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Proving F'(x)= f(x) using the definition of integral?! 
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#1
Feb709, 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! 


#2
Feb709, 12:40 PM

P: 448

Why is this in the Topology & Geometry forum?



#3
Feb709, 12:46 PM

P: 15

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


#4
Feb709, 02:05 PM

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PF Gold
P: 39,568

Proving F'(x)= f(x) using the definition of integral?!
I'll move this to Calculus



#5
Feb709, 02:18 PM

Math
Emeritus
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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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