Theorem: If $f$ is continuous on $[a, b]$ and $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$ for all $x \in (a, b)$.

Proof: We must show that $\lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x)$

By definition of $F$: $F(x+h) - F(x) = \int_a^{x+h} f(t)\,dt - \int_a^x f(t)\,dt$

Using the additivity property of integrals: $F(x+h) - F(x) = \int_x^{x+h} f(t)\,dt$

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

Since $f$ is continuous at $x$, for any $\epsilon > 0$, there exists $\delta > 0$ such that $|t - x| < \delta \implies |f(t) - f(x)| < \epsilon$

For $|h| < \delta$ and $t \in [x, x+h]$ (assuming $h > 0$; the case $h < 0$ is similar): $f(x) - \epsilon < f(t) < f(x) + \epsilon$

Integrating these inequalities: $\int_x^{x+h} [f(x) - \epsilon]\,dt < \int_x^{x+h} f(t)\,dt < \int_x^{x+h} [f(x) + \epsilon]\,dt$ $h(f(x) - \epsilon) < \int_x^{x+h} f(t)\,dt < h(f(x) + \epsilon)$

Dividing by $h > 0$: $f(x) - \epsilon < \frac{1}{h}\int_x^{x+h} f(t)\,dt < f(x) + \epsilon$

This shows: $\left|\frac{F(x+h) - F(x)}{h} - f(x)\right| < \epsilon$

Since $\epsilon$ was arbitrary, we conclude: $\lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x)$

Therefore $F'(x) = f(x)$. $\square$

Theorem: If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$, then $\int_a^b f(x)\,dx = F(b) - F(a)$.

Proof Sketch: By Part 1, we know $G(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$.

Since $F$ is also an antiderivative of $f$, we have $F'(x) = G'(x) = f(x)$ for all $x \in (a, b)$.

By the theorem on uniqueness of antiderivatives, $F(x) = G(x) + C$ for some constant $C$.

Evaluating at $x = a$: $F(a) = G(a) + C = \int_a^a f(t)\,dt + C = 0 + C = C$

Therefore $C = F(a)$, so $F(x) = G(x) + F(a)$.

Evaluating at $x = b$: $F(b) = G(b) + F(a) = \int_a^b f(t)\,dt + F(a)$

Rearranging: $\int_a^b f(t)\,dt = F(b) - F(a) \quad \square$