Fix $x \in (a, b)$. For small $h \neq 0$ with $x + h \in (a, b)$,

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

By the Extreme Value Theorem, $f$ attains its minimum $m$ and maximum $M$ on $[x, x+h]$ (or $[x+h, x]$ if $h < 0$). Then

$m h \leq \int_x^{x+h} f(t) \, dt \leq M h,$

so $m \leq \frac{1}{h} \int_x^{x+h} f(t) \, dt \leq M$.

As $h \to 0$, by continuity of $f$, both $m \to f(x)$ and $M \to f(x)$. By squeeze theorem,

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

Thus $F'(x) = f(x)$.

Let $G(x) = \int_a^x f(t) \, dt$. By FTC Part I, $G'(x) = f(x)$.

Since $F' = f$ and $G' = f$, we have $(F - G)' = 0$. By the Mean Value Theorem, $F - G$ is constant on $[a, b]$. Thus $F(x) - G(x) = C$ for all $x \in [a, b]$.

Evaluating at $x = a$: $F(a) - G(a) = F(a) - 0 = C$, so $C = F(a)$.

Evaluating at $x = b$:

$F(b) - G(b) = C = F(a),$

so $G(b) = F(b) - F(a)$. But $G(b) = \int_a^b f(t) \, dt$, thus

$\int_a^b f(x) \, dx = F(b) - F(a).$