Explanation

(a)
Let $f(x) = \int_{0}^{x}{te^{t^{2}}}dt$
$\Rightarrow \ \ \ \ \ f^{'}(x) = \lbrack xe^{x^{2}}\rbrack$ [by Leibnitz rule]
For maxima or minima, put $f^{'}(x) = 0\ \Rightarrow \ \ \ x = 0$
Now, $f^{''}(x) = e^{x^{2}} + 2x\ e^{x^{2}}$
$f^{''}(0) = 1 > 0$
$\therefore f(x)$ is minimum at $x = 0$
$\therefore\ \ \ \ \ \ f(0) = 0$