Answer
$1-e^{-1/2}$
Work Step by Step
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
$$
\int_0^1 xe^{ -x^{2}/2} dx=-\int_0^{-1/2} e^{ u} du\\
=-\left[e^{u}\right]_0^{-1/2}=-(e^{-1/2}-1)=1-e^{-1/2}
.
$$
