Fix $x \in (a, b)$ and let $h \neq 0$ be small enough that $x + h \in (a, b)$.

Step 1: Express the difference quotient as an integral.

By definition,

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

Thus the difference quotient is

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

(If $h < 0$, the integral is from $x+h$ to $x$, and the $1/h$ factor accounts for the sign.)

Step 2: Bound the integrand using continuity.

Since $f$ is continuous on the compact interval $[x, x+h]$ (or $[x+h, x]$ if $h < 0$), by the Extreme Value Theorem, $f$ attains its minimum $m$ and maximum $M$ on this interval. That is,

$m = \min_{t \in [x, x+h]} f(t) \quad \text{and} \quad M = \max_{t \in [x, x+h]} f(t).$

By monotonicity of the integral,

$m \cdot h \leq \int_x^{x+h} f(t) \, dt \leq M \cdot h \quad \text{(assuming } h > 0\text{)}.$

Dividing by $h > 0$ (or by $h < 0$ and reversing inequalities),

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

Step 3: Take the limit as $h \to 0$.

As $h \to 0$, the interval $[x, x+h]$ shrinks to $\{x\}$. By continuity of $f$ at $x$, both the minimum $m$ and maximum $M$ on $[x, x+h]$ approach $f(x)$:

$\lim_{h \to 0} m = \lim_{h \to 0} M = f(x).$

By the squeeze theorem,

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

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