If $f(x)$ is continuous on $[a,b]$ and $\mu\leq f(x) \leq M$, $x\in [a,b]$, prove that
$\frac{\mu}{2}(b-a)^2 \leq \int_a^b\left[\int_a^xf(t)dt\right]dx \leq \frac{M}{2}(b-a)^2$
Is it possible to use mean value theorem on $g(x)=\int_a^xf(t)dt$ to prove the above?
One could apply the mean-value theorem for integrals: For each $x$, $$ g(x)=\int_a^x f(t)dt = (x-a) f(c_x) $$ for some $c_x \in (a, x)$, and therefore $$(x-a) \mu \le g(x) \le (x-a)M \, .$$ Integrating this relationship over the interval $[a,b]$ gives the desired result.
But the mean-value theorem is not really needed, just replace $f(t)$ by its upper (resp. lower) bound: $$ \int_a^b\left[\int_a^xf(t)\,dt\right]dx \le \int_a^b\left[\int_a^x M \,dt\right]dx = M \frac{(b-a)^2}{2} $$