Indefinite Integrals Exercises

Answer

As b approaches -∞ the quantity e^b approaches 0, since a very negative power of e is the same thing as 1 divided by a very large power of e:

This means

This integral converges to 1.

Answer

This integral converges to .

Answer

Since b^1/2 approaches ∞ as b approaches ∞, this integral diverges.

Answer

Since we're told p ≠ 1, we know that the antiderivative of 

is

(if p = 1 were an option, the antiderivative could be ln x instead). So when we work out the integral, we get

Since p > 1, the exponent p – 1 is greater than 0. As b approaches ∞ the quantity b^{p – 1} will grow larger and larger, so

will approach zero. This means the integral will converge to

Since p > 1 this quantity is positive, which is reassuring since the integral refers to an area above the x-axis.

Answer

Since we're told p ≠ 1, we know that the antiderivative of

can't be ln x, but is

The integral works out mostly the same way it did in the previous problem:

However, things are a bit different for the final steps. Since p < 1, the exponent p – 1 is negative. This means that as b approaches ∞ the quantity b^{p – 1} is getting smaller and smaller approaching zero, so the quantity

is approaching -∞. This integral diverges.

Answer

False. If f (x) converges to a non-zero number L as x approaches ∞, then the improper integral won't converge. The integral could converge if L = 0, but even then it wouldn't have to.

Answer

True. If the limit of f ( x ) as x approaches ∞ doesn't exist, then that improper integral can't hope to exist either.

Answer

False. It's possible to have f (x) to converge to 0 as x approaches ∞ but to also have this improper integral diverge. As an example, take

Then , but

diverges.

Answer

True. If the limit exists but doesn't equal zero, we determined in question 1 that the improper integral diverges. If the limit doesn't exist, we determined in question 2 that the improper integral diverges.

Answer

First split up the integral into two improper integrals with only one badly-behaved limit each:

Now work out each of those improper integrals.

Since c is approaching -∞, the quantity e^c is approaching 0. So

Now for the other integral.

This limit diverges. Since one of the two improper integrals diverges, the original integral diverges also.

If we had chosen to evaluate the integral

first, we wouldn't have had to bother with the other one at all. If we had only drawn a picture of the function first, we could have seen from the graph that

would diverge! The moral of the story: look at the graph of the function before you start integrating.

If you did look at a graph first and did this the more efficient way, give yourself a pat on the back. Or a carrot.

Answer

We have to split up the integral into two improper integrals with only one badly-behaved limit of integration each:

Evaluate the first improper integral:

Looking at a graph of the function arctan x we can see that as b approaches -∞, arctan b approaches :

This means

Now for the other integral. We look at the graph of arctan x to figure out what arctan b approaches as b approaches ∞.

Putting these results together,

It's a good sign that we got a positive number, since the area between the graph of  and the x-axis all lies on top of the x-axis.