Answer
$\frac{1}{2}-\frac{1}{2e}$
Work Step by Step
$\int_{0}^{1}x~e^{-x^2}~dx$
Let $u = -x^2$
$\frac{du}{dx} = -2x$
$dx = -\frac{du}{2x}$
When $x = 0$, then $u = 0$
When $x = 1$, then $u = -1$
$\int_{0}^{-1} x~e^u(-\frac{du}{2x})$
$= \int_{-1}^{0} \frac{1}{2}~e^u~du$
$= \frac{1}{2}~(e^u~\vert_{-1}^{0}~)$
$= \frac{1}{2}~(e^0-e^{-1})$
$= \frac{1}{2}~(1-\frac{1}{e})$
$= \frac{1}{2}-\frac{1}{2e}$