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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- Aug 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 - Aug 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 $\displaystyle u = e^{2x}$

$\displaystyle u = x^2 \Leftrightarrow u' = 2x$

$\displaystyle v' = e^{2x} \Leftrightarrow v = \dfrac{1}{2}e^{2x}$

$\displaystyle \dfrac{1}{2}x^2e^{2x} - \int \dfrac{1}{2}e^{2x} \cdot 2x = \dfrac{1}{2}x^2e^{2x} - \int xe^{2x}$

Now do integration by parts again. - Aug 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.

- Aug 14th 2011, 04:45 PMHallsofIvyRe: Trying another integral of parts
Yes, the chain rule: If u= 2x, then $\displaystyle f(x)= e^{2x}= f(u)= e^u$

$\displaystyle \frac{df}{dx}= \frac{df}{du}{\frac{du}{dx}= e^{u}(2)= 2e^{2x}$

But the**integral**goes the other way. To integrate $\displaystyle \int e^{2x}dx$, let u= 2x so that du= 2dx, $\displaystyle dx= \frac{1}{2}du$ so the integral becomes $\displaystyle \frac{1}{2}\int e^u du= \frac{1}{2}e^u+ C= \frac{1}{2}e^{2x}+ C$.

Differentiating $\displaystyle e^{2x}$ requires, as you say, a multiplication by 2.**Integration**, being the inverse of differentiation, requires a division by 2. - Aug 14th 2011, 05:00 PMAlso sprach ZarathustraRe: Trying another integral of parts