Substitution method with Integration by Parts?

[csinger1]

Substitution method with Integration by Parts?

Homework Statement

Evaluate the integral...
∫x^3[e^(-x^2)]dx

Homework Equations

∫udv=uv-∫vdu

The Attempt at a Solution

I first tried using integration by parts setting u and dv equal to anything and everything. This seemed to make it more complicated so I decided to use the substitution method setting y=-x^2 and dy=-2xdx and finally -1/2dy=xdx to give me
1/2∫ye^ydy
From here i used integration by parts with u=y, du=dy, v=e^y and dv=e^y dy to get
1/2(ye^y-∫e^y dy) = 1/2(ye^y-e^y)
and then
1/2[-x^2e^(-x^2)-e^(-x^2)]
This problem got really confusing with all the variables and I just want to make sure that I'm not way off base with my methods and solution.
Thanks in advance for any input!

 

[Dick]

Homework Statement

Evaluate the integral...
∫x^3[e^(-x^2)]dx

Homework Equations

∫udv=uv-∫vdu

The Attempt at a Solution

I first tried using integration by parts setting u and dv equal to anything and everything. This seemed to make it more complicated so I decided to use the substitution method setting y=-x^2 and dy=-2xdx and finally -1/2dy=xdx to give me
1/2∫ye^ydy
From here i used integration by parts with u=y, du=dy, v=e^y and dv=e^y dy to get
1/2(ye^y-∫e^y dy) = 1/2(ye^y-e^y)
and then
1/2[-x^2e^(-x^2)-e^(-x^2)]
This problem got really confusing with all the variables and I just want to make sure that I'm not way off base with my methods and solution.
Thanks in advance for any input!

Looks right to me.

 

[csinger1]

That's a relief! I spent a loooonnnggg time working this problem. Thanks!

 

[Mark44]

Once you have gotten an antiderivative, it's a good practice to check it by differentiating it. If your work is correct, the derivative of your answer should be the original integrand.