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Die Heuning Pot Literature Guide
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Introduction to Indefinite Integrals - At A Glance:

Improper integrals with badly-behaved functions are deceptive.

They look like normal definite integrals

,

but somewhere in the interval from a to b, possibly at one of the endpoints, there will be a vertical asymptote where the function f tends towards ∞ or -∞.

In fact, the function may have more than one vertical asymptote in the interval of integration.

Because of all these possibilities for bad behavior, we recommend graphing functions and looking at their formulas before trying to integrate. Be on the lookout for bad behavior! You want to catch all the places where the function is badly-behaved.

To evaluate this type of improper integral, the first thing we have to do is figure out where the function is badly behaved. Then we can write the improper integral as the limit it really is, work out the integral inside the limit, and finally evaluate the limit.

Badly-Behaved at One Endpoint

We'll start with improper integrals where the function is badly behaved at one limit of integration, but nowhere else.

Every other improper integral with a badly-behaved function can be broken down into integrals like this.

Convergence and Divergence

There's weird stuff going on again. If we take p > 1, the integral

diverges. If we take p < 1, the integral converges. This is the opposite of what happened with the other type of improper integral.

To take an example, on the interval [0,1] the area between  and the x-axis is infinite but the area between  and the x-axis is finite. The graphs don't even look that different. It's perfectly normal if this makes your brain bend a little.

When talking about things "converging" or "diverging," make sure you say whether you're talking about a function or an integral. The function

 diverges as x approaches 0. However, the integral

converges. This means we can have a function diverge but a closely associated improper integral diverge. It's also possible for both the function and an improper integral to diverge. The function  diverges as x approaches 0, and so does its improper integral

Badly-Behaved in the Middle

If a function is badly-behaved in between its limits of integration, we split the improper integral into two pieces. We do this by making a new endpoint at the value of x where the function is badly-behaved. Suppose that

is an improper integral because f has a vertical asymptote at x = b:

Then we split the integral into two new improper integrals:

The function f is badly behaved at just one limit of integration for each of these improper integrals.

When we do this, it's important to remember that

are both improper integrals, and thus are both limits. Both of these limits must exist in order for

to exist. Said another way, if either

diverges, then

diverges also.

There's both good news and bad news here. The good news is that if we find one of the new improper integrals diverges, we're done with the problem. The bad news is that if both of the new improper integrals converge, we have to work out both of them! This is the same good news / bad news we got for the other type of integral.

Example 1

Determine whether

converges or diverges. If it converges, find its value. Graph the region whose area is represented by this integral.


Example 2

Determine whether

converges or diverges. If it converges, find its value. Graph the region whose weighted area is represented by this integral.


Example 3

Determine whether

converges or diverges.


Exercise 1

Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.

Exercise 2

Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.

Exercise 3

Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.

Exercise 4

Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.

 for p > 1

Exercise 5

Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.

 for p < 1

Exercise 6

Determine whether each statement is true or false. Explain your answers.

If  does not exist, then

must diverge.

Exercise 7

If  does not exist, then

must converge.

Exercise 8

Determine whether the integral converges or diverges, and find its value if it converges.

Exercise 9

Determine whether the integral converges or diverges, and find its value if it converges.

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