If $f(t)$ is an integrable function over a finite interval $I$, then the integral from any fixed number $a \in I$ to another number $x \in I$ defines a new function $F$ whose value at $x$ is

$F(x)$ = $\int _{a}^{x}f(t)dt$

The above equation defines a way to define new functions, but its importance now is the connection it makes between integrals and derivatives.

We prove the fundamental theorem by applying the definition of the derivative directly to the function $F(x)$, when $x$ and $x+h$ are in [a, b]. this means writing out the difference quotient

$\frac{F(x+h)-F(x)}{h}$...........(1)

and showing that its limit as $h \to 0$ is the number $f(x)$ for each $x$ in [a, b].

When we replace $F(x+h)$ and $F(x)$ by their defining integrals, we can write

$F(x+h)-F(x)$ = $\int _{a}^{x+h}f(t)dt-\int _{a}^{x}f(t)dt$

by simplifying the right hand side by adding rule for integrals we will have

so that the equation (1) become

$\frac{F(x+h)-F(x)}{h}$ =

$\frac{1}{h}$$[F(x+h)-F(x)]$

= $\frac{1}{h}$$\int _{x}^{x+h}f(t)dt$

According to mean value theorem for definite integrals, the value of the last expression in the above equation is one of the values taken on by $f$ in the interval between $x$ and $x+h$, that is for some number $c$ in this interval ,

$\frac{1}{h}$$\int _{x}^{x+h}f(t)dt$ = $f(c)$

As $h \to 0$,$x+h$ approaches $x$, forcing $c$ to approach $x$. since $f$ is continuous at $x$, $f(c)$ approaches $f(x)$:

$\lim_{h \to 0}f(c)$ = $f(x)$

from the definition of the derivative we have

$\frac{dF}{dx}$ = $\lim_{h \to 0}$$\frac{F(x+h)-F(x)}{h}$

= $\lim_{h \to 0}$$\int _{x}^{x+h}f(t)dt$

= $f(x)$

Hence we have

$\frac{dF}{dx}$ = $\frac{d}{dx}$$\int_{a}^{x}f(t)dt$ = $f(x)$