$\int\limits_0^1 x{e^{{x^2}}}dx$

Let $I = \int\limits_0^1 {x{e^{{x^2}}}dx} = \cfrac{1}{2}\int\limits_0^1 {2x{e^{{x^2}}}dx}$

Put ${x^2} = t$ $\Rightarrow$ $2xdx = dt$

When $x = 0,t = 0$ and when $x = 1,t = 1$

$\therefore$ $I = \cfrac{1}{2}\int\limits_0^1 {{e^t}dt} = \left[ {\cfrac{1}{2}{e^t}} \right]_0^1 = \cfrac{1}{2}\left[ {{e^1} - {e^0}} \right] = \cfrac{1}{2}\left[ {e - 1} \right]$