Improper Integral of $xe^{-x}$.

Question

I was working on this problem but I didn't get the right answer, though I can't find my mistake.

Here is the question and my attempt:

$\int_a^\infty xe^{-x}dx$ evaluate.

$\lim_{b\to \infty} \int_a^b xe^{-x}dx$

then using parts, by letting $u = x$ and $dv/dx = e^{-x}$,

I got:

=$\lim_{b\to \infty}(-e^{-x})_a^b + \int_a^b e^{-x}dx$

=$\lim_{b\to \infty}((-e^{-x})_a^b + -e^{-b}+e^{-a})$

=$\lim_{b\to \infty}(-e^{-b}+e^{-a} -e^{-b}+e^{-a})$

=$2e^{-a}$

Answer 1

Integrating by parts gives $$\int xe^{-x}\,dx=-xe^{-x}+e^{-x}.$$

Answer 2

i will have to do it. $$\int_a^\infty x e^{-x} \, dx = \int_a^\infty x d \left(-e^{-x}\right) = -xe^{-x}\big|a^\infty + \int_a^\infty e^{-x} \, dx = ae^{-a} -e^{-x}\big|_a^\infty = (a+1)e^{-a}.$$