1, and y = x2 − 2 are shown.
Figure 4.85 The family of antiderivatives of 2x2x consists of all functions of the form x2+C,x2+C, where CC is any real number.

For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For example, for n1,n1,

xndx=xn+1n+1+C,xndx=xn+1n+1+C,

which comes directly from

ddx(xn+1n+1)=(n+1)xnn+1=xn.ddx(xn+1n+1)=(n+1)xnn+1=xn.

This fact is known as the power rule for integrals.

Theorem 4.15

Power Rule for Integrals

For n1,n1,

xndx=xn+1n+1+C.xndx=xn+1n+1+C.

Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the indefinite integrals for several common functions. A more complete list appears in Appendix B.

Differentiation Formula Indefinite Integral
ddx(k)=0ddx(k)=0 kdx=kx0dx=kx+Ckdx=kx0dx=kx+C
ddx(xn)=nxn1ddx(xn)=nxn1 xndn=xn+1n+1+Cxndn=xn+1n+1+C for n1n1
ddx(ln|x|)=1xddx(ln|x|)=1x 1xdx=ln|x|+C1xdx=ln|x|+C
ddx(ex)=exddx(ex)=ex exdx=ex+Cexdx=ex+C
ddx(sinx)=cosxddx(sinx)=cosx cosxdx=sinx+Ccosxdx=sinx+C
ddx(cosx)=sinxddx(cosx)=sinx sinxdx=cosx+Csinxdx=cosx+C
ddx(tanx)=sec2xddx(tanx)=sec2x sec2xdx=tanx+Csec2xdx=tanx+C
ddx(cscx)=cscxcotxddx(cscx)=cscxcotx cscxcotxdx=cscx+Ccscxcotxdx=cscx+C
ddx(secx)=secxtanxddx(secx)=secxtanx secxtanxdx=secx+Csecxtanxdx=secx+C
ddx(cotx)=csc2xddx(cotx)=csc2x csc2xdx=cotx+Ccsc2xdx=cotx+C
ddx(sin−1x)=11x2ddx(sin−1x)=11x2 11x2=sin−1x+C11x2=sin−1x+C
ddx(tan−1x)=11+x2ddx(tan−1x)=11+x2 11+x2dx=tan−1x+C11+x2dx=tan−1x+C
ddx(sec−1|x|)=1xx21ddx(sec−1|x|)=1xx21 1xx21dx=sec−1|x|+C1xx21dx=sec−1|x|+C
Table 4.13 Integration Formulas

From the definition of indefinite integral of f,f, we know

f(x)dx=F(x)+Cf(x)dx=F(x)+C

if and only if FF is an antiderivative of f.f. Therefore, when claiming that

f(x)dx=F(x)+Cf(x)dx=F(x)+C

it is important to check whether this statement is correct by verifying that F(x)=f(x).F(x)=f(x).

Example 4.51

Verifying an Indefinite Integral

Each of the following statements is of the form f(x)dx=F(x)+C.f(x)dx=F(x)+C. Verify that each statement is correct by showing that F(x)=f(x).F(x)=f(x).

  1. (x+ex)dx=x22+ex+C(x+ex)dx=x22+ex+C
  2. xexdx=xexex+Cxexdx=xexex+C

Solution

  1. Since
    ddx(x22+ex+C)=x+ex,ddx(x22+ex+C)=x+ex,

    the statement
    (x+ex)dx=x22+ex+C(x+ex)dx=x22+ex+C

    is correct.
    Note that we are verifying an indefinite integral for a sum. Furthermore, x22x22 and exex are antiderivatives of xx and ex,ex, respectively, and the sum of the antiderivatives is an antiderivative of the sum. We discuss this fact again later in this section.
  2. Using the product rule, we see that
    ddx(xexex+C)=ex+xexex=xex.ddx(xexex+C)=ex+xexex=xex.

    Therefore, the statement
    xexdx=xexex+Cxexdx=xexex+C

    is correct.
    Note that we are verifying an indefinite integral for a product. The antiderivative xexexxexex is not a product of the antiderivatives. Furthermore, the product of antiderivatives, x2ex/2x2ex/2 is not an antiderivative of xexxex since
    ddx(x2ex2)=xex+x2ex2xex.ddx(x2ex2)=xex+x2ex2xex.

    In general, the product of antiderivatives is not an antiderivative of a product.
Checkpoint 4.50

Verify that xcosxdx=xsinx+cosx+C.xcosxdx=xsinx+cosx+C.

In Table 4.13, we listed the indefinite integrals for many elementary functions. Let’s now turn our attention to evaluating indefinite integrals for more complicated functions. For example, consider finding an antiderivative of a sum f+g.f+g. In Example 4.51a. we showed that an antiderivative of the sum x+exx+ex is given by the sum (x22)+ex(x22)+ex—that is, an antiderivative of a sum is given by a sum of antiderivatives. This result was not specific to this example. In general, if FF and GG are antiderivatives of any functions ff and g,g, respectively, then

ddx(F(x)+G(x))=F(x)+G(x)=f(x)+g(x).ddx(F(x)+G(x))=F(x)+G(x)=f(x)+g(x).

Therefore, F(x)+G(x)F(x)+G(x) is an antiderivative of f(x)+g(x)f(x)+g(x) and we have

(f(x)+g(x))dx=F(x)+G(x)+C.(f(x)+g(x))dx=F(x)+G(x)+C.

Similarly,

(f(x)g(x))dx=F(x)G(x)+C.(f(x)g(x))dx=F(x)G(x)+C.

In addition, consider the task of finding an antiderivative of kf(x),kf(x), where kk is any real number. Since

ddx(kf(x))=kddxF(x)=kf(x)ddx(kf(x))=kddxF(x)=kf(x)

for any real number k,k, we conclude that

kf(x)dx=kF(x)+C.kf(x)dx=kF(x)+C.

These properties are summarized next.

Theorem 4.16

Properties of Indefinite Integrals

Let FF and GG be antiderivatives of ff and g,g, respectively, and let kk be any real number.

Sums and Differences

(f(x)±g(x))dx=F(x)±G(x)+C(f(x)±g(x))dx=F(x)±G(x)+C

Constant Multiples

kf(x)dx=kF(x)+Ckf(x)dx=kF(x)+C

From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with antiderivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (see Example 4.51b. for an example involving an antiderivative of a product.) We look at and address integrals involving these more complicated functions in Introduction to Integration. In the next example, we examine how to use this theorem to calculate the indefinite integrals of several functions.

Example 4.52

Evaluating Indefinite Integrals

Evaluate each of the following indefinite integrals:

  1. (5x37x2+3x+4)dx(5x37x2+3x+4)dx
  2. x2+4x3xdxx2+4x3xdx
  3. 41+x2dx41+x2dx
  4. tanxcosxdxtanxcosxdx

Solution

  1. Using Properties of Indefinite Integrals, we can integrate each of the four terms in the integrand separately. We obtain
    (5x37x2+3x+4)dx=5x3dx7x2dx+3xdx+4dx.(5x37x2+3x+4)dx=5x3dx7x2dx+3xdx+4dx.

    From the second part of Properties of Indefinite Integrals, each coefficient can be written in front of the integral sign, which gives
    5x3dx7x2dx+3xdx+4dx=5x3dx7x2dx+3xdx+41dx.5x3dx7x2dx+3xdx+4dx=5x3dx7x2dx+3xdx+41dx.
    Using the power rule for integrals, we conclude that
    (5x37x2+3x+4)dx=54x473x3+32x2+4x+C.(5x37x2+3x+4)dx=54x473x3+32x2+4x+C.
  2. Rewrite the integrand as
    x2+4x3x=x2x+4x3x=0.x2+4x3x=x2x+4x3x=0.

    Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
    (x+4x2/3)dx=xdx+4x−2/3dx=12x2+41(−23)+1x(−2/3)+1+C=12x2+12x1/3+C.(x+4x2/3)dx=xdx+4x−2/3dx=12x2+41(−23)+1x(−2/3)+1+C=12x2+12x1/3+C.
  3. Using Properties of Indefinite Integrals, write the integral as
    411+x2dx.411+x2dx.

    Then, use the fact that tan−1(x)tan−1(x) is an antiderivative of 1(1+x2)1(1+x2) to conclude that
    41+x2dx=4tan−1(x)+C.41+x2dx=4tan−1(x)+C.
  4. Rewrite the integrand as
    tanxcosx=sinxcosxcosx=sinx.tanxcosx=sinxcosxcosx=sinx.

    Therefore,
    tanxcosx=sinx=cosx+C.tanxcosx=sinx=cosx+C.
Checkpoint 4.51

Evaluate (4x35x2+x7)dx.(4x35x2+x7)dx.

Initial-Value Problems

We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.

A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation

dydx=f(x)dydx=f(x)
4.9

is a simple example of a differential equation. Solving this equation means finding a function yy with a derivative f.f. Therefore, the solutions of Equation 4.9 are the antiderivatives of f.f. If FF is one antiderivative of f,f, every function of the form y=F(x)+Cy=F(x)+C is a solution of that differential equation. For example, the solutions of

dydx=6x2dydx=6x2

are given by

y=6x2dx=2x3+C.y=6x2dx=2x3+C.

Sometimes we are interested in determining whether a particular solution curve passes through a certain point (x0,y0)(x0,y0)—that is, y(x0)=y0.y(x0)=y0. The problem of finding a function yy that satisfies a differential equation

dydx=f(x)dydx=f(x)
4.10

with the additional condition

y(x0)=y0y(x0)=y0
4.11

is an example of an initial-value problem. The condition y(x0)=y0y(x0)=y0 is known as an initial condition. For example, looking for a function yy that satisfies the differential equation

dydx=6x2dydx=6x2

and the initial condition

y(1)=5y(1)=5

is an example of an initial-value problem. Since the solutions of the differential equation are y=2x3+C,y=2x3+C, to find a function yy that also satisfies the initial condition, we need to find CC such that y(1)=2(1)3+C=5.y(1)=2(1)3+C=5. From this equation, we see that C=3,C=3, and we conclude that y=2x3+3y=2x3+3 is the solution of this initial-value problem as shown in the following graph.

The graphs for y = 2x3 + 6, y = 2x3 + 3, y = 2x3, and y = 2x3 − 3 are shown.
Figure 4.86 Some of the solution curves of the differential equation dydx=6x2dydx=6x2 are displayed. The function y=2x3+3y=2x3+3 satisfies the differential equation and the initial condition y(1)=5.y(1)=5.

Example 4.53

Solving an Initial-Value Problem

Solve the initial-value problem

dydx=sinx,y(0)=5.dydx=sinx,y(0)=5.

Solution

First we need to solve the differential equation. If dydx=sinx,dydx=sinx, then

y=sin(x)dx=cosx+C.y=sin(x)dx=cosx+C.

Next we need to look for a solution yy that satisfies the initial condition. The initial condition y(0)=5y(0)=5 means we need a constant CC such that cosx+C=5.cosx+C=5. Therefore,

C=5+cos(0)=6.C=5+cos(0)=6.

The solution of the initial-value problem is y=cosx+6.y=cosx+6.

Checkpoint 4.52

Solve the initial value problem dydx=3x−2,y(1)=2.dydx=3x−2,y(1)=2.

Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function v(t)v(t) is the derivative of a position function s(t),s(t), and the acceleration a(t)a(t) is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.

Example 4.54

Decelerating Car

A car is traveling at the rate of 8888 ft/sec (60(60 mph) when the brakes are applied. The car begins decelerating at a constant rate of 1515 ft/sec2.

  1. How many seconds elapse before the car stops?
  2. How far does the car travel during that time?

Solution

  1. First we introduce variables for this problem. Let tt be the time (in seconds) after the brakes are first applied. Let a(t)a(t) be the acceleration of the car (in feet per seconds squared) at time t.t. Let v(t)v(t) be the velocity of the car (in feet per second) at time t.t. Let s(t)s(t) be the car’s position (in feet) beyond the point where the brakes are applied at time t.t.
    The car is traveling at a rate of 88ft/sec.88ft/sec. Therefore, the initial velocity is v(0)=88v(0)=88 ft/sec. Since the car is decelerating, the acceleration is
    a(t)=−15ft/s2.a(t)=−15ft/s2.

    The acceleration is the derivative of the velocity,
    v(t)=−15.v(t)=−15.

    Therefore, we have an initial-value problem to solve:
    v(t)=−15,v(0)=88.v(t)=−15,v(0)=88.

    Integrating, we find that
    v(t)=−15t+C.v(t)=−15t+C.

    Since v(0)=88,C=88.v(0)=88,C=88. Thus, the velocity function is
    v(t)=−15t+88.v(t)=−15t+88.

    To find how long it takes for the car to stop, we need to find the time tt such that the velocity is zero. Solving −15t+88=0,−15t+88=0, we obtain t=8815t=8815 sec.
  2. To find how far the car travels during this time, we need to find the position of the car after 88158815 sec. We know the velocity v(t)v(t) is the derivative of the position s(t).s(t). Consider the initial position to be s(0)=0.s(0)=0. Therefore, we need to solve the initial-value problem
    s(t)=−15t+88,s(0)=0.s(t)=−15t+88,s(0)=0.

    Integrating, we have
    s(t)=152t2+88t+C.s(t)=152t2+88t+C.

    Since s(0)=0,s(0)=0, the constant is C=0.C=0. Therefore, the position function is
    s(t)=152t2+88t.s(t)=152t2+88t.

    After t=8815t=8815 sec, the position is s(8815)258.133s(8815)258.133 ft.
Checkpoint 4.53

Suppose the car is traveling at the rate of 4444 ft/sec. How long does it take for the car to stop? How far will the car travel?

Section 4.10 Exercises

For the following exercises, show that F(x)F(x) are antiderivatives of f(x).f(x).

465.

F(x)=5x3+2x2+3x+1,f(x)=15x2+4x+3F(x)=5x3+2x2+3x+1,f(x)=15x2+4x+3

466.

F(x)=x2+4x+1,f(x)=2x+4F(x)=x2+4x+1,f(x)=2x+4

467.

F(x)=x2ex,f(x)=ex(x2+2x)F(x)=x2ex,f(x)=ex(x2+2x)

468.

F(x)=cosx,f(x)=sinxF(x)=cosx,f(x)=sinx

469.

F(x)=ex,f(x)=exF(x)=ex,f(x)=ex

For the following exercises, find the antiderivative of the function.

470.

f(x)=1x2+xf(x)=1x2+x

471.

f(x)=ex3x2+sinxf(x)=ex3x2+sinx

472.

f(x)=ex+3xx2f(x)=ex+3xx2

473.

f(x)=x1+4sin(2x)f(x)=x1+4sin(2x)

For the following exercises, find the antiderivative F(x)F(x) of each function f(x).f(x).

474.

f(x)=5x4+4x5f(x)=5x4+4x5

475.

f(x)=x+12x2f(x)=x+12x2

476.

f(x)=1xf(x)=1x

477.

f(x)=(x)3f(x)=(x)3

478.

f(x)=x1/3+(2x)1/3f(x)=x1/3+(2x)1/3

479.

f(x)=x1/3x2/3f(x)=x1/3x2/3

480.

f(x)=2sin(x)+sin(2x)f(x)=2sin(x)+sin(2x)

481.

f(x)=sec2(x)+1f(x)=sec2(x)+1

482.

f(x)=sinxcosxf(x)=sinxcosx

483.

f(x)=sin2(x)cos(x)f(x)=sin2(x)cos(x)

484.

f(x)=0f(x)=0

485.

f(x)=12csc2(x)+1x2f(x)=12csc2(x)+1x2

486.

f(x)=cscxcotx+3xf(x)=cscxcotx+3x

487.

f(x)=4cscxcotxsecxtanxf(x)=4cscxcotxsecxtanx

488.

f(x)=8secx(secx4tanx)f(x)=8secx(secx4tanx)

489.

f(x)=12e−4x+sinxf(x)=12e−4x+sinx

For the following exercises, evaluate the integral.

490.

(−1)dx(−1)dx

491.

sinxdxsinxdx

492.

(4x+x)dx(4x+x)dx

493.

3x2+2x2dx3x2+2x2dx

494.

(secxtanx+4x)dx(secxtanx+4x)dx

495.

(4x+x4)dx(4x+x4)dx

496.

(x−1/3x2/3)dx(x−1/3x2/3)dx

497.

14x3+2x+1x3dx14x3+2x+1x3dx

498.

(ex+ex)dx(ex+ex)dx

For the following exercises, solve the initial value problem.

499.

f(x)=x−3,f(1)=1f(x)=x−3,f(1)=1

500.

f(x)=x+x2,f(0)=2f(x)=x+x2,f(0)=2

501.

f(x)=cosx+sec2(x),f(π4)=2+22f(x)=cosx+sec2(x),f(π4)=2+22

502.

f(x)=x38x2+16x+1,f(0)=0f(x)=x38x2+16x+1,f(0)=0

503.

f(x)=2x2x22,f(1)=0f(x)=2x2x22,f(1)=0

For the following exercises, find two possible functions ff given the second- or third-order derivatives.

504.

f(x)=x2+2f(x)=x2+2

505.

f(x)=exf(x)=ex

506.

f(x)=1+xf(x)=1+x

507.

f(x)=cosxf(x)=cosx

508.

f(x)=8e−2xsinxf(x)=8e−2xsinx

509.

A car is being driven at a rate of 4040 mph when the brakes are applied. The car decelerates at a constant rate of 1010 ft/sec2. How long before the car stops?

510.

In the preceding problem, calculate how far the car travels in the time it takes to stop.

511.

You are merging onto the freeway, accelerating at a constant rate of 1212 ft/sec2. How long does it take you to reach merging speed at 6060 mph?

512.

Based on the previous problem, how far does the car travel to reach merging speed?

513.

A car company wants to ensure its newest model can stop in 88 sec when traveling at 7575 mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

514.

A car company wants to ensure its newest model can stop in less than 450450 ft when traveling at 6060 mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.

For the following exercises, find the antiderivative of the function, assuming F(0)=0.F(0)=0.

515.

[T] f(x)=x2+2f(x)=x2+2

516.

[T] f(x)=4xxf(x)=4xx

517.

[T] f(x)=sinx+2xf(x)=sinx+2x

518.

[T] f(x)=exf(x)=ex

519.

[T] f(x)=1(x+1)2f(x)=1(x+1)2

520.

[T] f(x)=e−2x+3x2f(x)=e−2x+3x2

For the following exercises, determine whether the statement is true or false. Either prove it is true or find a counterexample if it is false.

521.

If f(x)f(x) is the antiderivative of v(x),v(x), then 2f(x)2f(x) is the antiderivative of 2v(x).2v(x).

522.

If f(x)f(x) is the antiderivative of v(x),v(x), then f(2x)f(2x) is the antiderivative of v(2x).v(2x).

523.

If f(x)f(x) is the antiderivative of v(x),v(x), then f(x)+1f(x)+1 is the antiderivative of v(x)+1.v(x)+1.

524.

If f(x)f(x) is the antiderivative of v(x),v(x), then (f(x))2(f(x))2 is the antiderivative of (v(x))2.(v(x))2.

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