# Improper Integral stuck

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