Question

how do I solve integral (x^2)*(e^-x)?
I put U = -x and it doesn't simplify anything

Answer 1

This has tobe done by integration by parts. Have you seen that before?

Answer 2

no I know how to separate integral(something + something) or trignometric stuff

Answer 3

never seen these kind of problem before

Answer 4

Ah. Yeah, this has to be done by parts. Basically the idea is you choose one part of your integral to be something you differentiate and one part to be something you integrate. The idea is you want to eventually take the derivative of something down to 0 or have it where after you integrate part, differentiate part, things cancel. That being said, this would be the formula for by parts:

\[u*\int\limits_{}^{}v - [\int\limits_{}^{}u'*\int\limits_{}^{}v]\]

U is the part of thefunction you want to differentiate, v is the part that you want to integrate. So basically that is u times integral of v, minus the whole quantity of derivative of u times integral v, all integrated once more.

For your problem, you want u to be x^2 and v to be e^(-x) This means u' is 2x and integral v is just -e^(-x). SO plugging all of this intot he formula I get

\[-x^{2}e^{-x} + \int\limits_{}^{}2xe^{-x}\] SO now we have another integral that we cant dsimply integrate. This means we do by parts again. If I let u = 2x, u' = 2, v once again will be e^(-x) and integral v is -e^(-x). As you can see, us doing these steps sis slowly eliminating that x^2 : ) So after by parts again we get:

\[-x^{2}e^{-x} + -2xe^{-x}+2\int\limits_{}^{}e^{-x}dx\]Now, the formula has minuses in front of those integrals, but I just factored out the negative that integral e^(-x) creates. And at the end of this last inegral I factored ourt that negative and 2. So finally we're just left with integral of e^(-x), which is easy, just -e^(-x). So now once I put it all together my answer is:

\[-x^{2}e^{-x}-2xe^{-x}-2e^{-x} + C\]