Answer
$$\int_0^1xe^{-x^2}\hspace{0.5mm}dx=\frac{e-1}{2e}$$
Work Step by Step
Let $u=-x^2$
Then $du=-2xdx$,
and therefore $\frac{du}{-2x}=dx.$
Substituting this back into our integral we get
$\int xe^{u}\frac{du}{-2x} = \int\frac{e^u}{-2}\hspace{0.5mm}du=-\frac{1}{2}\int e^u\hspace{0,5mm}du=-\frac{1}{2}\left(e^u+C\right)$
$=-\frac{1}{2}\left(e^{-x^2}+C\right)$
Plugging in our integration limits and ignoring our integration constant we get
$-\frac{1}{2}e^{-x^2}\bigg\vert_0^1=-\frac{1}{2}e^{-1}-\left(-\frac{1}{2}e^0\right)=-\frac{1}{2}e^{-1}+\frac{1}{2}$
=$\frac{1}{2}-\frac{1}{2e}=\frac{e}{2e}-\frac{1}{2e}=\frac{e-1}{2e}$
