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

  1. Feb 7, 2009 #1
    Hey guys,:smile:
    Can you help me prove this?:confused:

    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!:wink:
     
  2. jcsd
  3. Feb 7, 2009 #2
    Why is this in the Topology & Geometry forum?
     
  4. Feb 7, 2009 #3
    sorry, i'm new here
    I'm gonna post it over there and delete this one if possible
     
  5. Feb 7, 2009 #4

    HallsofIvy

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    I'll move this to Calculus
     
  6. Feb 7, 2009 #5

    HallsofIvy

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    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].
     
    Last edited: Feb 7, 2009
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