How do I find the integral #int(x*e^-x)dx# ?
1 Answer
#int xe^(-x) dx = -xe^(-x) - e^(-x) + C#
Process:
#int x e^(-x) dx =# ?
This integral will require integration by parts. Keep in mind the formula:
#int u dv = uv - int v du#
We will let
Therefore,
#v = int e^(-x) dx#
let#q = -x# .thus,
#dq = -dx#
We will rewrite the integral, adding two negatives to accommodate
#v = -int -e^(-x) dx#
Written in terms of
#v = -int e^(q) dq#
Therefore,
#v = -e^(q)#
Substituting back for
#v = -e^(-x)#
Now, looking back at the IBP's formula, we have everything we need to start substituting:
#int xe^(-x) dx = x*(-e^(-x)) - int -e^(-x) dx#
Simplify, canceling the two negatives:
#int xe^(-x) dx = -xe^(-x) + int e^(-x) dx#
That second integral should be easy to solve - it's equal to
#int xe^(-x) dx = -xe^(-x) - e^(-x) + C#