Calculus AB and Calculus BC

CHAPTER 7 Applications of Integration to Geometry

D. IMPROPER INTEGRALS

There are two classes of improper integrals:

(1) those in which at least one of the limits of integration is infinite (the interval is not bounded); and

(2) those of the type Image where f (x) has a point of discontinuity (becoming infinite) at x = c, a Image c Image b (the function is not bounded).

Illustrations of improper integrals of class (1) are:

Image

The following improper integrals are of class (2):

Image

Sometimes an improper integral belongs to both classes. Consider, for example,

Image

In each case, the interval is not bounded and the integrand fails to exist at some point on the interval of integration.

Note, however, that each integral of the following set is proper:

Image

The integrand, in every example above, is defined at each number on the interval of integration.

Improper integrals of class (1), where the interval is not bounded, are handled as limits:

Image

where f is continuous on [a,b]. If the limit on the right exists, the improper integral on the left is said to converge to this limit; if the limit on the right fails to exist, we say that the improper integral diverges (or is meaningless).

The evaluation of improper integrals of class (1) is illustrated in Examples 17–23.

EXAMPLE 17

Find Image

SOLUTION: Image The given integral thus converges to 1. In Figure N7–22 we interpret Image as the area above the x-axis, under the curve of y = 3, and bounded at the left by the vertical line x = 1.

Image

FIGURE N7–22

BC ONLY

EXAMPLE 18

Image

It can be proved that Image converges if p > 1 but diverges if p Image 1. Figure N7–23 gives a geometric interpretation in terms of area of Image Only the first-quadrant area under Image bounded at the left by x = 1 exists. Note that

Image

FIGURE N7–23

EXAMPLE 19

Image

EXAMPLE 20

Image

BC ONLY

EXAMPLE 21

Image

EXAMPLE 22

Image

Thus, this improper integral diverges.

EXAMPLE 23

Image Since this limit does not exist (sin b takes on values between −1 and 1 as b → ∞), it follows that the given integral diverges.

Note, however, that it does not become infinite; rather, it diverges by oscillation.

Improper integrals of class (2), where the function has an infinite discontinuity, are handled as follows.

To investigate Image where f becomes infinite at x = a, we define Image to be Image The given integral then converges or diverges according to whether the limit does or does not exist. If f has its discontinuity at b, we define Image to be Image again, the given integral converges or diverges as the limit does or does not exist. When, finally, the integrand has a discontinuity at an interior point c on the interval of integration (a < c < b), we let

Image

Now the improper integral converges only if both of the limits exist. If either limit does not exist, the improper integral diverges.

The evaluation of improper integrals of class (2) is illustrated in Examples 24–31.

BC ONLY

EXAMPLE 24

Find Image

SOLUTION: Image

In Figure N7–24 we interpret this integral as the first-quadrant area under Image and to the left of x = 1.

Image

FIGURE N7–24

EXAMPLE 25

Does Image converge or diverge?

SOLUTION: Image

Therefore, this integral diverges.

It can be shown that Image(a > 0) converges if p < 1 but diverges if p Image 1. Figure N7–25 shows an interpretation of Image in terms of areas where Image 1, and 3. Only the first-quadrant area under Image to the left of x = 1 exists.

Note that

Image

BC ONLY

Image

FIGURE N7–25

EXAMPLE 26

Image

EXAMPLE 27

Image

This integral diverges.

EXAMPLE 28

Image

BC ONLY

EXAMPLE 29

Image

Neither limit exists; the integral diverges.

NOTE: This example demonstrates how careful one must be to notice a discontinuity at an interior point. If it were overlooked, one might proceed as follows:

Image

Since this integrand is positive except at zero, the result obtained is clearly meaningless. Figure N7–26 shows the impossibility of this answer.

Image

FIGURE N7–26

THE COMPARISON TEST

We can often determine whether an improper integral converges or diverges by comparing it to a known integral on the same interval. This method is especially helpful when it is not easy to actually evaluate the appropriate limit by finding an antiderivative for the integrand. There are two cases.

(1) Convergence. If on the interval of integration f (x) ≤ g(x) and Image is known to converge, then Image also converges. For example, consider Image We know that Image converges. Since Image the improper integral Image must also converge.

(2) Divergence. If on the interval of integration f (x) ≥ g(x) and Image is known to diverge, then Image also diverges. For example, consider Image We know that Image diverges. Since sec x ≥ 1, it follows that Image hence the improper integral Image must also diverge.

BC ONLY

EXAMPLE 30

Determine whether or not Image converges.

SOLUTION: Although there is no elementary function whose derivative is e−x2, we can still show that the given improper integral converges. Note, first, that if x Image 1 then x2 Image x, so that −x2 Imagex and ex2 Image e−x. Furthermore,

Image

Since Image converges and Image dx converges by the Comparison Test.

EXAMPLE 31

Show that Image converges.

SOLUTION: Image

we will use the Comparison Test to show that both of these integrals converge. Since if 0 < x Image 1, then x + x4 > x and Image it follows that

Image

We know that Image converges; hence Image must converge.

Further, if x Image 1 then x + x4 Image x4 and Image so

Image

We know that Image converges, hence Image also converges.

Thus the given integral, Image converges.

NOTE: Examples 32 and 33 involve finding the volumes of solids. Both lead to improper integrals.

BC ONLY

EXAMPLE 32

Find the volume, if it exists, of the solid generated by rotating the region in the first quadrant bounded above by Image at the left by x = 1, and below by y = 0, about the x-axis.

Image

FIGURE N7–27

SOLUTION:

About the x-axis.

Disk.

Image

BC ONLY

EXAMPLE 33

Find the volume, if it exists, of the solid generated by rotating the region in the first quadrant bounded above by Image at the left by x = 1, and below by y = 0, about the y-axis.

Image

FIGURE N7–28

SOLUTION:

About the y-axis.

Shell.

ΔV = 2πxy Δx = 2π Δx.

Note that Image diverges to infinity.

Chapter Summary

In this chapter, we have reviewed how to find areas and volumes using definite integrals. We’ve looked at area under a curve and between two curves. We’ve reviewed volumes of solids with known cross sections, and the methods of disks and washers for finding volumes of solids of revolution.

For BC Calculus students, we’ve applied these techniques to parametrically defined functions and polar curves and added methods for finding lengths of arc. We’ve also looked at improper integrals and tests for determining convergence and divergence.

No question requiring the use of shells will appear on the AP exam.