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

$$F^{\prime }(x)=\underset{h\to 0}{lim}\frac{F(x+h)-F(x)}{h}=\underset{h\to 0}{lim}f({\xi }_h)=f(x)$$

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
Canonical name	ProofOfTheFundamentalTheoremOfCalculus
Date of creation	2013-03-22 13:45:37
Last modified on	2013-03-22 13:45:37
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	10
Author	paolini (1187)
Entry type	Proof
Classification	msc 26-00