comparision test

**Abbas** · December 31st 2009, 06:17 AM

Use a comparison to determine whether the integral converges or diverges:

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

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

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

it is pi^2x in the denomenator

**tonio** · December 31st 2009, 07:53 AM

Originally Posted by Abbas

Use a comparison to determine whether the integral converges or diverges:

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

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

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

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

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

Tonio

it is pi^2x in the denomenator

.

**Abbas** · December 31st 2009, 11:32 AM

ThanX alot ..
the second prblems limits ( 2 , infinite )

**tonio** · December 31st 2009, 06:16 PM

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

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