Question:
$\int_{0}^{1} x e^{x^{2}} d x$
Solution:
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
$I=\mathrm{F}(1)-\mathrm{F}(0)$
$=\frac{1}{2} e-\frac{1}{2} e^{0}$
$=\frac{1}{2}(e-1)$