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

[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?
Thank you in advance guys!

 

[shoehorn]

Why is this in the Topology & Geometry forum?

 

[irresistible]

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

 

[HallsofIvy]

I'll move this to Calculus

 

[HallsofIvy]

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).