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

by irresistible
Tags: integrable, integral, prove
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irresistible
#1
Feb7-09, 12:25 PM
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[tex]\in[/tex] [a,b].
Prove from the definition of the integral that;

F(b)-F(a) =[tex]\int[/tex] 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!
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shoehorn
#2
Feb7-09, 12:40 PM
P: 448
Why is this in the Topology & Geometry forum?
irresistible
#3
Feb7-09, 12:46 PM
P: 15
sorry, i'm new here
I'm gonna post it over there and delete this one if possible

HallsofIvy
#4
Feb7-09, 02:05 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Proving F'(x)= f(x) using the definition of integral?!

I'll move this to Calculus
HallsofIvy
#5
Feb7-09, 02:18 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Now: choose any given value for [itex]x_0[/itex]. [itex]F(x_0)= F(a)+ \int_a^{x_0} f(t)dt[/itex]. For any h> 0, [itex]F(x_0+ h)= \int_a^{x_0+h}f(t)dt[/itex] and [itex] F(x_0+h)- F(x_0)= \int_{x_0}^{x_0+ h} f(t)dt[/itex].

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


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