Can we relax hypothesis of Fundamental theorem of calculus? Let $F$ is continuous  $[a,b]$ and differentiable on $[a,b]$ and $F'(x)=f(x)$ for $x\in [a,b]$. Assume that $f$ is Riemann integrable. Then Fundamental theorem of calculus say that
$$F(x)-F(a)=\int_{a}^x f(t) dt$$

My Question is: Can we say that $$F(x)-F(a)=\int_{a}^x f(t) dt$$ hold true if we  remove the assumption that $F$ is differentiable at  $a$ and $b$.

My thoughts: The proof uses the mean value theorem to prove the theorem but mean value requires only that $F$ is continuous $[a,b]$ and differentiable on $(a,b)$.

Apostol gives the theorem in following manner

FTC: Let $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $ and let $g:(a, b) \to\mathbb {R} $ be such that $g'(x) =f(x) $ for all $x\in(a, b) $. Then the limits $$\lim_{x\to a^{+} } g(x), \lim_{x\to b^{-}} g(x) $$ exist and we have $$\int_{a} ^{b} f(x) \, dx=\lim_{x\to b^-} g(x) - \lim_{x\to a^+} g(x) $$

Thus essentially you don't need the $F$ in your question to be differentiable (or even continuous or defined) at end points $a, b$.

On request of user @sani via  comment I give a proof of the above mentioned theorem.
Let $$F(x) =\int_{a} ^{x} f(t) \, dt\tag{1}$$ Since $f$ is Riemann integrable on $[a, b] $ it is bounded on $[a, b] $ and let $M$ be an upper bound for $|f|$ on $[a, b] $. Then $$|F(x+h) - F(x) |=\left|\int_x^{x+h} f(t) \, dt\right|\leq M|h|$$ if both $x, x+h$ lie in $[a, b] $. This proves that $F$ is continuous on $[a, b] $.
Consider $g$ defined on $(a, b) $ such that $g'(x) =f(x) $ on $(a, b) $. Let $c\in(a, b) $. By the usual FTC we have $$g(x) =g(c) +\int_{c} ^{x} f(t) \, dt$$ for all $x\in(a, b) $ and using $(1)$ we can write above equation as $$g(x) =g(c) +F(x) - F(c) \tag{2}$$ Since $F$ is continuous on $[a, b] $ we can see the limits of RHS of $(2)$ as $x\to a^+$ and as $x\to b^{-} $ exist and we have $$\lim_{x\to a^+} g(x) =g(c) +F(a) - F(c) $$ and $$\lim_{x\to b^-} g(x) =g(c) +F(b) - F(c) $$ Subtracting these two equations we get $$F(b) - F(a) =\lim_{x\to b^-} g(x) - \lim_{x\to a^+} g(x) $$ Note that $F(a) =0$ and $F(b) =\int_a^b f(x) \, dx$ via definition $(1)$ and the proof for the above mentioned theorem is complete.
