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

Answer 1

Ezra Adams

#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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Answer 2

Cameron Alexander

To evaluate the integral ( \int_0^\infty x e^{-x} , dx ), use integration by parts. Let ( u = x ) and ( dv = e^{-x} , dx ). Then, ( du = dx ) and ( v = -e^{-x} ).

Apply the integration by parts formula: [ \int u , dv = uv - \int v , du ]

[ \int_0^\infty x e^{-x} , dx = \left[-xe^{-x}\right]_0^\infty + \int_0^\infty e^{-x} , dx ]

Evaluate the first term at the limits: [ \lim_{x \to \infty} -xe^{-x} - (-0e^0) ]

[ = 0 + 0 = 0 ]

Now evaluate the remaining integral: [ \int_0^\infty e^{-x} , dx = \lim_{b \to \infty} \int_0^b e^{-x} , dx ]

[ = \lim_{b \to \infty} \left[-e^{-x}\right]_0^b ]

[ = \lim_{b \to \infty} (-e^{-b} - (-e^0)) ]

[ = \lim_{b \to \infty} (-e^{-b} - (-1)) ]

[ = 1 ]

Therefore, the integral ( \int_0^\infty x e^{-x} , dx ) converges and its value is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.