Integral sign with lower limit of 0 and upper limit of 1

x^2e^2x dx

So I found u = e^2x v=1/2(x)^3

du=2e^2x dv= x^2 dx

answer is 1/3(x^3e^2x)-1/12(x^4)(2e^2x) +c.

NOT sure what to do with the limits.

Do I have this correct above??

thanks again

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- August 14th 2011, 02:11 PMbradycatTrying another integral of parts
Integral sign with lower limit of 0 and upper limit of 1

x^2e^2x dx

So I found u = e^2x v=1/2(x)^3

du=2e^2x dv= x^2 dx

answer is 1/3(x^3e^2x)-1/12(x^4)(2e^2x) +c.

NOT sure what to do with the limits.

Do I have this correct above??

thanks again - August 14th 2011, 02:24 PMe^(i*pi)Re: Trying another integral of parts
You need to swap your u and v around. You'll end up in a loop if you continue to set

Now do integration by parts again. - August 14th 2011, 02:37 PMbradycatRe: Trying another integral of parts
Ok looking at this, question. When I read online on how to find the deriviative of e^2x is was 2 in front, why half> Just asking, then what I read was wrong.

- August 14th 2011, 04:45 PMHallsofIvyRe: Trying another integral of parts
Yes, the chain rule: If u= 2x, then

But the**integral**goes the other way. To integrate , let u= 2x so that du= 2dx, so the integral becomes .

Differentiating requires, as you say, a multiplication by 2.**Integration**, being the inverse of differentiation, requires a division by 2. - August 14th 2011, 05:00 PMAlso sprach ZarathustraRe: Trying another integral of parts