it is actually integrate from 0 to -infin, but I understand that part. When I see xe^x I integrate it and get (1/2x^2)(e^x). My solutions manuel says it is (x-1)(e^x). I don't get the same answer when using my work, what am I doing wrong?

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- Sep 3rd 2009, 01:16 PMjebckrIntegrate xe^x dx
it is actually integrate from 0 to -infin, but I understand that part. When I see xe^x I integrate it and get (1/2x^2)(e^x). My solutions manuel says it is (x-1)(e^x). I don't get the same answer when using my work, what am I doing wrong?

- Sep 3rd 2009, 01:22 PMMatt Westwood
Integrate by parts: $\displaystyle \int u dv = uv - \int v du$.

Here you set $\displaystyle u = x, dv = e^x$ and so $\displaystyle uv = x e^x $ and $\displaystyle \int v du = \int e^x = e^x$.

Hence, as they say, the result. - Sep 3rd 2009, 01:29 PMjebckr
So how does (xe^x)-(e^x) become (x-1)e^x? I'm sure I'm missing some simple rule here...

- Sep 3rd 2009, 01:35 PMcgiulz
Yes:

(xe^x)-(e^x)

= xe^x - e^x

= e^x(x - 1)

= (x - 1)e^x