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

by irresistible
Tags: integrable, integral, prove
 P: 15 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!
 P: 444 Why is this in the Topology & Geometry forum?
 P: 15 sorry, i'm new here I'm gonna post it over there and delete this one if possible
PF Patron
 PF Patron Sci Advisor Thanks Emeritus P: 38,387 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 $f(x_0)$.