(related to Theorem: Indefinite Integral, Antiderivative)
We want to show that on the close interval \([a,x]\), the indefinite integral \(F(x):=\int_a^x f(t)dt\) of a continuous real function \(f\) is differentiable, and that its derivative is \(F'=f\). Note, that continuity is a sufficient condition for \(f\) to be Riemann integrable. Therefore, the integral \(F(x):=\int_a^x f(t)dt\) exists.
Let \(h\neq 0\). By definition, the derivative is given by
\[F'(x):=\lim_{h\to 0}\frac {F(x+h)-F(x)}{h}.\]
It follows from the linearity of the Riemann integral. \[\frac {F(x+h)-F(x)}{h}=\frac 1h\left(\int_a^{x+h}f(t)dt-\int_a^xf(t)dt\right)=\frac 1h\int_x^{x+h}f(t)dt.\]
Because \(f\) is continuous, the mean value theorem for Riemann integrals tells us that there exists \(\xi_h\in[x,x+h]\) (or \(\xi_h\in[x+h,x]\) for \(h < 0\)) with
\[\int_x^{x+h}f(t)dt=hf(\xi_h).\]
Note that \(\xi_h\) tends to \(x\), as \(h\) tends to \(0\). Because \(f\) is continuous, we get the required result: \[F'(x)=\lim_{h\to 0}\frac 1h\int_x^{x+h}f(t)dt=\lim_{h\to 0}\frac 1h(hf(\xi_h))=f(x).\]