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How do you evaluate the integral #int xe^-x dx# from 0 to #oo#?

1 Answer

Sep 24, 2016

1

Explanation:

#I (alpha) = int_0^oo e^(-alpha x) dx = [-1/alpha e^(- alpha x)]_0^oo= 1/alpha#

#(dI)/(d alpha) = int_0^oo -x e^(-alpha x) dx = -1/alpha^2#

#implies int_0^oo x e^(-alpha x) dx = 1/alpha^2 #

#implies int_0^oo x e^(-x) dx = 1#

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