Wikipedia proof Fundamental theorem of calculus

FTC part I says that

"If $f: [a,b]\to \mathbb R$ is continuous then $F(x) = \int_a^x f(t) dt$ is continuous on $[a,b]$ and differentiable with $F'(x) = f(x)$ on $(a,b)$."

Proof of fact that $F$ is differentiable is here where mean value theorem is used:

"According to the mean value theorem for integration, there exists a $c$ in $[x_1, x_1 + \Delta x]$ such that $$\int_{x_1}^{x_1 + \Delta x} f(t) dt = f(c) \Delta x$$"

But mean value theorem says if $F$ is continuous on $[x_1, x_1 + \Delta x]$ and differentiable on $(x_1, x_1 + \Delta x)$ then there is such $c$. But that $F$ is differentiable on $(x_1, x_1 + \Delta x)$ is what this proof is proving therefore we can't assume it.

Is this proof on Wikipedia wrong? If not, why can we apply MVT (mean value theorem) to show that $F$ is differentiable when we need $F$ to be differentiable in order to apply MVT? Thank you.

No.
In the proof of the mean value theorem for integration we use the intermediate value theorem and not the mean value theorem.

• @blue: The MVT that you are used to requires differentiability, but there is an analogous result for integrals, where $F$ plays the role of $f'$ in MVT, and so $\int_b^aF(t)dt$ plays the role of $f$. (The proof on Wikipedia for MVT$\int$ does not rely on the FToC, so this is a valid maneuver) – Eric Stucky Apr 26 '13 at 10:47