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
and the
x -axis over the interval
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
We can integrate
for any value of
so it is reasonable to look at the behavior of this integral as we substitute larger values of
[link] shows that
may be interpreted as area for various values of
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.
Definition
Let
be continuous over an interval of the form
Then
provided this limit exists.
Let
be continuous over an interval of the form
Then
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.
Let
be continuous over
Then
provided that
and
both converge. If either of these two integrals diverge, then
diverges. (It can be shown that, in fact,
for any value of
In our first example, we return to the question we posed at the start of this section: Is the area between the graph of
and the
-axis over the interval
finite or infinite?
Finding an area
Determine whether the area between the graph of
and the
x -axis over the interval
is finite or infinite.
We first do a quick sketch of the region in question, as shown in the following graph.
We can see that the area of this region is given by
Then we have
Since the improper integral diverges to
the area of the region is infinite.