Question

Evaluate the definite integral ${\int }_0^{1}x{e}^{x^2}dx$

Answer 1

Let $I={\int }_0^{1}x{e}^{x^2}dx$
Put $x^2=t\Rightarrow 2xdx=dt$
As $x\to 0,t\to 0$ and as $x\to 1,t\to 1$,
$\therefore I=\frac{1}{2}{\int }_0^1{e}^{t}dt$
$\Rightarrow \frac{1}{2}\int e^{t}dt=\frac{1}{2}{e}^t=F\left(t\right)$
By second fundamental theorem of calculus, we obtain
$I=F\left(1\right)-F\left(0\right)$
$=\frac{1}{2}e-\frac{1}{2}{e}^0$
$=\frac{1}{2}(e-1)$