Improper Integral

1. Nov 2, 2007

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????

2. Nov 2, 2007

3. Nov 2, 2007

rocomath

yes do it by parts

4. Nov 3, 2007

Antineutron

does the integral really converge to 1/25?

5. Nov 3, 2007

6. Nov 3, 2007

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?

7. Nov 3, 2007

l46kok

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.

8. Nov 3, 2007

HallsofIvy

Staff Emeritus
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