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 $\int_a^{\infty} x e^{-(x-a)} dx$
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$\begingroup$
calculus
integration
definite-integrals
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0Great. So now what? – 2012-04-28
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0By the way, the answer is not $(a+1)$. – 2012-04-28
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0Now that the post is modified, the answer is $(a+1)$. – 2012-04-28
1 Answers
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$$\int x e^{-(x-a)} = e^a\int xe^{-x}$$
$$\int xe^{-x} = -xe^{-x} - \int -e^{-x} +c= -e^{-x}(x+1) +c$$ (applying Integration by parts)
so the final answer is, $$-e^{a-x}(x+1) +c.e^{a}$$
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0Confusing a *number* and a *function*. – 2012-04-28
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0I changed the function to x-a in the power term – 2012-04-28
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0Evaluating from a to infinity the answer is a+1. – 2012-04-28
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0lord12: Then you can copy the answer above, with this modification in mind, and see what happens. – 2012-04-28