How do you integrate $int x^2 e^(-x)dx$ using integration by parts?

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Answer 1 · 2018-04-11T07:49:38

$intx^2e^(-x)dx=-e^(-x)(x^2+2x+2)+C$

Integration by parts says that:
$intv(du)/(dx)=uv-intu(dv)/(dx)$

$u=x^2;(du)/(dx)=2x$
$(dv)/(dx)=e^(-x);v=-e^(-x)$

$intx^2e^(-x)dx=-x^2e^(-x)-int-2xe^(-2x)dx$

Now we do this:
$int-2xe^(-2x)dx$

$u=2x;(du)/(dx)=2$
$(dv)/(dx)=-e^(-x);v=e^(-x)$

$int-2xe^(-x)dx=2xe^(-x)-int2e^(-x)dx=2xe^(-x)+2e^(-x)$

$intx^2e^(-x)dx=-x^2e^(-x)-(2xe^(-x)+2e^(-x))=-x^2e^(-x)-2xe^(-x)-2e^(-x)+C=-e^(-x)(x^2+2x+2)+C$

How do you integrate int x^2 e^(-x)dxint x^2 e^(-x)dx using integration by parts?