Answer
$$\frac{{1 - {e^{ - 3}}}}{6}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x{e^{ - 3{x^2}}}} dx \cr
& {\text{Integrate using the substitution method}} \cr
& u = - 3{x^2},{\text{ }}du = - 6xdx,{\text{ }}dx = - \frac{1}{{6x}}du \cr
& {\text{The new limits of integration are}} \cr
& x = 1 \to u = - 3 \cr
& x = 0 \to u = 0 \cr
& {\text{Substitute and integrate}} \cr
& \int_0^1 {x{e^{ - 3{x^2}}}} dx = \int_0^{ - 3} {x{e^u}} \left( { - \frac{1}{{6x}}} \right)du \cr
& = - \frac{1}{6}\int_0^{ - 3} {{e^u}} du \cr
& = - \frac{1}{6}\left[ {{e^u}} \right]_0^{ - 3} \cr
& = - \frac{1}{6}\left[ {{e^{ - 3}} - {e^0}} \right] \cr
& = - \frac{1}{6}\left[ {{e^{ - 3}} - 1} \right] \cr
& = \frac{{1 - {e^{ - 3}}}}{6} \cr} $$
