Proving F'(x)= f(x) using the definition of integral?

1. Feb 7, 2009

irresistible

Hey guys,
Can you help me prove this?

Suppose that f:[a.b] -> R is integrable and that F:[a,b]->R is a differentiable function such thet F'(x)= f(x) for all x$$\in$$ [a,b].
Prove from the definition of the integral that;

F(b)-F(a) =$$\int$$ f(x) dx ( integral going from a to b)

I can prove this using the Fundamental theorem of calculus;however, this question specifically asks that we use the definition of integral to prove this:

I'm thinking that I have to use the "partition" prepositions to prove this.
Any ideas?

2. Feb 7, 2009

shoehorn

Why is this in the Topology & Geometry forum?

3. Feb 7, 2009

irresistible

sorry, i'm new here
I'm gonna post it over there and delete this one if possible

4. Feb 7, 2009

HallsofIvy

Staff Emeritus
I'll move this to Calculus

5. Feb 7, 2009

HallsofIvy

Staff Emeritus
Now: choose any given value for $x_0$. $F(x_0)= F(a)+ \int_a^{x_0} f(t)dt$. For any h> 0, $F(x_0+ h)= \int_a^{x_0+h}f(t)dt$ and $F(x_0+h)- F(x_0)= \int_{x_0}^{x_0+ h} f(t)dt$.

By the "integral mean value theorem" (that's where you use the definition of "integral"), there exist an $\overline{x}$, between $x_0$ and $x_0+ h$ such that $\int_{x_0}^{x_0+ h} f(t)dt= f(\overline{x})((x_0+h)- x_0)= f(\overline{x}h$. Then $F(x_0+h)- F(x_0)= f(\overline{x})h$ and
$$\frac{F(x_0+h)- F(x_0)}{h}= f(\overline{x})$$
Taking the limit as h goes to 0, since $\overline{x}$ must always be between $x_0$ and $x_0+ h$, $f(\overline{x})$ goes to f(x). That is, $dF/dx$, at $x= x_0$ is $f(x_0)$.

Last edited: Feb 7, 2009