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 xLaTeX Code: \\in [a,b].
Prove from the definition of the integral that;

F(b)-F(a) =LaTeX Code: \\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!

 

[CompuChip]

You could start by considering an equidistant partition
$$\mathcal P = \{ x_0 = a, x_1 = a + \Delta, x_2 = a + 2 \Delta, \cdots, x_N = a + N \Delta \}$$ where $$\Delta = \frac{b - a}{N}$$
and writing down the left and right Riemann sums. Then I think it's fairly straightforward manipulations of series and limit taking.

 

[HallsofIvy]

I presume the "definition of the integral" you are talking about is "$\int_a^b f(x)dx$ is the area of the region bounded by y= f(x), y= 0, x= a, x= b" (I am assuming here that f(x) is greater than or equal to 0 and that f is continuous. You may want to generalize that.)

Let u be the minimum value of f(x) on [a,b] and U be the maximum value. Then that area lies between the areas of the two rectangles $u(b-a)\le \int_a^b f(x)dx\le U(b-a)$ since f is continuous there must exist a value $\overline{x}$, between a and b, such that $\int_a^b f(x)dx= f(\overline{x})(b-a)$. (That's the "integral mean value theorem".)

Now define $$F(x)= \int_a^x f(t)dt$$. Then $$F(x+h)= \int_a^{x+h} f(t)dt$$ and $$F(x+h)- F(x)= \int_a^{x+h}f(t)dt- \int_a^x f(t)dt= \int_x^{x+h}f(t)dt$$

Applying the integral mean value theorem to that, for every non-zero h there must exist $\overline{x}$ between x and x+ h such that $$F(x+h)- F(x)= f(\overline{x})((x+h)- x)= f(\overline{x})h$$ and so
$$\frac{F(x+h)- F(x)}{h}= f(\overline{x})$$
Since $\overline{x}$ always lies between x and x+h, as h goes to 0, $\overline{x}$ goes to x and, since f is continuous we have
$$\lim_{h\rightarrow 0}\frac{F(x+h)- F(x)}{h}= F'(x)= f(x)$$