Answer
$$\int_0^1xe^{-x^2}dx=\frac{e-1}{2e}$$
Work Step by Step
To evaluate the integral
$$\int_0^1xe^{-x^2}dx$$
we will use substitution $-x^2=t$ which gives us $-2xdx=dt\Rightarrow xdx=-\frac{dt}{2}$. The integration bounds would be: for $x=0$ we have $t=0$ and for $x=1$ we have $t=-1$. Putting this into the integral we get:
$$\int_0^1xe^{-x^2}dx=\int_0^{-1}e^t\Big(-\frac{dt}{2}\Big)=\frac{1}{2}\int_{-1}^0e^tdt=\frac{1}{2}\left.e^t\right|_{-1}^0=\frac{1}{2}(e^0-e^{-1})=\frac{1}{2}(1-\frac{1}{e})=\frac{e-1}{2e}$$
