3.7 Improper Integrals

Learning Objectives

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

Is the area between the graph of \ds f\left(x\right)=\frac{1}{x} and the x-axis over the interval \ds \left[1,\infty \right) 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 \ds \int\limits_{a}^{\infty }f\left(x\right)\,dx ? We know how to evaluate \ds \int\limits_{a}^{t}f\left(x\right)\,dx for any value of \ds t, so it is reasonable to look at the behavior of this integral as we substitute larger values of \ds t. In Figure 1 below \ds \int\limits_{a}^{t}f\left(x\right)\,dx is interpreted as area below the graph of f(x) over an interval [a,t] for various values of \ds 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 1. 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 \ds f\left(x\right) be continuous over an interval \ds \left[a,\infty \right). Then

    \ds \int\limits_{a}^{\infty }f\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx ,

    provided this limit exists.
  2. Let \ds f\left(x\right) be continuous over an interval of the form \ds \left( - \infty ,b\right]. Then

    \ds \int\limits_{ - \infty }^{b}f\left(x\right)\,dx =\underset{t\to - \infty }{\text{lim}}\int\limits_{t}^{b}f\left(x\right)\,dx ,

    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 \ds f\left(x\right) be continuous over \ds \left( - \infty ,\infty \right). We define \ds \int\limits_{ - \infty }^{\infty }f\left(x\right)\,dx as

    \ds \int\limits_{ - \infty }^{\infty }f\left(x\right)\,dx =\int\limits_{ - \infty }^{0}f\left(x\right)\,dx +\int\limits_{0}^{\infty }f\left(x\right)\,dx ,

    provided that both \ds \int\limits_{ - \infty }^{0}f\left(x\right)\,dx and \ds \int\limits_{0}^{\infty }f\left(x\right)\,dx converge.

    If either of these two integrals is divergent, then \ds \int\limits_{ - \infty }^{\infty }f\left(x\right)\,dx diverges. (It can be shown that, in fact, \ds \int\limits_{ - \infty }^{\infty }f\left(x\right)\,dx =\int\limits_{ - \infty }^{a}f\left(x\right)\,dx +\int\limits_{a}^{\infty }f\left(x\right)\,dx for any value of \ds a.\right))

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of \ds f\left(x\right)=\frac{1}{x} and the \ds x-axis over the interval \ds \left[1,\infty \right) finite or infinite?

Finding an Area

Determine whether the area between the graph of \ds f\left(x\right)=\frac{1}{x} and the x-axis over the interval \ds \left[1,\infty \right) 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 2. We can find the area between the curve \ds f\left(x\right)=1\text{/}x and the x-axis on an infinite interval.

We can see that the area of this region is given by \ds A=\int\limits_{1}^{\infty }\frac{1}{x}\,dx . Then we have

\ds \begin{array}{ccccc}\hfill A&\ds =\int\limits_{1}^{\infty }\frac{1}{x}\,dx \hfill &\ds &\ds &\ds \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\int\limits_{1}^{t}\frac{1}{x}\,dx \hfill &\ds &\ds &\ds \text{Rewrite the improper integral as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\text{ln}|x|\Big|_1^t\hfill &\ds &\ds &\ds \text{Find the antiderivative.}\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left(\text{ln}|t|-\text{ln}\phantom{\rule{0.1em}{0ex}}(1)\right)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =\infty .\hfill &\ds &\ds &\ds \text{Evaluate the limit.}\hfill \end{array}

Since the improper integral diverges to \ds \infty , the area of the region is infinite.

Finding a Volume

Find the volume of the solid obtained by revolving the region bounded by the graph of \ds f\left(x\right)=\frac{1}{x} and the x-axis over the interval \ds \left[1,\infty \right) about the \ds x-axis.

Solution

The solid is shown in Figure 3 below. Using the disk method, we see that the volume V is

\ds V=\pi \int\limits_{1}^{\infty }\frac{1}{{x}^{2}}\,dx .
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. The solid of revolution can be generated by rotating an infinite area about the x-axis.

Then we have

\ds \begin{array}{ccccc}\hfill V&\ds =\pi \int\limits_{1}^{\infty }\frac{1}{{x}^{2}}\,dx \hfill &\ds &\ds &\ds \\[5mm]\ds &\ds =\pi \underset{t\to \infty }{\text{lim}}\int\limits_{1}^{t}\frac{1}{{x}^{2}}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\pi \underset{t\to \infty }{\text{lim}}\left(-\frac{1}{x}\right)\Big|_1^t\hfill &\ds &\ds &\ds \text{Find the antiderivative.}\hfill \\[5mm]\ds &\ds =\pi \underset{t\to \infty }{\text{lim}}\left( - \phantom{\rule{0.2em}{0ex}}\frac{1}{t}+1\right)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =\pi .\hfill &\ds &\ds &\ds \end{array}

The improper integral converges to \ds \pi . Therefore, the volume of the solid of revolution is \ds \pi .

In conclusion, although the area of the region between the x-axis and the graph of \ds f\left(x\right)=\frac1x over the interval \ds \left[1,\infty \right) 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.

You can read more about Gabriel’s Horn on Wikipedia.

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 4. (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 \ds k, the probability that \ds X, the time between events, is between \ds a and \ds b is given by

\ds P\left(a\le x\le b\right)=\int\limits_{a}^{b}f\left(x\right)\,dx \phantom{\rule{0.2em}{0ex}}\text{where}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=\left\{\begin{array}{ll}0\phantom{\rule{0.2em}{0ex}}\ &\text{if}\ \phantom{\rule{0.1em}{0ex}}x<0\\[3mm]\ds k{e}^{ - kx}\ &\text{if}\ \phantom{\rule{0.1em}{0ex}}x\ge 0\end{array}.

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

\ds P\left(a\le x\le b\right)=\int\limits_{a}^{b}f\left(x\right)\,dx \phantom{\rule{0.2em}{0ex}}\text{where}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=\left\{\begin{array}{ll}0\ \phantom{\rule{0.2em}{0ex}}&\text{if}\ \phantom{\rule{0.1em}{0ex}}x<0\\[3mm]\ds 3{e}^{-3x}\ &\text{if}\ \phantom{\rule{0.1em}{0ex}}x\ge 0\end{array}.

To answer the question, we must compute \ds P\left(X\ge 8\right)=\int\limits_{8}^{\infty }3{e}^{-3x}\,dx 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:

\ds \begin{array}{cc}\ds \hfill P\left(X\ge 8\right)&\ds =\int\limits_{8}^{\infty }3{e}^{-3x}\,dx \hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\int\limits_{8}^{t}3{e}^{-3x}\,dx \hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}{ - {e}^{-3x}}\Big|_{8}^{t}\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left( - {e}^{-3t}+{e}^{-24}\right)\hfill \\[5mm]\ds &\ds =e^{-24}\approx 3.8\phantom{\rule{0.2em}{0ex}}\cdot\phantom{\rule{0.2em}{0ex}}{10}^{-11}.\hfill \end{array}

The value \ds 3.8\phantom{\rule{0.2em}{0ex}}\cdot\phantom{\rule{0.2em}{0ex}}{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 that the changes were effective.

Evaluating an Improper Integral over an Infinite Interval

Evaluate \ds \int\limits_{ - \infty }^{0}\frac{1}{{x}^{2}+4}\,dx . State whether the improper integral converges or diverges.

Solution

Begin by rewriting \ds \int\limits_{ - \infty }^{0}\frac{1}{{x}^{2}+4}\,dx as a limit using the definition. We have:

\ds \begin{array}{ccccc}\hfill \ds\int\limits_{ - \infty }^{0}\frac{1}{{x}^{2}+4}\,dx &\ds =\underset{t\to - \infty }{\text{lim}}\int\limits_{t}^{0}\frac{1}{{x}^{2}+4}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to - \infty }{\text{lim}}\frac{1}{2}{\text{tan}}^{-1}\left(\frac{x}{2}\right)\Big|_t^0\hfill &\ds &\ds &\ds \text{Find the antiderivative.}\hfill \\[5mm]\ds &\ds =\frac{1}{2}\underset{t\to - \infty }{\text{lim}}\left({\text{tan}}^{-1}(0)-{\text{tan}}^{-1}\left(\frac{t}{2}\right)\right)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =\frac{\pi }{4}.\hfill &\ds &\ds &\ds \text{Evaluate the limit and simplify.}\hfill \end{array}

The improper integral converges to \ds \frac{\pi }{4}.

Evaluating an Improper Integral over \ds \left( - \infty ,\infty \right)

Evaluate \ds \int\limits_{ - \infty }^{\infty }x{e}^{x}\,dx . State whether the improper integral is convergent or divergent.

Solution

Start by splitting up the integral:

\ds \int\limits_{ - \infty }^{\infty }x{e}^{x}\,dx =\int\limits_{ - \infty }^{0}x{e}^{x}\,dx +\int\limits_{0}^{\infty }x{e}^{x}\,dx .

If either \ds \int\limits_{ - \infty }^{0}x{e}^{x}\,dx or \ds \int\limits_{0}^{\infty }x{e}^{x}\,dx diverges, then \ds \int\limits_{ - \infty }^{\infty }x{e}^{x}\,dx diverges. Compute each integral separately. For the first integral,

\ds \begin{array}{ccccc}\hfill \ds\int\limits_{ - \infty }^{0}x{e}^{x}\,dx &\ds =\underset{t\to - \infty }{\text{lim}}\int\limits_{t}^{0}x{e}^{x}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to - \infty }{\text{lim}}\left(x{e}^{x}-{e}^{x}\right)\Big|_t^0\hfill &\ds &\ds &\ds \begin{array}{l}\text{Use integration by parts to find the}\hfill \\[5mm]\ds \text{antiderivative. (Here}\phantom{\rule{0.2em}{0ex}}u=x\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dv={e}^{x}.)\hfill \end{array}\hfill \\[10mm]\ds &\ds =\underset{t\to - \infty }{\text{lim}}\left(-1-t{e}^{t}+{e}^{t}\right)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =-1.\hfill &\ds &\ds &\ds \begin{array}{c}\text{Evaluate the limit.}\phantom{\rule{0.2em}{0ex}}\mathit{\text{Note:}}\phantom{\rule{0.2em}{0ex}}\underset{t\to - \infty }{\text{lim}}t{e}^{t}\phantom{\rule{0.2em}{0ex}}\text{is}\hfill \\[5mm]\ds \text{is of indeterminate form}\phantom{\rule{0.2em}{0ex}}0\cdot \infty .\phantom{\rule{0.2em}{0ex}}\text{Thus,}\hfill \\[5mm]\ds \underset{t\to - \infty }{\text{lim}}t{e}^{t}=\underset{t\to - \infty }{\text{lim}}\frac{t}{{e}^{ - t}}=\underset{t\to - \infty }{\text{lim}}\ \frac{1}{-{e}^{ - t}}\hfill \\[5mm]=\underset{t\to - \infty }{\text{lim}}-{e}^{t}=0\phantom{\rule{0.2em}{0ex}}\ds \text{by the L'Hôpital's Rule.}\hfill \end{array}\hfill \end{array}

The first improper integral converges. For the second integral,

\ds \begin{array}{ccccc}\hfill \ds\int\limits_{0}^{\infty }x{e}^{x}\,dx &\ds =\underset{t\to \infty }{\text{lim}}\int\limits_{0}^{t}x{e}^{x}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left(x{e}^{x}-{e}^{x}\right)\Big|_0^t\hfill &\ds &\ds &\ds \text{Find the antiderivative.}\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left(t{e}^{t}-{e}^{t}+1\right)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left(\left(t-1\right){e}^{t}+1\right)\hfill &\ds &\ds &\ds \text{Rewrite.}\phantom{\rule{0.2em}{0ex}}\left(t{e}^{t}-{e}^{t}\phantom{\rule{0.2em}{0ex}}\text{is indeterminate.)}\hfill \\[5mm]\ds &\ds =\infty .\hfill &\ds &\ds &\ds \text{Evaluate the limit.}\hfill \end{array}

Thus, \ds \int\limits_{0}^{\infty }x{e}^{x}\,dx diverges. Since this integral diverges, \ds \int\limits_{ - \infty }^{\infty }x{e}^{x}\,dx diverges as well.

Evaluate \ds \int\limits_{-3}^{\infty }{e}^{ - x}\,dx . State whether the improper integral converges or diverges.

Answer

\ds {e}^{3}, converges

Hint

\ds \int\limits_{-3}^{\infty }{e}^{ - x}\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{-3}^{t}{e}^{ - x}\,dx

Integrating a Discontinuous Function

We now examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form \ds \int\limits_{a}^{b}f\left(x\right)\,dx , where \ds f\left(x\right) is continuous over \ds \left[a,b\right) and discontinuous at \ds b. Since the function \ds f\left(x\right) is continuous over \ds \left[a,t\right] for all \ds t satisfying \ds a<t<b, the integral \ds \int\limits_{a}^{t}f\left(x\right)\,dx is defined for all such values of \ds t. Hence, it makes sense to consider the values of \ds \int\limits_{a}^{t}f\left(x\right)\,dx as \ds t approaches \ds b for \ds a<t<b. That is, we define \ds \int\limits_{a}^{b}f\left(x\right)\,dx =\underset{t\to {b}^{-}}{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx , provided this limit exists. The figure below illustrates \ds \int\limits_{a}^{t}f\left(x\right)\,dx as areas of regions for values of \ds t approaching \ds 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 5. 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 \ds \int\limits_{a}^{b}f\left(x\right)\,dx , where \ds f\left(x\right) is continuous over \ds \left(a,b\right] and discontinuous at \ds a. We now proceed with a formal definition.

Definition

  1. Let \ds f\left(x\right) be continuous over \ds \left[a,b\right). Then,

    \ds \int\limits_{a}^{b}f\left(x\right)\,dx =\underset{t\to {b}^{-}}{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx .
  2. Let \ds f\left(x\right) be continuous over \ds \left(a,b\right]. Then,

    \ds \int\limits_{a}^{b}f\left(x\right)\,dx =\underset{t\to {a}^{+}}{\text{lim}}\int\limits_{t}^{b}f\left(x\right)\,dx .

    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 \ds f\left(x\right) is continuous over \ds \left[a,b\right] except at a point \ds c in \ds \left(a,b\right), then we define \ds\int\limits_a^b f(x)\,dx as

    \ds \int\limits_{a}^{b}f\left(x\right)\,dx =\int\limits_{a}^{c}f\left(x\right)\,dx +\int\limits_{c}^{b}f\left(x\right)\,dx ,

    provided both \ds \int\limits_{a}^{c}f\left(x\right)\,dx and \ds \int\limits_{c}^{b}f\left(x\right)\,dx converge. If either of these two integrals diverges, then \ds \int\limits_{a}^{b}f\left(x\right)\,dx diverges.

The following examples demonstrate the application of this definition.

Integrating a Discontinuous Integrand

Evaluate \ds \int\limits_{0}^{4}\frac{1}{\sqrt{4-x}}\,dx , if possible. State whether the integral converges or diverges.

Solution

The function \ds f\left(x\right)=\frac{1}{\sqrt{4-x}} is continuous over \ds \left[0,4\right) and discontinuous at 4. Using the above definition, we rewrite \ds \int\limits_{0}^{4}\frac{1}{\sqrt{4-x}}\,dx as a limit:

\ds \begin{array}{ccccc}\hfill \ds\int\limits_{0}^{4}\frac{1}{\sqrt{4-x}}\,dx &\ds =\underset{t\to {4}^{-}}{\text{lim}}\int\limits_{0}^{t}\frac{1}{\sqrt{4-x}}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to {4}^{-}}{\text{lim}}\big(-2\sqrt{4-x}\big)\Big|_0^t\hfill &\ds &\ds &\ds \text{Find the antiderivative.}\hfill \\[5mm]\ds &\ds =\underset{t\to {4}^{-}}{\text{lim}}\big(-2\sqrt{4-t}+4\big)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =4.\hfill &\ds &\ds &\ds \text{Evaluate the limit.}\hfill \end{array}

Because the limit exists, the improper integral converges.

Integrating a Discontinuous Integrand

Evaluate \ds \int\limits_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx . State whether the integral converges or diverges.

Solution

Since \ds f\left(x\right)=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x) is continuous over \ds \left(0,2\right] and is discontinuous at zero, we can rewrite the integral as a limit using the definition of improper integral of the corresponding type:

\ds \begin{array}{ccccc}\hfill \ds\int\limits_{0}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx &\ds =\underset{t\to {0}^{+}}{\text{lim}}\int\limits_{t}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to {0}^{+}}{\text{lim}}\left(\frac{1}{2}{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)-\frac{1}{4}{x}^{2}\right)\Big|_t^2\hfill &\ds &\ds &\ds \begin{array}{c}\text{Evaluate}\phantom{\rule{0.2em}{0ex}}{\int }^{\text{​}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx \phantom{\rule{0.2em}{0ex}}\text{using integration by parts}\hfill \\[5mm]\ds \text{with}\phantom{\rule{0.2em}{0ex}}u=\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}dv=x.\hfill \end{array}\hfill \\[5mm]\ds &\ds =\underset{t\to {0}^{+}}{\text{lim}}\left(2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1-\frac{1}{2}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}(t)+\frac{1}{4}{t}^{2}\right).\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2-1.\hfill &\ds &\ds &\ds \begin{array}{c}\text{Evaluate the limit.}\phantom{\rule{0.2em}{0ex}}\underset{t\to {0}^{+}}{\text{lim}}{t}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}(t)\phantom{\rule{0.2em}{0ex}}\text{is indeterminate.}\hfill \\[5mm]\ds \text{To evaluate it, rewrite as a quotient and apply}\hfill \\[5mm]\ds \text{L'Hôpital's rule.}\hfill \end{array}\hfill \end{array}

The improper integral converges.

Integrating a Discontinuous Integrand

Evaluate \ds \int\limits_{-1}^{1}\frac{1}{{x}^{3}}\,dx . State whether the improper integral converges or diverges.

Solution

Since \ds f\left(x\right)=\frac1{{x}^{3}} is continuous at every point of [-1,1] except zero, we use the corresponding definition to write

\ds \int\limits_{-1}^{1}\frac{1}{{x}^{3}}\,dx =\int\limits_{-1}^{0}\frac{1}{{x}^{3}}\,dx +\int\limits_{0}^{1}\frac{1}{{x}^{3}}\,dx .

Our integral converges if both integrals on the right converge. If either of the two integrals on the right diverges, then the original integral diverges as well. Begin with \ds \int\limits_{-1}^{0}\frac{1}{{x}^{3}}\,dx :

\ds \begin{array}{ccccc}\hfill \ds\int\limits_{-1}^{0}\frac{1}{{x}^{3}}\,dx &\ds =\underset{t\to {0}^{-}}{\text{lim}}\int\limits_{-1}^{t}\frac{1}{{x}^{3}}\,dx \hfill &\ds &\ds &\ds \text{Rewrite as a limit.}\hfill \\[5mm]\ds &\ds =\underset{t\to {0}^{-}}{\text{lim}}\left(-\frac{1}{2{x}^{2}}\right)\Big|_{-1}^t\hfill &\ds &\ds &\ds \text{Find the antiderivative.}\hfill \\[5mm]\ds &\ds =\underset{t\to {0}^{-}}{\text{lim}}\left(-\frac{1}{2{t}^{2}}+\frac{1}{2}\right)\hfill &\ds &\ds &\ds \text{Evaluate the antiderivative.}\hfill \\[5mm]\ds &\ds =-\infty .\hfill &\ds &\ds &\ds \text{Evaluate the limit.}\hfill \end{array}

Therefore, \ds \int\limits_{-1}^{0}\frac{1}{{x}^{3}}\,dx diverges, and hence \ds \int\limits_{-1}^{1}\frac{1}{{x}^{3}}\,dx diverges regardless of the behavior of \ds \int\limits_{0}^{1}\frac{1}{{x}^{3}}\,dx.

Evaluate \ds \int\limits_{0}^{1}\frac{1}{(1-x)^{3\text{/}2}}\,dx . State whether the integral converges or diverges.

Answer

\ds \infty , diverges

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 \ds f\left(x\right) and \ds g\left(x\right) satisfying \ds 0\le f\left(x\right)\le g\left(x\right) for \ds x\ge a. In this case, we may view integrals of these functions over intervals of the form \ds \left[a,t\right] as areas, so by the comparison property for definite integrals, we have the relationship

\ds 0\le \int\limits_{a}^{t}f\left(x\right)\,dx \le \int\limits_{a}^{t}g\left(x\right)\,dx \phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}t\ge a.
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 6. If \ds 0\le f\left(x\right)\le g\left(x\right) for \ds x\ge a, then for \ds t\ge a, \ds \int\limits_{a}^{t}f\left(x\right)\,dx \le \int\limits_{a}^{t}g\left(x\right)\,dx .

Thus, if\ds \int\limits_{a}^{\infty }f\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx =\infty, then

\ds \int\limits_{a}^{\infty }g\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}g\left(x\right)\,dx\ge \underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx=\infty as well.

That is, if the area of the region between the graph of \ds f\left(x\right) and the x-axis over \ds \left[a,\infty \right) is infinite, then the area of the region between the graph of \ds g\left(x\right) and the x-axis over \ds \left[a,\infty \right) is infinite too.

On the other hand, if

\ds \int\limits_{a}^{\infty }g\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}g\left(x\right)\,dx =L for some real number \ds L, then

\ds \int\limits_{a}^{\infty }f\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx must converge to some value less than or equal to \ds L, since \ds \int\limits_{a}^{t}f\left(x\right)\,dx increases as \ds t increases and \ds \int\limits_{a}^{t}f\left(x\right)\,dx \le L for all \ds t\ge a.

That is, if the area of the region between the graph of \ds g\left(x\right) and the x-axis over \ds \left[a,\infty \right) is finite, then the area of the region between the graph of \ds f\left(x\right) and the x-axis over \ds \left[a,\infty \right) is also finite.
These conclusions are summarized in the following theorem.

Comparison Theorem

Let \ds f\left(x\right) and \ds g\left(x\right) be continuous over \ds \left[a,\infty \right). Assume that \ds 0\le f\left(x\right)\le g\left(x\right) for \ds x\ge a.

  1. If \ds \int\limits_{a}^{\infty }f\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx =\infty , then \ds \int\limits_{a}^{\infty }g\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}g\left(x\right)\,dx =\infty .
  2. If \ds \int\limits_{a}^{\infty }g\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}g\left(x\right)\,dx =L, where \ds L is a real number, then \ds \int\limits_{a}^{\infty }f\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx =M for some real number \ds M\le L.

Applying the Comparison Theorem

Use the comparison theorem to show that \ds \int\limits_{1}^{\infty }\frac{1}{x{e}^{x}}\,dx converges.

Solution

The integrand is continuous over [1,\infty) and we can also see that  for x>1, \ds 0\le \frac{1}{x{e}^{x}}\le \frac{1}{{e}^{x}}={e}^{ - x},

so if \ds \int\limits_{1}^{\infty }{e}^{ - x}\,dx converges, then so does \ds \int\limits_{1}^{\infty }\frac{1}{x{e}^{x}}\,dx . To evaluate \ds \int\limits_{1}^{\infty }{e}^{ - x}\,dx , first rewrite it as a limit:

\ds \begin{array}{cc}\ds \hfill \int\limits_{1}^{\infty }{e}^{ - x}\,dx &\ds =\underset{t\to \infty }{\text{lim}}\int\limits_{1}^{t}{e}^{ - x}\,dx \hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left( - {e}^{ - x}\right)\Big|_1^t\hfill \\[5mm]\ds &\ds =\underset{t\to \infty }{\text{lim}}\left( - {e}^{ - t}+{e}^{1}\right)\hfill \\[5mm]\ds &\ds ={e}.\hfill \end{array}

Since the limit is finite, \ds \int\limits_{1}^{\infty }{e}^{ - x}\,dx converges, and hence, by the comparison theorem, so does \ds \int\limits_{1}^{\infty }\frac{1}{x{e}^{x}}\,dx .

Applying the Comparison Theorem

Use the comparison theorem to show that \ds \int\limits_{1}^{\infty }\frac{1}{{x}^{p}}\,dx diverges for all \ds p<1.

Solution

First we note that \ds\frac1{x^p} is continuous over [1,\infty). If \ds p<1, then \ds \frac1x\le \frac1{x^p} for all \ds x\in\left[1,\infty \right). We already showed that \ds \int\limits_{1}^{\infty }\frac{1}{x}\,dx =\infty . Therefore, by the comparison theorem,\ds \int\limits_{1}^{\infty }\frac{1}{{x}^{p}}\,dx diverges for all \ds p<1.

Use the comparison theorem to show that \ds \int\limits_{e}^{\infty }\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}{x}\,dx diverges.

Hint

\ds \frac{1}{x}\le \frac{\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}{x} on \ds \left[e,\infty \right)

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

\ds L\left\{f\left(t\right)\right\}=F\left(s\right)=\int\limits_{0}^{\infty }{e}^{ - st}f\left(t\right)dt.

Note that the input to a Laplace transform is a function of time, \ds f\left(t\right), and the output is a function of frequency, \ds F\left(s\right). Although many real-world examples require the use of complex numbers (involving the imaginary number \ds i=\sqrt{-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 \ds f\left(t\right)=t. We have

\ds L\left\{t\right\}=\int\limits_{0}^{\infty }t{e}^{ - st}dt.

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

\ds L\left\{t\right\}=\int\limits_{0}^{\infty }t{e}^{ - st}dt=\underset{z\to \infty }{\text{lim}}\int\limits_{0}^{z}t{e}^{ - st}dt.

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

\ds \begin{array}{cccccccc}\hfill u&\ds =\hfill &\ds t\hfill &\ds &\ds &\ds \hfill dv&\ds =\hfill &\ds {e}^{ - st}dt\hfill \\[2mm]\ds \hfill du&\ds =\hfill &\ds dt\hfill &\ds &\ds &\ds \hfill v&\ds =\hfill &\ds -\frac{1}{s}{e}^{ - st}.\hfill \end{array}

Then we obtain

\ds \begin{array}{cc}\ds \hfill \underset{z\to \infty }{\text{lim}}\int\limits_{0}^{z}t{e}^{ - st}dt&\ds =\underset{z\to \infty }{\text{lim}}\left[{\left[-\frac{t}{s}{e}^{ - st}\right]}\Big|_{0}^{z}+\frac{1}{s}\int\limits_{0}^{z}{e}^{ - st}dt\right]\hfill \\[5mm]\ds &\ds =\underset{z\to \infty }{\text{lim}}\left[\left[-\frac{z}{s}{e}^{ - sz}+\frac{0}{s}{e}^{-0s}\right]+\frac{1}{s}\int\limits_{0}^{z}{e}^{ - st}dt\right]\hfill \\[5mm]\ds &\ds =\underset{z\to \infty }{\text{lim}}\left[\left[-\frac{z}{s}{e}^{ - sz}+0\right]-\frac{1}{s}{\left[\frac{{e}^{ - st}}{s}\right]}\Big|_{0}^{z}\right]\hfill \\[5mm]\ds &\ds =\underset{z\to \infty }{\text{lim}}\left[\left[-\frac{z}{s}{e}^{ - sz}\right]-\frac{1}{{s}^{2}}\left[{e}^{ - sz}-1\right]\right]\hfill \\[5mm]\ds &\ds =\underset{z\to \infty }{\text{lim}}\left[-\frac{z}{s{e}^{sz}}\right]-\underset{z\to \infty }{\text{lim}}\left[\frac{1}{{s}^{2}{e}^{sz}}\right]+\underset{z\to \infty }{\text{lim}}\frac{1}{{s}^{2}}\hfill \\[5mm]\ds &\ds =0-0+\frac{1}{{s}^{2}}\hfill \\[5mm]\ds &\ds =\frac{1}{{s}^{2}}.\hfill \end{array}
Therefore, \ds L\{t\}=\frac1{s^2}.
  1. Calculate the Laplace transform of \ds f\left(t\right)=1.
  2. Calculate the Laplace transform of \ds f\left(t\right)={e}^{-3t}.
  3. Calculate the Laplace transform of \ds f\left(t\right)={t}^{2}. (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

    \ds L\left\{f\left(t\right)\right\}=\int\limits_{0}^{\infty }{e}^{ - st}f\left(t\right)dt=\underset{z\to \infty }{\text{lim}}\int\limits_{0}^{z}{e}^{ - st}f\left(t\right)dt.
  4. Use integration by parts to evaluate \ds \underset{z\to \infty }{\text{lim}}\int\limits_{0}^{z}{e}^{ - st}f\left(t\right)dt. (Let \ds u=f\left(t\right) and \ds dv={e}^{ - st}dt.)
     After integrating by parts and evaluating the limit, you should see that

    \ds L\left\{f\left(t\right)\right\}=\frac{f\left(0\right)}{s}+\frac{1}{s}\left[L\left\{{f}^{\prime }\left(t\right)\right\}\right].


     Then,

    \ds L\left\{{f}^{\prime }\left(t\right)\right\}=sL\left\{f\left(t\right)\right\}-f\left(0\right).


     It follows that 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 \ds f\left(t\right) and its antiderivative are related. Let \ds g\left(t\right)=\int\limits_{0}^{t}f\left(u\right)du. Then,

    \ds L\left\{g\left(t\right)\right\}=\int\limits_{0}^{\infty }{e}^{ - st}g\left(t\right)dt=\underset{z\to \infty }{\text{lim}}\int\limits_{0}^{z}{e}^{ - st}g\left(t\right)dt.
  5. Use integration by parts to evaluate \ds \underset{z\to \infty }{\text{lim}}\int\limits_{0}^{z}{e}^{ - st}g\left(t\right)dt. (Let \ds u=g\left(t\right) and \ds dv={e}^{ - st}dt. Note that from the way we have defined \ds g\left(t\right), \ds du=f\left(t\right)dt.\right))
     As you might expect, you should see that

    \ds L\left\{g\left(t\right)\right\}=\frac{1}{s}\cdot L\left\{f\left(t\right)\right\}.


    That is, integration in the time domain simplifies to division by s in the frequency domain.

Key Concepts

  • Integrals of functions over infinite intervals are defined in terms of limits.
  • Integrals of functions over an interval for which the function has a discontinuity at an endpoint may be defined in terms of limits.
  • The convergence or divergence of an improper integral may be determined by comparing it with the value of an improper integral for which the convergence or divergence is known.

Key Equations

  • Improper integrals
    \ds \begin{array}{c}\ds\int\limits_{a}^{\infty }f\left(x\right)\,dx =\underset{t\to \infty }{\text{lim}}\int\limits_{a}^{t}f\left(x\right)\,dx \hfill \\[5mm]\ds \int\limits_{ - \infty }^{b}f\left(x\right)\,dx =\underset{t\to - \infty }{\text{lim}}\int\limits_{t}^{b}f\left(x\right)\,dx \hfill \\[5mm]\ds \int\limits_{ - \infty }^{\infty }f\left(x\right)\,dx =\int\limits_{ - \infty }^{0}f\left(x\right)\,dx +\int\limits_{0}^{\infty }f\left(x\right)\,dx \hfill \end{array}

Exercises

Evaluate the following integrals, if possible. If the integral diverges, answer “divergent.”

1. \ds \int\limits_{2}^{4}\frac{\,dx }{{\left(x-3\right)}^{2}}

Answer

divergent

2. \ds \int\limits_{0}^{\infty }\frac{1}{4+{x}^{2}}\,dx

3. \ds \int\limits_{0}^{2}\frac{1}{\sqrt{4-{x}^{2}}}\,dx

Answer

\ds \frac{\pi }{2}

4. \ds \int\limits_{1}^{\infty }\frac{1}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}\,dx

5. \ds \int\limits_{1}^{\infty }x{e}^{ - x}\,dx

Answer

\ds \frac{2}{e}

6. \ds \int\limits_{ - \infty }^{\infty }\frac{x}{{x}^{2}+1}\,dx

7. Without integrating, determine whether the integral \ds \int\limits_{1}^{\infty }\frac{1}{\sqrt{{x}^{3}+1}}\,dx converges or diverges by comparing the integrand with \ds\frac{1}{\sqrt{{x}^{3}}}.

Answer

converges

8. Without integrating, determine whether the integral \ds \int\limits_{1}^{\infty }\frac{1}{\sqrt{x+1}}\,dx converges or diverges.

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

9. \ds \int\limits_{0}^{\infty }{e}^{ - x}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx

Answer

Converges to 1/2.

10. \ds \int\limits_{1}^{\infty }\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}{x}\,dx

11. \ds \int\limits_{0}^{1}\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}{\sqrt{x}}\,dx

Answer

−4

12. \ds \int\limits_{0}^{1}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx

13. \ds \int\limits_{ - \infty }^{\infty }\frac{1}{4{x}^{2}+1}\,dx

Answer

\ds \frac{\pi}2

14. \ds \int\limits_{1}^{5}\frac{\,dx }{\sqrt{x-1}}

15. \ds \int\limits_{-2}^{2}\frac{\,dx }{{\left(1+x\right)}^{2}}

Answer

diverges

16. \ds \int\limits_{0}^{\infty }{e}^{ - x}\,dx

17. \ds \int\limits_{0}^{\infty }\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\,dx

Answer

diverges

18. \ds \int\limits_{ - \infty }^{\infty }\frac{{e}^{x}}{1+{e}^{2x}}\,dx

19. \ds \int\limits_{0}^{1}\frac{\,dx }{\sqrt[3]{x}}

Answer

\ds\frac32

20. \ds \int\limits_{0}^{2}\frac{\,dx }{x\sqrt x}

21. \ds \int\limits_{-1}^{2}\frac{\,dx }{{x}^{3}}

Answer

diverges

22. \ds \int\limits_{0}^{1}\frac{\,dx }{({1-{x}^{2}})^{3\text{/}2}}

23. \ds \int\limits_{0}^{3}\frac{1}{x-1}\,dx

Answer

diverges

24. \ds \int\limits_{1}^{\infty }\frac{x+1}{{x}^{5}}\,dx

25. \ds \int\limits_{3}^{5}\frac{5}{{\left(x-4\right)}^{6}}\,dx

Answer

diverges

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.

26. \ds \int\limits_{1}^{\infty }\frac{\,dx }{{x}^{2}+4x}; compare with \ds \int\limits_{1}^{\infty }\frac{\,dx }{{x}^{2}}.

27. \ds \int\limits_{1}^{\infty }\frac{\,dx }{\sqrt{x}+1}; compare with \ds \int\limits_{1}^{\infty }\frac{\,dx }{2\sqrt{x}}.

Answer

Both integrals diverge.

Use the comparison theorem to determine whether the given improper integral converges or diverges.

28. \ds \int\limits_{1}^{\infty }\frac{4+\cos(x)}{x^3+2x}\,dx

29. \ds \int\limits_{2}^{\infty}\frac{(x^2+1)^{5}}{(2x-1)^{11}}\,dx

Answer

diverges

30. \ds \int\limits_{1}^{\infty}\frac{e^{\sin(x)}+x^2}{\sqrt{x^5-x}}\,dx

31. \ds \int\limits_{3}^{\infty}\frac{x\ln(x)}{x^4+2}\,dx

(Hint: compare \ln(x) to x.)

Answer

converges

32. \ds \int\limits_{2}^{\infty}\frac{x^5+\cot^{-1}(x)}{(\sqrt[3] x-1)^{17}}\,dx

33. \ds \int\limits_{0}^{\infty}\frac{x^7}{(x^2+2)(x^3+3)(x^4+4)}\,dx

Answer

converges

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

34. \ds \int\limits_{1}^{\infty }\frac{\,dx }{{x}^{e}}

35. \ds \int\limits_{0}^{1}\frac{\,dx }{{x}^{\pi }}

Answer

diverges

36. \ds \int\limits_{0}^{1}\frac{\,dx }{\sqrt[5]{1-x}}

37. \ds \int\limits_{1}^{3}\frac{x}{x-2}\,dx

Answer

diverges

38. \ds \int\limits_{ - \infty }^{0}\frac{2x^3+x}{{x}^{4}+x^2+1}\,dx

39. \ds \int\limits_{-1}^{0}\frac{x^2}{\sqrt{1-{x}^{2}}}\,dx

Answer

\ds \frac{\pi}4

40. \ds \int\limits_{0}^{1}\frac{\sqrt{\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}}{x}\,dx

41. \ds \int\limits_{0}^{\pi\text{/}2}\frac{1}{\cos(x)}\,dx

Answer

diverges

42. \ds \int\limits_{0}^{\infty }\frac{2x+1}{{3}^{x}}\,dx

43. \ds \int\limits_{ - \infty }^{1}\frac{x}{{\left({x}^{2}+1\right)}^{2}}\,dx

Answer

\ds-\frac14

44. \ds \int\limits_{0}^{\infty }\frac1{e^{x+1}}\,dx

45. \ds \int\limits_{0}^{9}\frac{\,dx }{\sqrt{9-x}}

Answer

6

46. \ds \int\limits_{-27}^{1}\frac{\,dx }{\sqrt[3]{{x}^{2}}}

47. \ds \int\limits_{0}^{1\text{/}2}\frac{\,dx }{\sqrt{1-4{x}^{2}}}

Answer

\ds \frac{\pi }{4}

48. \ds \int\limits_{6}^{12}\frac{dt}{t\sqrt{{t}^{2}-36}}

49. \ds \int\limits_{0}^{4}x\phantom{\rule{0.1em}{0ex}}\text{ln}\left(4x\right)\,dx

Answer

\ds 8\phantom{\rule{0.1em}{0ex}}\text{ln}\left(16\right)-4

50. \ds \int\limits_{0}^{3}\frac{x}{\sqrt{9-{x}^{2}}}\,dx

51. Evaluate \ds \int\limits_{2}^{\infty }\frac{\,dx }{{\left({x}^{2}-1\right)}^{3\text{/}2}}.

Answer

\ds -1+\frac{2}{\sqrt{3}}

52. Find the area of the region in the first quadrant between the curve \ds y={e}^{-6x} and the x-axis.

53. Find the area of the region bounded by the curve \ds y=\frac{7}{{x}^{2}}, the x-axis, and on the left by \ds x=1.

Answer

7

54. Find the area under the curve \ds y=\frac{1}{{\left(x+1\right)}^{3\text{/}2}}, bounded on the left by \ds x=3.

55. Find the area under \ds y=\frac{5}{1+{x}^{2}} in the first quadrant.

Answer

\ds \frac{5\pi }{2}

56. Find the volume of the solid generated by revolving the region under the curve \ds y=\frac{3}{x+1} from \ds x=1 to \ds x=\infty about the x-axis.

57. Find the volume of the solid generated by revolving the region under the curve \ds y=6{e}^{-2x} in the first quadrant about the y-axis.

Answer

\ds 3\pi

58. Find the volume of the solid generated by revolving the region under the curve \ds y=3{e}^{ - x} in the first quadrant about the x-axis.

The Laplace transform of a continuous function over the interval \ds \left[0,\infty \right) is defined by \ds F\left(s\right)=\int\limits_{0}^{\infty }{e}^{ - sx}f\left(x\right)\,dx (see the Student Project). 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.

59. \ds f\left(x\right)=1

Answer

\ds \frac{1}{s},s\symbol{"3E}0

60. \ds f\left(x\right)=x

61. \ds f\left(x\right)=\text{cos}\left(2x\right)

Answer

\ds \frac{s}{{s}^{2}+4},s\symbol{"3E}0

62. \ds f\left(x\right)={e}^{ax}

A function is a probability density function if it satisfies the following definition: \ds \int\limits_{ - \infty }^{\infty }f\left(t\right)dt=1. The probability that a random variable x lies between a and b is given by \ds P\left(a\le x\le b\right)=\int\limits_{a}^{b}f\left(t\right)dt.

63. Show that

\ds f\left(x\right)=\left\{\begin{array}{ll}0\phantom{\rule{0.1em}{0ex}}\ &\text{if}\ \phantom{\rule{0.1em}{0ex}}(x)<0\\[2mm]\ds 7{e}^{-7x}\ &\text{if}\ \phantom{\rule{0.1em}{0ex}}(x)\ge 0\end{array}

is a probability density function.

Answer

\ds \int\limits_{ - \infty }^{\infty }f\left(t\right)dt=1.

64. Using the function defined in the preceding problem, find the probability that x is between 0 and 1.

Glossary

improper integral
an integral over an infinite interval or an integral of a function containing an infinite discontinuity on the interval; an improper integral is defined in terms of a limit. The improper integral converges if this limit is a finite real number; otherwise, the improper integral diverges

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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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