Using the mean value theorem for integrals to finish the proof of FTC

Let $f(t)$ be continuous on $[a,b]$.
For each $x$ in $[a,b]$, define $F(x)$ by the formula:

$$F(x) = \int_a^x f(t) dt.$$

To finsh the proof of FTC, we must prove that $F'(x) = f(x)$.
We do this by calculating the derivative of $F$ from first principles.
For the proof, we use the Mean Value Theorem for integrals of continuous functions:

$$\int_p^q f(t) dt = f(\xi)(q -p),$$

for some $\xi$ between $p$ and $q$.
Put $p = x$ and $q = x+h$, so that $q - p = h$.
Then we have, by the Mean Value Theorem for Integrals, for any $x$ and any $h$, such that $x$ and $x + h$ lie in the interval $[a,b]$:

$$\int_x^{x+ h} f(t) dt = f(\xi(h)) h,$$

where $\xi(h)$ lies between $x$ and $x + h$.
Then we have, for any $x$ and any $h \ne 0$, such that $x$ and $x + h$ lie in the interval $[a,b]$:

$$\frac{F(x + h) - F(x)}{h} = \frac{1}{h}\left(\int_a^{x + h} f(t) dt - \int_a^x f(t) dt\right) = \frac{1}{h}\int_x^{x + h} f(t) dt = f(\xi(h)).$$

Now let $h \rightarrow 0$.
Then $x + h\rightarrow x$, so by squeeze, since $\xi(h)$ always lies between $x$ and $x + h$, we have: $\xi(h) \rightarrow x$ also.
Since $f$ is continuous, we then have $f(\xi(h)) \rightarrow f(x)$.
So we have proved that for any $x$ in $[a,b]$:

$$\lim_{h\rightarrow 0} \frac{F(x + h) - F(x)}{h} = f(x).$$

But the left hand side of this equation is just the definition of $F'(x)$.
So we have proved that $F'(x)$ exists and is equal to $f(x)$, for any $x$ in $[a,b]$, as required.
This completes the proof of FTC.