Show that

Let $I=\int_{0}^{1} x e^{x^{2}} d x$

Put $x^{2}=t \Rightarrow 2 x d x=d t$

As $x \rightarrow 0, t \rightarrow 0$ and as $x \rightarrow 1, t \rightarrow 1$

$\therefore I=\frac{1}{2} \int_{0}^{1} e^{t} d t$

$\frac{1}{2} \int e^{t} d t=\frac{1}{2} e^{t}=\mathrm{F}(t)$

By second fundamental theorem of calculus, we obtain

$I=\mathrm{F}(1)-\mathrm{F}(0)$

$=\frac{1}{2} e-\frac{1}{2} e^{0}$

$=\frac{1}{2}(e-1)$