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...
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