# comparision test

• Dec 31st 2009, 06:17 AM
Abbas
comparision test
Use a comparison to determine whether the integral converges or diverges:

A $\displaystyle \int \frac{x^2e^x}{lnx}dx =$ from 2 to infinite

This integral diverges , but how can I show that using the comparison test?

B $\displaystyle \int \frac{1000}{pi^(2x)+x^2}dx =$

it is pi^2x in the denomenator
• Dec 31st 2009, 07:53 AM
tonio
Quote:

Originally Posted by Abbas
Use a comparison to determine whether the integral converges or diverges:

A $\displaystyle \int \frac{x^2e^x}{lnx}dx =$ from 2 to infinite

This integral diverges , but how can I show that using the comparison test?

We have $\displaystyle \frac{x^2e^x}{\ln x}\geq e^x$ and clearly $\displaystyle \int\limits_2^\infty e^x\,dx$ diverges.

B $\displaystyle \int \frac{1000}{pi^(2x)+x^2}dx =$

What are the limits here?? $\displaystyle \frac{1000}{\pi^2x+x^2}\le \frac{1000}{x^2}$ , and $\displaystyle \int\limits_1^\infty \frac{1000}{x^2}\,dx =\lim_{b\to \infty}-\frac{1000}{b}+1000=1000$ , so the integral converges (unless the lower limit is not 1 but something else...)

Tonio

it is pi^2x in the denomenator

.
• Dec 31st 2009, 11:32 AM
Abbas
ThanX alot ..
the second prblems limits ( 2 , infinite )
• Dec 31st 2009, 06:16 PM
tonio
Quote:

Originally Posted by Abbas
ThanX alot ..
the second prblems limits ( 2 , infinite )

It's just the same: the integral still converges as can easily be checked. The problem arises when the lower limit is zero.

Tonio