I cannot integrate $\int_a^{\infty} x e^{-(x-a)} dx$. I know the answer should be $(a+1)$ but when I use integration by parts I do not get that answer. Note that $a$ is a constant.
Evaluating $\int_a^{\infty} x e^{-(x-a)} dx$
1
$\begingroup$
calculus
integration
definite-integrals
-
0Now that the post is modified, the answer is $(a+1)$. – 2012-04-28
1 Answers
3
$\int x e^{-(x-a)} = e^a\int xe^{-x}$
$\int xe^{-x} = -xe^{-x} - \int -e^{-x} +c= -e^{-x}(x+1) +c$ (applying Integration by parts)
so the final answer is, $-e^{a-x}(x+1) +c.e^{a}$
-
0lord12: Then you can copy the answer above, with this modification in mind, and see what happens. – 2012-04-28