integrate xe^(2x) dx

Question

anonymous · Answer

$$\int\limits_{}^{}x e^{2x}$$dx

abb0t · Answer

Are you familiar with u-substitution?
You are going to apply that to this problem.

abb0t · Answer

u = 2x
du = xdx

anonymous · Answer

yes I am, I figured i would need it but was still a little confused

anonymous · Answer

but the x doesn't cancel out

anonymous · Answer

$$\int\limits_{}^{}xe^u du/2$$

abb0t · Answer

notice how you have du = xdx which you DO have.
$$\int\limits e^u(xdx) <-- du$$

anonymous · Answer

wouldn't du = 2dx though?

abb0t · Answer

Oh, wait, i think you might want to use integration by parts. sorry.

anonymous · Answer

by using integration by parts let u = x

anonymous · Answer

I don't think I've learned that yet, hmm. I was trying to find the volume of something using the cylindrical shell method and my integral ended up being $$2\pi \int\limits_{}^{}2x+xe ^{2x} dx$$

anonymous · Answer

i just realized I should be using disk method since it's rotating about horizontal line, sorry! I'll give you best answer anyway since you were helping

abb0t · Answer

$$\int\limits fg'dx= fg-\int\limits f'gdx$$
u = f(x)    v = g(x)
du = f'(x)dx     dv = g'(x)dx
$$\int\limits udv = uv- \int\limits vdu$$