Recall that a continuous function is Riemann integrable, so the integral $$ F(x) = \int_c^x f(t)\, dt $$ is well defined.

Consider the increment of $F$ : $$ F(x+h)-F(x) = \int_c^{x+h} f(t)\, dt - \int_c^x f(t)\, dt = \int_x^{x+h} f(t)\, dt $$ (we have used the linearity of the integral with respect to the function and the additivity with respect to the domain).

Now let $M$ be the maximum of $f$ on $[x,x+h]$ and $m$ be the minimum. Clearly we have $$ m h \le \int_x^{x+h} f(t)\, dt \le M h $$ (this is due to the monotonicity of the integral with respect to the integrand) which can be written as $$ \frac{F(x+h)-F(x)}{h} = \frac{\int_{x}^{x+h}f(t)\, dt}{h} \in [m,M] $$

Since $f$ is continuous, by the mean-value theorem, there exists $\xi_h\in [x,x+h]$ such that $f(\xi_h) = \frac{F(x+h)-F(x)}{h}$ so that $$ F'(x)= \lim_{h\to 0} \frac{ F(x+h)-F(x)}{h} = \lim_{h\to 0} f(\xi_h) = f(x) $$ since $\xi_h\to x$ as $h\to 0$ . This proves the first part of the theorem.

For the second part suppose that $G$ is any antiderivative of $f$ , i.e. $G'=f$ . Let $F$ be the integral function $$ F(x)= \int_a^x f(t) \, dt. $$ We have just proven that $F'=f$ . So $F'(x)=G'(x)$ for all $x\in [a,b]$ or, which is the same, $(G-F)'=0$ . This means that $G-F$ is constant on $[a,b]$ that is, there exists $k$ such that $G(x)=F(x)+k$ . Since $F(a)=0$ we have $G(a)=k$ and hence $G(x)=F(x)+G(a)$ for all $x\in[a,b]$ . Thus $$ \int_a^b f(t)\, dt = F(b) = G(b) - G(a). $$