According to the MVT for integrals, for $f:[a,b] \to \mathbb{R}$ continuous there is $c\in [a,b]$ such that $\int\limits_a^bf(x)\,\mathrm{d}x=f(c)(b-a)$ (one proof uses IVT for the continuous function $f$, see https://en.wikipedia.org/wiki/Mean_value_theorem#Mean_value_theorems_for_definite_integrals).
If one uses though the MVT for derivatives for the differentiable function $F(x)=\int\limits_a^xf(t)\,\mathrm{d}t$, there comes a $c\in (a,b)$ such that $$F'(c)(b-a)=F(b)-F(a) \iff \int\limits_a^bf(x)\,\mathrm{d}x=f(c)(b-a).$$
In the first case $c\in [a,b]$ and in the second one, $c\in (a,b).$ How come this ambiguity? Is there a way in the first case to guarantee that $c$ belongs in $(a,b)$?
Note. Graphically, I noticed that if $\int\limits_a^bf(x)\,\mathrm{d}x=f(a)(b-a)$, then for some $c\in (a,b)$ we must also have $\int\limits_a^bf(x)\,\mathrm{d}x=f(c)(b-a).$ Is that correct in general?
The value $c$ can always be taken in $(a,b)$. Let $\overline{f} = \frac{1}{b-a} \int_a^b f(x) dx$. Then $$ \int_a^b (f(x) - \overline{f}) dx = 0,$$ hence $f - \overline{f}$ has zero mean over the intrval $[a,b]$. If $f \equiv \overline{f}$, then you can take $c$ as any point in $(a,b)$. Otherwise, by splitting where $f > \overline{f}$ and $f < \overline{f}$, by continuity there must be a point $c \in (a,b)$ such that $f(c) - \overline{f} = 0$.