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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:

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

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

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

=$2e^{-a}$

Tim Raczkowski
  • 20,046
user2250537
  • 1,101

2 Answers2

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

Tim Raczkowski
  • 20,046
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i will have to do it. $$\int_a^\infty x e^{-x} \, dx = \int_a^\infty x d \left(-e^{-x}\right) = -xe^{-x}\big|a^\infty + \int_a^\infty e^{-x} \, dx = ae^{-a} -e^{-x}\big|_a^\infty = (a+1)e^{-a}.$$

abel
  • 29,170
  • I believe the OP was only asking for his mistake to be found. Looking at his work, seems he could get the correct answer on his own, once the mistake had been pinpointed. – Tim Raczkowski Mar 16 '15 at 15:21