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

  1. Feb 7, 2009 #1
    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 xLaTeX Code: \\in [a,b].
    Prove from the definition of the integral that;

    F(b)-F(a) =LaTeX Code: \\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!
  2. jcsd
  3. Feb 8, 2009 #2


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    You could start by considering an equidistant partition
    [tex]\mathcal P = \{ x_0 = a, x_1 = a + \Delta, x_2 = a + 2 \Delta, \cdots, x_N = a + N \Delta \} [/tex] where [tex]\Delta = \frac{b - a}{N}[/tex]
    and writing down the left and right Riemann sums. Then I think it's fairly straightforward manipulations of series and limit taking.
  4. Feb 8, 2009 #3


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    I presume the "definition of the integral" you are talking about is "[itex]\int_a^b f(x)dx[/itex] is the area of the region bounded by y= f(x), y= 0, x= a, x= b" (I am assuming here that f(x) is greater than or equal to 0 and that f is continuous. You may want to generalize that.)

    Let u be the minimum value of f(x) on [a,b] and U be the maximum value. Then that area lies between the areas of the two rectangles [itex]u(b-a)\le \int_a^b f(x)dx\le U(b-a)[/itex] since f is continuous there must exist a value [itex]\overline{x}[/itex], between a and b, such that [itex]\int_a^b f(x)dx= f(\overline{x})(b-a)[/itex]. (That's the "integral mean value theorem".)

    Now define [tex]F(x)= \int_a^x f(t)dt[/tex]. Then [tex]F(x+h)= \int_a^{x+h} f(t)dt[/tex] and [tex]F(x+h)- F(x)= \int_a^{x+h}f(t)dt- \int_a^x f(t)dt= \int_x^{x+h}f(t)dt[/tex]

    Applying the integral mean value theorem to that, for every non-zero h there must exist [itex]\overline{x}[/itex] between x and x+ h such that [tex]F(x+h)- F(x)= f(\overline{x})((x+h)- x)= f(\overline{x})h[/tex] and so
    [tex]\frac{F(x+h)- F(x)}{h}= f(\overline{x})[/tex]
    Since [itex]\overline{x}[/itex] always lies between x and x+h, as h goes to 0, [itex]\overline{x}[/itex] goes to x and, since f is continuous we have
    [tex]\lim_{h\rightarrow 0}\frac{F(x+h)- F(x)}{h}= F'(x)= f(x)[/tex]
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