We often use integrals of the functions , for various values of *p*, to help determine whether other integrals converge or diverge.

You already did the work to show this, so we'll just summarize the results. Assuming *p* is greater than 0 (because otherwise the exponents do weird things),

- converges if
*p* < 1 and diverges otherwise. - converges if
*p* > 1 and diverges otherwise.

This is often called the *p*-test for improper integrals.

## Practice:

Use the <em>p</em>-test to determine if converges or diverges. | |

Since the integrand is of the form where *p* > 1 and the interval of integration is [0,1], this integral diverges. Piece of cake, right? The hard part is remembering when you want *p* > 1 and when you want *p* < 1. There's a great trick for that: you can get by with remembering just one integral. The integral converges because gets close to the *x*-axis quickly as *x* approaches ∞. If you can remember this, then you can remember that because you have an example where *p* = 2 is greater than 1 and the integral converges. That means We switch between *p* > 1 and *p* < 1 when we change the interval of integration from [1,∞) to [0,1], so this means and That's the whole *p*-test, and all we had to remember was that the integral converges. We recommend taking your time when using the *p*-test. It can be easy to get mixed up, even if you know what you're doing! | |

Use the *p*-test to determine if the integral converges or diverges.

Answer

The integral

diverges because the integrand is of the form with *p* > 1 and we're integrating from 0 to 1.

Use the *p*-test to determine if the integral converges or diverges.

Answer

The integral

is of the form

with *p* < 1, so this integral converges.

Use the *p*-test to determine if the integral converges or diverges.

Answer

The integral

converges because the integrand is of the form with *p* > 1 and we're integrating from 1 to ∞.

Use the *p*-test to determine if the integral converges or diverges.

Answer

The integral is of the form

where *p* < 1, so this integral diverges.