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...

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- Apr 30th 2011, 06:55 AMKumaintegration 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, 07:08 AMe^(i*pi)
Your u and v subs are fine.

What did you do next?

$\displaystyle \int udv = uv - \int vdu$

$\displaystyle -0.5xe^{-2} + 0.5 \int e^{-2x}$ - Apr 30th 2011, 07:14 AMKuma
= 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, 07:19 AMe^(i*pi)
That's right (and it does go to 0. For the first term e^(-2x) will shrink faster than x will increase)

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

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