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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.

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

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  • $\begingroup$ @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) $\endgroup$ – Eric Stucky Apr 26 '13 at 10:47

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