## Calculus (3rd Edition)

$1-e^{-1/2}$
Recall that $(e^x)'=e^x$ Let $u= -x^{2}/2$, then $du=-xdx$ and $u$ takes the values from $0$ to $-1/2$ and hence we have $$\int_0^1 xe^{ -x^{2}/2} dx=-\int_0^{-1/2} e^{ u} du\\ =-\left[e^{u}\right]_0^{-1/2}=-(e^{-1/2}-1)=1-e^{-1/2} .$$