Proof of the fundamental theorem

The proof starts from the difference quotient of the integral function and uses the integral mean value theorem to recognise, in the limit, the value $f(x)$.

Proof

We compute the difference quotient of $F$: $\frac{F(x+h)-F(x)}{h} = \frac{1}{h}\left[\int_a^{x+h}f(t)\,dt - \int_a^x f(t)\,dt\right] = \frac{1}{h}\int_x^{x+h}f(t)\,dt.$

Now $\dfrac{1}{h}\displaystyle\int_x^{x+h}f(t)\,dt$ is exactly the integral mean of $f$ over the interval $[x, x+h]$. By the integral mean value theorem (just proved), there exists a point $c_h\in[x, x+h]$ such that: $\frac{1}{h}\int_x^{x+h}f(t)\,dt = f(c_h).$

When $h\to 0$, the interval $[x, x+h]$ “collapses” and $c_h\to x$. Since $f$ is continuous, $f(c_h)\to f(x)$. Hence: $F'(x) = \lim_{h\to 0}\frac{F(x+h)-F(x)}{h} = \lim_{h\to 0}f(c_h) = f(x). \qquad \blacksquare$

Links

Topics: Calculus theorems
Concepts: Integral function · Integral mean value theorem · Fundamental theorem of calculus
Skills: Proving · Integrating