# proof of the fundamental theorem of calculus

Recall that a continuous function is Riemann integrable on every interval $[c,x]$, 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).

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

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

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. $G^{\prime}=f$. Let $F$ be the integral function

 $F(x)=\int_{a}^{x}f(t)\,dt.$

We have just proven that $F^{\prime}=f$. So $F^{\prime}(x)=G^{\prime}(x)$ for all $x\in[a,b]$ or, which is the same, $(G-F)^{\prime}=0$. 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

 $\int_{a}^{b}f(t)\,dt=F(b)=G(b)-G(a).$
Title proof of the fundamental theorem of calculus ProofOfTheFundamentalTheoremOfCalculus 2013-03-22 13:45:37 2013-03-22 13:45:37 paolini (1187) paolini (1187) 10 paolini (1187) Proof msc 26-00