# Thread: Trying another integral of parts

1. ## Trying 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

NOT sure what to do with the limits.

Do I have this correct above??
thanks again

2. ## 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 $u = e^{2x}$

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

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

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

Now do integration by parts again.

3. ## Re: 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.

4. ## Re: 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.
Yes, the chain rule: If u= 2x, then $f(x)= e^{2x}= f(u)= e^u$
$\frac{df}{dx}= \frac{df}{du}{\frac{du}{dx}= e^{u}(2)= 2e^{2x}$

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

Differentiating $e^{2x}$ requires, as you say, a multiplication by 2. Integration, being the inverse of differentiation, requires a division by 2.

5. ## Re: Trying 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

$\int x^2e^{2x}dx=Ax^2e^{2x}+Bxe^{2x}+Ce^{2x}+D$
Take derivative operation on both sides and find $A,B,C$ and $D$.