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Calculus Volume 2

3.7 Improper Integrals

Calculus Volume 23.7 Improper Integrals
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  1. Preface
  2. 1 Integration
    1. Introduction
    2. 1.1 Approximating Areas
    3. 1.2 The Definite Integral
    4. 1.3 The Fundamental Theorem of Calculus
    5. 1.4 Integration Formulas and the Net Change Theorem
    6. 1.5 Substitution
    7. 1.6 Integrals Involving Exponential and Logarithmic Functions
    8. 1.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  3. 2 Applications of Integration
    1. Introduction
    2. 2.1 Areas between Curves
    3. 2.2 Determining Volumes by Slicing
    4. 2.3 Volumes of Revolution: Cylindrical Shells
    5. 2.4 Arc Length of a Curve and Surface Area
    6. 2.5 Physical Applications
    7. 2.6 Moments and Centers of Mass
    8. 2.7 Integrals, Exponential Functions, and Logarithms
    9. 2.8 Exponential Growth and Decay
    10. 2.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  4. 3 Techniques of Integration
    1. Introduction
    2. 3.1 Integration by Parts
    3. 3.2 Trigonometric Integrals
    4. 3.3 Trigonometric Substitution
    5. 3.4 Partial Fractions
    6. 3.5 Other Strategies for Integration
    7. 3.6 Numerical Integration
    8. 3.7 Improper Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  5. 4 Introduction to Differential Equations
    1. Introduction
    2. 4.1 Basics of Differential Equations
    3. 4.2 Direction Fields and Numerical Methods
    4. 4.3 Separable Equations
    5. 4.4 The Logistic Equation
    6. 4.5 First-order Linear Equations
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  6. 5 Sequences and Series
    1. Introduction
    2. 5.1 Sequences
    3. 5.2 Infinite Series
    4. 5.3 The Divergence and Integral Tests
    5. 5.4 Comparison Tests
    6. 5.5 Alternating Series
    7. 5.6 Ratio and Root Tests
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Chapter Review Exercises
  7. 6 Power Series
    1. Introduction
    2. 6.1 Power Series and Functions
    3. 6.2 Properties of Power Series
    4. 6.3 Taylor and Maclaurin Series
    5. 6.4 Working with Taylor Series
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  8. 7 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 7.1 Parametric Equations
    3. 7.2 Calculus of Parametric Curves
    4. 7.3 Polar Coordinates
    5. 7.4 Area and Arc Length in Polar Coordinates
    6. 7.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 3.7.1. Evaluate an integral over an infinite interval.
  • 3.7.2. Evaluate an integral over a closed interval with an infinite discontinuity within the interval.
  • 3.7.3. Use the comparison theorem to determine whether a definite integral is convergent.

Is the area between the graph of f(x)=1xf(x)=1x and the x-axis over the interval [1,+)[1,+) finite or infinite? If this same region is revolved about the x-axis, is the volume finite or infinite? Surprisingly, the area of the region described is infinite, but the volume of the solid obtained by revolving this region about the x-axis is finite.

In this section, we define integrals over an infinite interval as well as integrals of functions containing a discontinuity on the interval. Integrals of these types are called improper integrals. We examine several techniques for evaluating improper integrals, all of which involve taking limits.

Integrating over an Infinite Interval

How should we go about defining an integral of the type a+f(x)dx?a+f(x)dx? We can integrate atf(x)dxatf(x)dx for any value of t,t, so it is reasonable to look at the behavior of this integral as we substitute larger values of t.t. Figure 3.17 shows that atf(x)dxatf(x)dx may be interpreted as area for various values of t.t. In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.
Figure 3.17 To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.

Definition

  1. Let f(x)f(x) be continuous over an interval of the form [a,+).[a,+). Then
    a+f(x)dx=limt+atf(x)dx,a+f(x)dx=limt+atf(x)dx,
    3.16

    provided this limit exists.
  2. Let f(x)f(x) be continuous over an interval of the form (,b].(,b]. Then
    bf(x)dx=limttbf(x)dx,bf(x)dx=limttbf(x)dx,
    3.17

    provided this limit exists.
    In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
  3. Let f(x)f(x) be continuous over (,+).(,+). Then
    +f(x)dx=0f(x)dx+0+f(x)dx,+f(x)dx=0f(x)dx+0+f(x)dx,
    3.18

    provided that 0f(x)dx0f(x)dx and 0+f(x)dx0+f(x)dx both converge. If either of these two integrals diverge, then +f(x)dx+f(x)dx diverges. (It can be shown that, in fact, +f(x)dx=af(x)dx+a+f(x)dx+f(x)dx=af(x)dx+a+f(x)dx for any value of a.)a.)

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of f(x)=1xf(x)=1x and the xx-axis over the interval [1,+)[1,+) finite or infinite?

Example 3.47

Finding an Area

Determine whether the area between the graph of f(x)=1xf(x)=1x and the x-axis over the interval [1,+)[1,+) is finite or infinite.

Solution

We first do a quick sketch of the region in question, as shown in the following graph.

This figure is the graph of the function y = 1/x. It is a decreasing function with a vertical asymptote at the y-axis. In the first quadrant there is a shaded region under the curve bounded by x = 1 and x = 4.
Figure 3.18 We can find the area between the curve f(x)=1/xf(x)=1/x and the x-axis on an infinite interval.

We can see that the area of this region is given by A=11xdx.A=11xdx. Then we have

A=11xdx=limt+1t1xdxRewrite the improper integral as a limit.=limt+ln|x||1tFind the antiderivative.=limt+(ln|t|ln1)Evaluate the antiderivative.=+.Evaluate the limit.A=11xdx=limt+1t1xdxRewrite the improper integral as a limit.=limt+ln|x||1tFind the antiderivative.=limt+(ln|t|ln1)Evaluate the antiderivative.=+.Evaluate the limit.

Since the improper integral diverges to +,+, the area of the region is infinite.

Example 3.48

Finding a Volume

Find the volume of the solid obtained by revolving the region bounded by the graph of f(x)=1xf(x)=1x and the x-axis over the interval [1,+)[1,+) about the xx-axis.

Solution

The solid is shown in Figure 3.19. Using the disk method, we see that the volume V is

V=π1+1x2dx.V=π1+1x2dx.
This figure is the graph of the function y = 1/x. It is a decreasing function with a vertical asymptote at the y-axis. The graph shows a solid that has been generated by rotating the curve in the first quadrant around the x-axis.
Figure 3.19 The solid of revolution can be generated by rotating an infinite area about the x-axis.

Then we have

V=π1+1x2dx=πlimt+1t1x2dxRewrite as a limit.=πlimt+1x|1tFind the antiderivative.=πlimt+(1t+1)Evaluate the antiderivative.=π.V=π1+1x2dx=πlimt+1t1x2dxRewrite as a limit.=πlimt+1x|1tFind the antiderivative.=πlimt+(1t+1)Evaluate the antiderivative.=π.

The improper integral converges to π.π. Therefore, the volume of the solid of revolution is π.π.

In conclusion, although the area of the region between the x-axis and the graph of f(x)=1/xf(x)=1/x over the interval [1,+)[1,+) is infinite, the volume of the solid generated by revolving this region about the x-axis is finite. The solid generated is known as Gabriel’s Horn.

Media

Visit this website to read more about Gabriel’s Horn.

Example 3.49

Chapter Opener: Traffic Accidents in a City

This is a picture of a city street with a traffic signal. The picture has very busy lanes of traffic in both directions.
Figure 3.20 (credit: modification of work by David McKelvey, Flickr)

In the chapter opener, we stated the following problem: Suppose that at a busy intersection, traffic accidents occur at an average rate of one every three months. After residents complained, changes were made to the traffic lights at the intersection. It has now been eight months since the changes were made and there have been no accidents. Were the changes effective or is the 8-month interval without an accident a result of chance?

Probability theory tells us that if the average time between events is k,k, the probability that X,X, the time between events, is between aa and bb is given by

P(axb)=abf(x)dxwheref(x)={0ifx<0kekxifx0.P(axb)=abf(x)dxwheref(x)={0ifx<0kekxifx0.

Thus, if accidents are occurring at a rate of one every 3 months, then the probability that X,X, the time between accidents, is between aa and bb is given by

P(axb)=abf(x)dxwheref(x)={0ifx<03e−3xifx0.P(axb)=abf(x)dxwheref(x)={0ifx<03e−3xifx0.

To answer the question, we must compute P(X8)=8+3e−3xdxP(X8)=8+3e−3xdx and decide whether it is likely that 8 months could have passed without an accident if there had been no improvement in the traffic situation.

Solution

We need to calculate the probability as an improper integral:

P(X8)=8+3e−3xdx=limt+8t3e−3xdx=limt+e−3x|8t=limt+(e−3t+e−24)3.8×10−11.P(X8)=8+3e−3xdx=limt+8t3e−3xdx=limt+e−3x|8t=limt+(e−3t+e−24)3.8×10−11.

The value 3.8×10−113.8×10−11 represents the probability of no accidents in 8 months under the initial conditions. Since this value is very, very small, it is reasonable to conclude the changes were effective.

Example 3.50

Evaluating an Improper Integral over an Infinite Interval

Evaluate 01x2+4dx.01x2+4dx. State whether the improper integral converges or diverges.

Solution

Begin by rewriting 01x2+4dx01x2+4dx as a limit using Equation 3.17 from the definition. Thus,

01x2+4dx=limxt01x2+4dxRewrite as a limit. =limt12tan−1x2|t0Find the antiderivative. =12limt(tan−10tan−1t2)Evaluate the antiderivative. =π4.Evaluate the limit and simplify.01x2+4dx=limxt01x2+4dxRewrite as a limit. =limt12tan−1x2|t0Find the antiderivative. =12limt(tan−10tan−1t2)Evaluate the antiderivative. =π4.Evaluate the limit and simplify.

The improper integral converges to π4.π4.

Example 3.51

Evaluating an Improper Integral on (,+)(,+)

Evaluate +xexdx.+xexdx. State whether the improper integral converges or diverges.

Solution

Start by splitting up the integral:

+xexdx=0xexdx+0+xexdx.+xexdx=0xexdx+0+xexdx.

If either 0xexdx0xexdx or 0+xexdx0+xexdx diverges, then +xexdx+xexdx diverges. Compute each integral separately. For the first integral,

0xexdx=limtt0xexdxRewrite as a limit.=limt(xexex)|t0Use integration by parts to find theantiderivative. (Hereu=xanddv=ex.)=limt(−1tet+et)Evaluate the antiderivative.=−1.Evaluate the limit. Note:limttetisindeterminate of the form0·.Thus,limttet=limttet=limt−1et=limtet=0byL’Hôpital’s Rule.0xexdx=limtt0xexdxRewrite as a limit.=limt(xexex)|t0Use integration by parts to find theantiderivative. (Hereu=xanddv=ex.)=limt(−1tet+et)Evaluate the antiderivative.=−1.Evaluate the limit. Note:limttetisindeterminate of the form0·.Thus,limttet=limttet=limt−1et=limtet=0byL’Hôpital’s Rule.

The first improper integral converges. For the second integral,

0+xexdx=limt+0txexdxRewrite as a limit.=limt+(xexex)|0tFind the antiderivative.=limt+(tetet+1)Evaluate the antiderivative.=limt+((t1)et+1)Rewrite.(tetetis indeterminate.)=+.Evaluate the limit.0+xexdx=limt+0txexdxRewrite as a limit.=limt+(xexex)|0tFind the antiderivative.=limt+(tetet+1)Evaluate the antiderivative.=limt+((t1)et+1)Rewrite.(tetetis indeterminate.)=+.Evaluate the limit.

Thus, 0+xexdx0+xexdx diverges. Since this integral diverges, +xexdx+xexdx diverges as well.

Checkpoint 3.27

Evaluate −3+exdx.−3+exdx. State whether the improper integral converges or diverges.

Integrating a Discontinuous Integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form abf(x)dx,abf(x)dx, where f(x)f(x) is continuous over [a,b)[a,b) and discontinuous at b.b. Since the function f(x)f(x) is continuous over [a,t][a,t] for all values of tt satisfying a<t<b,a<t<b, the integral atf(x)dxatf(x)dx is defined for all such values of t.t. Thus, it makes sense to consider the values of atf(x)dxatf(x)dx as tt approaches bb for a<t<b.a<t<b. That is, we define abf(x)dx=limtbatf(x)dx,abf(x)dx=limtbatf(x)dx, provided this limit exists. Figure 3.21 illustrates atf(x)dxatf(x)dx as areas of regions for values of tt approaching b.b.

This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and increasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. There is also a vertical asymptote at x = b. The region in the first curve is small, and progressively gets wider under the second and third graph as t gets further from a, and closer to b on the x-axis.
Figure 3.21 As t approaches b from the left, the value of the area from a to t approaches the area from a to b.

We use a similar approach to define abf(x)dx,abf(x)dx, where f(x)f(x) is continuous over (a,b](a,b] and discontinuous at a.a. We now proceed with a formal definition.

Definition

  1. Let f(x)f(x) be continuous over [a,b).[a,b). Then,
    abf(x)dx=limtbatf(x)dx.abf(x)dx=limtbatf(x)dx.
    3.19
  2. Let f(x)f(x) be continuous over (a,b].(a,b]. Then,
    abf(x)dx=limta+tbf(x)dx.abf(x)dx=limta+tbf(x)dx.
    3.20

    In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
  3. If f(x)f(x) is continuous over [a,b][a,b] except at a point cc in (a,b),(a,b), then
    abf(x)dx=acf(x)dx+cbf(x)dx,abf(x)dx=acf(x)dx+cbf(x)dx,
    3.21

    provided both acf(x)dxacf(x)dx and cbf(x)dxcbf(x)dx converge. If either of these integrals diverges, then abf(x)dxabf(x)dx diverges.

The following examples demonstrate the application of this definition.

Example 3.52

Integrating a Discontinuous Integrand

Evaluate 0414xdx,0414xdx, if possible. State whether the integral converges or diverges.

Solution

The function f(x)=14xf(x)=14x is continuous over [0,4)[0,4) and discontinuous at 4. Using Equation 3.19 from the definition, rewrite 0414xdx0414xdx as a limit:

0414xdx=limt40t14xdxRewrite as a limit.=limt4(−24x)|0tFind the antiderivative.=limt4(−24t+4)Evaluate the antiderivative.=4.Evaluate the limit.0414xdx=limt40t14xdxRewrite as a limit.=limt4(−24x)|0tFind the antiderivative.=limt4(−24t+4)Evaluate the antiderivative.=4.Evaluate the limit.

The improper integral converges.

Example 3.53

Integrating a Discontinuous Integrand

Evaluate 02xlnxdx.02xlnxdx. State whether the integral converges or diverges.

Solution

Since f(x)=xlnxf(x)=xlnx is continuous over (0,2](0,2] and is discontinuous at zero, we can rewrite the integral in limit form using Equation 3.20:

02xlnxdx=limt0+t2xlnxdxRewrite as a limit.=limt0+(12x2lnx14x2)|t2Evaluatexlnxdxusing integration by partswithu=lnxanddv=x.=limt0+(2ln2112t2lnt+14t2).Evaluate the antiderivative.=2ln21.Evaluate the limit.limt0+t2lntis indeterminate.To evaluate it, rewrite as a quotient and applyL’Hôpital’s rule.02xlnxdx=limt0+t2xlnxdxRewrite as a limit.=limt0+(12x2lnx14x2)|t2Evaluatexlnxdxusing integration by partswithu=lnxanddv=x.=limt0+(2ln2112t2lnt+14t2).Evaluate the antiderivative.=2ln21.Evaluate the limit.limt0+t2lntis indeterminate.To evaluate it, rewrite as a quotient and applyL’Hôpital’s rule.

The improper integral converges.

Example 3.54

Integrating a Discontinuous Integrand

Evaluate −111x3dx.−111x3dx. State whether the improper integral converges or diverges.

Solution

Since f(x)=1/x3f(x)=1/x3 is discontinuous at zero, using Equation 3.21, we can write

−111x3dx=−101x3dx+011x3dx.−111x3dx=−101x3dx+011x3dx.

If either of the two integrals diverges, then the original integral diverges. Begin with −101x3dx:−101x3dx:

−101x3dx=limt0−1t1x3dxRewrite as a limit.=limt0(12x2)|−1tFind the antiderivative.=limt0(12t2+12)Evaluate the antiderivative.=+.Evaluate the limit.−101x3dx=limt0−1t1x3dxRewrite as a limit.=limt0(12x2)|−1tFind the antiderivative.=limt0(12t2+12)Evaluate the antiderivative.=+.Evaluate the limit.

Therefore, −101x3dx−101x3dx diverges. Since −101x3dx−101x3dx diverges, −111x3dx−111x3dx diverges.

Checkpoint 3.28

Evaluate 021xdx.021xdx. State whether the integral converges or diverges.

A Comparison Theorem

It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with another carefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider two continuous functions f(x)f(x) and g(x)g(x) satisfying 0f(x)g(x)0f(x)g(x) for xaxa (Figure 3.22). In this case, we may view integrals of these functions over intervals of the form [a,t][a,t] as areas, so we have the relationship

0atf(x)dxatg(x)dxforta.0atf(x)dxatg(x)dxforta.
This figure has two graphs. The graphs are f(x) and g(x). The first graph f(x) is a decreasing, non-negative function with a horizontal asymptote at the x-axis. It has a sharper bend in the curve compared to g(x). The graph of g(x) is a decreasing, non-negative function with a horizontal asymptote at the x-axis.
Figure 3.22 If 0f(x)g(x)0f(x)g(x) for xa,xa, then for ta,ta, atf(x)dxatg(x)dx.atf(x)dxatg(x)dx.

Thus, if

a+f(x)dx=limt+atf(x)dx=+,a+f(x)dx=limt+atf(x)dx=+,

then

a+g(x)dx=limt+atg(x)dx=+a+g(x)dx=limt+atg(x)dx=+ as well. That is, if the area of the region between the graph of f(x)f(x) and the x-axis over [a,+)[a,+) is infinite, then the area of the region between the graph of g(x)g(x) and the x-axis over [a,+)[a,+) is infinite too.

On the other hand, if

a+g(x)dx=limt+atg(x)dx=La+g(x)dx=limt+atg(x)dx=L for some real number L,L, then

a+f(x)dx=limt+atf(x)dxa+f(x)dx=limt+atf(x)dx must converge to some value less than or equal to L,L, since atf(x)dxatf(x)dx increases as tt increases and atf(x)dxLatf(x)dxL for all ta.ta.

If the area of the region between the graph of g(x)g(x) and the x-axis over [a,+)[a,+) is finite, then the area of the region between the graph of f(x)f(x) and the x-axis over [a,+)[a,+) is also finite.

These conclusions are summarized in the following theorem.

Theorem 3.7

A Comparison Theorem

Let f(x)f(x) and g(x)g(x) be continuous over [a,+).[a,+). Assume that 0f(x)g(x)0f(x)g(x) for xa.xa.

  1. If a+f(x)dx=limt+atf(x)dx=+,a+f(x)dx=limt+atf(x)dx=+, then a+g(x)dx=limt+atg(x)dx=+.a+g(x)dx=limt+atg(x)dx=+.
  2. If a+g(x)dx=limt+atg(x)dx=L,a+g(x)dx=limt+atg(x)dx=L, where LL is a real number, then a+f(x)dx=limt+atf(x)dx=Ma+f(x)dx=limt+atf(x)dx=M for some real number ML.ML.

Example 3.55

Applying the Comparison Theorem

Use a comparison to show that 1+1xexdx1+1xexdx converges.

Solution

We can see that

01xex1ex=ex,01xex1ex=ex,

so if 1+exdx1+exdx converges, then so does 1+1xexdx.1+1xexdx. To evaluate 1+exdx,1+exdx, first rewrite it as a limit:

1+exdx=limt+1texdx=limt+(ex)|t1=limt+(et+e1)=e1.1+exdx=limt+1texdx=limt+(ex)|t1=limt+(et+e1)=e1.

Since 1+exdx1+exdx converges, so does 1+1xexdx.1+1xexdx.

Example 3.56

Applying the Comparison Theorem

Use the comparison theorem to show that 1+1xpdx1+1xpdx diverges for all p<1.p<1.

Solution

For p<1,p<1, 1/x1/(xp)1/x1/(xp) over [1,+).[1,+). In Example 3.47, we showed that 1+1xdx=+.1+1xdx=+. Therefore, 1+1xpdx1+1xpdx diverges for all p<1.p<1.

Checkpoint 3.29

Use a comparison to show that e+lnxxdxe+lnxxdx diverges.

Student Project

Laplace Transforms

In the last few chapters, we have looked at several ways to use integration for solving real-world problems. For this next project, we are going to explore a more advanced application of integration: integral transforms. Specifically, we describe the Laplace transform and some of its properties. The Laplace transform is used in engineering and physics to simplify the computations needed to solve some problems. It takes functions expressed in terms of time and transforms them to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solve problems in the frequency domain are much simpler than those required in the time domain.

The Laplace transform is defined in terms of an integral as

L{f(t)}=F(s)=0estf(t)dt.L{f(t)}=F(s)=0estf(t)dt.

Note that the input to a Laplace transform is a function of time, f(t),f(t), and the output is a function of frequency, F(s).F(s). Although many real-world examples require the use of complex numbers (involving the imaginary number i=−1),i=−1), in this project we limit ourselves to functions of real numbers.

Let’s start with a simple example. Here we calculate the Laplace transform of f(t)=tf(t)=t. We have

L{t}=0testdt.L{t}=0testdt.

This is an improper integral, so we express it in terms of a limit, which gives

L{t}=0testdt=limz0ztestdt.L{t}=0testdt=limz0ztestdt.

Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to t, so we treat the variable s as a constant. We have

u=tdv=estdtdu=dtv=1sest.u=tdv=estdtdu=dtv=1sest.

Then we obtain

limz0ztestdt=limz[[tsest]|0z+1s0zestdt]=limz[[zsesz+0se−0s]+1s0zestdt]=limz[[zsesz+0]1s[ests]|0z]=limz[[zsesz]1s2[esz1]]=limz[zsesz]limz[1s2esz]+limz1s2=00+1s2=1s2.limz0ztestdt=limz[[tsest]|0z+1s0zestdt]=limz[[zsesz+0se−0s]+1s0zestdt]=limz[[zsesz+0]1s[ests]|0z]=limz[[zsesz]1s2[esz1]]=limz[zsesz]limz[1s2esz]+limz1s2=00+1s2=1s2.
  1. Calculate the Laplace transform of f(t)=1.f(t)=1.
  2. Calculate the Laplace transform of f(t)=e−3t.f(t)=e−3t.
  3. Calculate the Laplace transform of f(t)=t2.f(t)=t2. (Note, you will have to integrate by parts twice.)
    Laplace transforms are often used to solve differential equations. Differential equations are not covered in detail until later in this book; but, for now, let’s look at the relationship between the Laplace transform of a function and the Laplace transform of its derivative.
    Let’s start with the definition of the Laplace transform. We have
    L{f(t)}=0estf(t)dt=limz0zestf(t)dt.L{f(t)}=0estf(t)dt=limz0zestf(t)dt.
  4. Use integration by parts to evaluate limz0zestf(t)dt.limz0zestf(t)dt. (Let u=f(t)u=f(t) and dv=estdt.)dv=estdt.)
    After integrating by parts and evaluating the limit, you should see that
    L{f(t)}=f(0)s+1s[L{f(t)}].L{f(t)}=f(0)s+1s[L{f(t)}].

    Then,
    L{f(t)}=sL{f(t)}f(0).L{f(t)}=sL{f(t)}f(0).

    Thus, differentiation in the time domain simplifies to multiplication by s in the frequency domain.
    The final thing we look at in this project is how the Laplace transforms of f(t)f(t) and its antiderivative are related. Let g(t)=0tf(u)du.g(t)=0tf(u)du. Then,
    L{g(t)}=0estg(t)dt=limz0zestg(t)dt.L{g(t)}=0estg(t)dt=limz0zestg(t)dt.
  5. Use integration by parts to evaluate limz0zestg(t)dt.limz0zestg(t)dt. (Let u=g(t)u=g(t) and dv=estdt.dv=estdt. Note, by the way, that we have defined g(t),g(t), du=f(t)dt.)du=f(t)dt.)
    As you might expect, you should see that
    L{g(t)}=1s·L{f(t)}.L{g(t)}=1s·L{f(t)}.

    Integration in the time domain simplifies to division by s in the frequency domain.

Section 3.7 Exercises

Evaluate the following integrals. If the integral is not convergent, answer “divergent.”

347.

24dx(x3)224dx(x3)2

348.

014+x2dx014+x2dx

349.

0214x2dx0214x2dx

350.

11xlnxdx11xlnxdx

351.

1xexdx1xexdx

352.

xx2+1dxxx2+1dx

353.

Without integrating, determine whether the integral 11x3+1dx11x3+1dx converges or diverges by comparing the function f(x)=1x3+1f(x)=1x3+1 with g(x)=1x3.g(x)=1x3.

354.

Without integrating, determine whether the integral 11x+1dx11x+1dx converges or diverges.

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.

355.

0excosxdx0excosxdx

356.

1lnxxdx1lnxxdx

357.

01lnxxdx01lnxxdx

358.

01lnxdx01lnxdx

359.

1x2+1dx1x2+1dx

360.

15dxx115dxx1

361.

−22dx(1+x)2−22dx(1+x)2

362.

0exdx0exdx

363.

0sinxdx0sinxdx

364.

ex1+e2xdxex1+e2xdx

365.

01dxx301dxx3

366.

02dxx302dxx3

367.

−12dxx3−12dxx3

368.

01dx1x201dx1x2

369.

031x1dx031x1dx

370.

15x3dx15x3dx

371.

355(x4)2dx355(x4)2dx

Determine the convergence of each of the following integrals by comparison with the given integral. If the integral converges, find the number to which it converges.

372.

1dxx2+4x;1dxx2+4x; compare with 1dxx2.1dxx2.

373.

1dxx+1;1dxx+1; compare with 1dx2x.1dx2x.

Evaluate the integrals. If the integral diverges, answer “diverges.”

374.

1dxxe1dxxe

375.

01dxxπ01dxxπ

376.

01dx1x01dx1x

377.

01dx1x01dx1x

378.

0dxx2+10dxx2+1

379.

−11dx1x2−11dx1x2

380.

01lnxxdx01lnxxdx

381.

0eln(x)dx0eln(x)dx

382.

0xexdx0xexdx

383.

x(x2+1)2dxx(x2+1)2dx

384.

0exdx0exdx

Evaluate the improper integrals. Each of these integrals has an infinite discontinuity either at an endpoint or at an interior point of the interval.

385.

09dx9x09dx9x

386.

−271dxx2/3−271dxx2/3

387.

03dx9x203dx9x2

388.

624dttt236624dttt236

389.

04xln(4x)dx04xln(4x)dx

390.

03x9x2dx03x9x2dx

391.

Evaluate .5tdx1x2..5tdx1x2. (Be careful!) (Express your answer using three decimal places.)

392.

Evaluate 14dxx21.14dxx21. (Express the answer in exact form.)

393.

Evaluate 2dx(x21)3/2.2dx(x21)3/2.

394.

Find the area of the region in the first quadrant between the curve y=e−6xy=e−6x and the x-axis.

395.

Find the area of the region bounded by the curve y=7x2,y=7x2, the x-axis, and on the left by x=1.x=1.

396.

Find the area under the curve y=1(x+1)3/2,y=1(x+1)3/2, bounded on the left by x=3.x=3.

397.

Find the area under y=51+x2y=51+x2 in the first quadrant.

398.

Find the volume of the solid generated by revolving about the x-axis the region under the curve y=3xy=3x from x=1x=1 to x=.x=.

399.

Find the volume of the solid generated by revolving about the y-axis the region under the curve y=6e−2xy=6e−2x in the first quadrant.

400.

Find the volume of the solid generated by revolving about the x-axis the area under the curve y=3exy=3ex in the first quadrant.

The Laplace transform of a continuous function over the interval [0,)[0,) is defined by F(s)=0esxf(x)dxF(s)=0esxf(x)dx (see the Student Project). This definition is used to solve some important initial-value problems in differential equations, as discussed later. The domain of F is the set of all real numbers s such that the improper integral converges. Find the Laplace transform F of each of the following functions and give the domain of F.

401.

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

402.

f(x)=xf(x)=x

403.

f(x)=cos(2x)f(x)=cos(2x)

404.

f(x)=eaxf(x)=eax

405.

Use the formula for arc length to show that the circumference of the circle x2+y2=1x2+y2=1 is 2π.2π.

A function is a probability density function if it satisfies the following definition: f(t)dt=1.f(t)dt=1. The probability that a random variable x lies between a and b is given by P(axb)=abf(t)dt.P(axb)=abf(t)dt.

406.

Show that f(x)={0ifx<07e−7xifx0f(x)={0ifx<07e−7xifx0 is a probability density function.

407.

Find the probability that x is between 0 and 0.3. (Use the function defined in the preceding problem.) Use four-place decimal accuracy.

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