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Mean value theorem for integral

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f is continuos on [a,b]. Show that there exists c in (a,b) such that the integral from a to b of f(x)dx equals (b-a)*f(c)

    2. Relevant equations

    3. The attempt at a solution
    Tried using the mean value theorem to come up with a solution by rearranging different variables, but they don't relate cause it's f(c) not f '(c).
  2. jcsd
  3. Dec 16, 2008 #2


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    Re: f(b)-f(a)=(b-a)f(c)

    You're not trying to show that [itex]f(b)-f(a)=(b-a)f(c)[/itex], you are trying to show [tex]\int_{a}^{b}f(x)dx=(b-a)f(c)[/tex].

    Hint: suppose that the antiderivative of [itex]f(x)[/itex] is [itex]F(x)[/itex]; what does the mean value theorem say about [itex]F(b)-F(a)[/itex]? What does the fundamental theorem of calculus say about [itex]F(b)-F(a)[/itex]?
  4. Dec 16, 2008 #3
    Re: f(b)-f(a)=(b-a)f(c)

    wow, ok definately didn't look at it good enough. thanks a lot for the hint
  5. Dec 17, 2008 #4
    Re: f(b)-f(a)=(b-a)f(c)

    Or you could think of it this way:

    SInce f is continuous on [a,b], then by the extreme value theorem it reaches its max and min value on that interval. That is there exist r,t such that f(r)=m, f(t)=M, where m and M are its smallest value and its greatest value on that interval.

    THat is

    [tex]m\leq f(x) \leq M[/tex] for all x in [a,b]

    Now integrating from a to be we get

    [tex]\int_{a}^bmdx\leq \int_{a}^bf(x)dx\leq \int_{a}^bMdx =>[/tex]

    [tex]m(b-a)\leq \int_{a}^bf(x)dx\leq (b-a)M[/tex]

    [tex]m\leq \frac{1}{b-a}\int_{a}^bf(x)dx\leq M[/tex]

    Now by the IVT there exists some number c on the interval (a,b) such that

    [tex] f(c)=\frac{1}{b-a}\int_{a}^bf(x)dx=>\int_{a}^bf(x)dx=(b-a)f(c)[/tex]

    Proof done!:approve:
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