We have, $\int_0^x t e^{t^2} d t$
Let $\quad t^2=z \Rightarrow 2 t d t=d z$
$f(x)=\int_0^{x^2} \frac{e^z}{2} d z=\frac{1}{2}\left[e^z\right]_0^{x^2}$
$\Rightarrow \quad f(x)=\frac{1}{2}\left[e^{x^2}-1\right]$
On differentiating both sides w.r.t. $x$,
$f^{\prime}(x)=\frac{1}{2}\left[2 x e^{x^2}\right]$
For maxima or minima, put $f^{\prime}(x)=0$.
$\Rightarrow \quad x=0$
$f^{\prime \prime}(x)=e^{x^2}+2 x e^{x^2}$
$f^{\prime \prime}(0)=1>0$
$\therefore f(x)$ is minimum at $x=0$
$\therefore \quad f(0)=\frac{1}{2}\left[e^0-1\right]=0$