Can someone explain the integral of xe^(-x) - A quick how to do it?

First, you need to set it up as a limit, because it's an improper integral:

int_0^infinity x/e^x dx = limit_(t -> infinity) int_0^t x/e^x dx

Now, we want to take the integral. You will use integration by parts. Let u=x, dv= e^-x dx. Then du=dx and v= -e^-x. So the integral becomes:

-xe^-x+int(e^-x) dx = -xe^-x-e^-x= -e^-x ( x+1)

Now, evaluate that from 0 to t. We get:

-e^-t (t+1) + 1

And take the limit as t goes to infinity:

lim_t-> infinity -(t+1)/e^t +1 = 1, because e^t is exponential and thus goes to infinity much faster than t+1.

So, the integral converges to 1.

Use integration by parts. Letting u=x and dv=e^(-x), int from 0 to inf of xe^(-x)=[-xe^(-x)-e^(-x)] evaluated from 0 to inf. Substituting and taking the limit yields 1.