# integration question

• Apr 30th 2011, 07:55 AM
Kuma
integration question
how to integrate this?

xe^-2x from 0 to infinity
i used ibp with dv = e^-2x and u = x
then v = -0.5e^-2x and du = 1

I'm not getting the right answer though...
• Apr 30th 2011, 08:08 AM
e^(i*pi)
Your u and v subs are fine.

What did you do next?

$\int udv = uv - \int vdu$

$-0.5xe^{-2} + 0.5 \int e^{-2x}$
• Apr 30th 2011, 08:14 AM
Kuma
= uv - int vdu
int vdu = 1/4e^-2x

so i get

-0.5xe^-2x - 0.25e^-2x | 0 to infinity
not 100% sure how to evaluate the first term at infinity though, i believe it goes to 0 but i might be wrong...
• Apr 30th 2011, 08:19 AM
e^(i*pi)
Quote:

Originally Posted by Kuma
= uv - int vdu
int vdu = 1/4e^-2x

so i get

-0.5xe^-2x - 0.25e^-2x | 0 to infinity
not 100% sure how to evaluate the first term at infinity though, i believe it goes to 0 but i might be wrong...

That's right (and it does go to 0. For the first term e^(-2x) will shrink faster than x will increase)

$0 - (-0.5 \cdot 0 \cdot 1 - 0.25 \cdot 1) = 0.25$

edit: wolfram agrees - http://www.wolframalpha.com/input/?i...0+and+infinity