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Mean value theorem for integrals

  1. Nov 5, 2007 #1
    Hi,

    I have this question which asks for a 2*pi periodic function on the reals, that is integrable on [-pi, pi] but fails to satisfy the mean value theorem for integrals.

    The question also says to help answer the above question, you may wish to show that the function:

    g: [0,1] -> reals given by,

    g(x) = sin(1/x) if x is non-zero
    g(x) = 0 if x=0


    Now I've shown that the above function is true, but I have no idea how to show that it fails to satisfy the mean value theorem or how it relates to a function that is 2*pi periodic and integrable on the given interval.

    Please help!
     
  2. jcsd
  3. Nov 6, 2007 #2

    Gib Z

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    Homework Helper

    What does the theorem say, and what conditions are required for it? Theres a certain equation related to the theorem, substitute g(x) into that equation.
     
  4. Nov 6, 2007 #3
    The theorem says if f:[a,b] -> Reals is a continious function, then there exists a "c" in [a,b] such that:

    [tex]{\frac {\int _{a}^{b}\!f \left( x \right) {dx}}{b-a}}=f \left( c
    \right) [/tex]

    I'm not sure what this other theorem is you speak of. How would I go about finding a function that meets the criteria in my original post but fails to satisfy the MVT above? Any ideas?
     
    Last edited: Nov 6, 2007
  5. Nov 6, 2007 #4

    HallsofIvy

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    Science Advisor

    He didn't mention any "other" theorem! But look at the hypotheses. If you are seeking an integrable function that FAILS to satisfy the mean value theorem, they you will need a function that FAILS to satisfy its hypotheses. I would recommend looking for a function that is NOT continuous at one or two points in [-pi, pi].

    Now, if you want a function that fails to satisfy the mean value theorem on ANY SUBINTERVAL that's harder- but your problem doesn't say that does it?
     
  6. Nov 6, 2007 #5
    I see what you mean, but why would it be harder for any subinterval?

    (sorry, I am just incredibly confused. :confused:)
     
  7. Nov 6, 2007 #6
    Because in that case there would have to be at least one point of discontinuity in every interval [a,b] for all a and b such that a<b.


    EDIT: Every interval that is a subset of the function's domain.
     
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