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  • 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 f ( x ) = 1 x and the x -axis over the interval [ 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 ) d x ? We can integrate a t f ( x ) d x for any value of t , so it is reasonable to look at the behavior of this integral as we substitute larger values of t . [link] shows that a t f ( x ) d x may be interpreted as area for various values of 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.
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 ) be continuous over an interval of the form [ a , + ) . Then
    a + f ( x ) d x = lim t + a t f ( x ) d x ,

    provided this limit exists.
  2. Let f ( x ) be continuous over an interval of the form ( , b ] . Then
    b f ( x ) d x = lim t t b f ( x ) d x ,

    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 ) be continuous over ( , + ) . Then
    + f ( x ) d x = 0 f ( x ) d x + 0 + f ( x ) d x ,

    provided that 0 f ( x ) d x and 0 + f ( x ) d x both converge. If either of these two integrals diverge, then + f ( x ) d x diverges. (It can be shown that, in fact, + f ( x ) d x = a f ( x ) d x + a + f ( x ) d x for any value of 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 ) = 1 x and the x -axis over the interval [ 1 , + ) finite or infinite?

Finding an area

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

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.
We can find the area between the curve f ( x ) = 1 / x and the x -axis on an infinite interval.

We can see that the area of this region is given by A = 1 1 x d x . Then we have

A = 1 1 x d x = lim t + 1 t 1 x d x Rewrite the improper integral as a limit. = lim t + ln | x | | 1 t Find the antiderivative. = lim t + ( ln | t | ln 1 ) Evaluate the antiderivative. = + . Evaluate the limit.

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

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Finding a volume

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

The solid is shown in [link] . Using the disk method, we see that the volume V is

V = π 1 + 1 x 2 d x .
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.
The solid of revolution can be generated by rotating an infinite area about the x -axis.

Then we have

V = π 1 + 1 x 2 d x = π lim t + 1 t 1 x 2 d x Rewrite as a limit. = π lim t + 1 x | 1 t Find the antiderivative. = π lim t + ( 1 t + 1 ) Evaluate the antiderivative. = π .

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

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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