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

  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. Why is this in the Topology & Geometry forum?
  4. sorry, i'm new here
    I'm gonna post it over there and delete this one if possible
  5. HallsofIvy

    HallsofIvy 41,265
    Staff Emeritus
    Science Advisor

    I'll move this to Calculus
  6. HallsofIvy

    HallsofIvy 41,265
    Staff Emeritus
    Science Advisor

    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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