Here, $f(x)=\left\{\begin{array}{cc}e^x, & 0 \leq x \leq 1 \\ 2-e^{x-1}, & 1 < x \leq 2 \\ x-e, & 2 < x \leq 3\end{array}\right.$ and $\quad g(x)=\int_0^x f(t) d t$
$$
\Rightarrow \quad g^{\prime}(x)=f(x)
$$
where, $g^{\prime}(x)=0$
$\Rightarrow \quad x=1+\log _e 2$
and $\quad x=e$.
Also, $\quad g(x)=\left\{\begin{array}{cc}-e^{x-1}, & 1 < x \leq 2 \\ 1, & 2 < x \leq 3\end{array}\right.$
$g\left(1+\log _e 2\right)=-e^{\log _e 2} < 0$, hence at $x=1+\log _e 2, g(x)$ has a local maximum.
Also, $g(e)=1>0$, hence at $x=e, g(x)$ has a local minima.
$\because f(x)$ is discontinuous at $x=1$, then we get
Local maxima at $x=1$ and local minima at $x=2$.