Calculate Improper Integral from 0 to Infinity: xe^(-5x)dx Solution

[frasifrasi]

For the integral from 0 to infinity of

xe^(-5x)dx...

I am getting as far as:

-1/5*x*e^(-5x) + 1/5*int of (e^(-5x)dx)

But I am getting stuck at this point. We are supposed to come out with 1/25 for the answer but how would I evaluate the "-1/5*x*e^(-5x)" since that is already out of the integral?

Thank you in advance.

 

[Avodyne]

See the first equation in
http://en.wikipedia.org/wiki/Integration_by_parts

 

[rocomath]

yes do it by parts

 

[Antineutron]

does the integral really converge to 1/25?

 

[Avodyne]

Yes, 1/25 is the answer.

 

[Antineutron]

How does it converge to 1/25? Do you have to evaluate the the result as a limit by applying the L'hopital rule? Or is there a better way of evaluating this integral?

 

[asd1249jf]

No, forget about the deadly L'hopital's rule, you only apply that on something that gives you 0/0 or infinity/infinity.

Your basic method of plugging in the limits of your integral, after you get the solution works here. Do you know what e^(-infinity) turns out to be?. That will simplify your solution.

 

[HallsofIvy]

Are you saying that you cannot evaluate xe^-5x at 0 and \infty? That's easy! at 0, you have 0*1= 0 and any polynomial time e^-x goes to 0 as x goes to \infty so that part is 0.

Of course, the integral of 1/5 e^-5x is -1/25 e^-5x. At x= \infty that is 0 and at x= 0, it is -1/25. The difference is 1/25