Mean value theorem for integrals

  • Thread starter titaniumx3
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  • #1
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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!
 

Answers and Replies

  • #2
Gib Z
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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.
 
  • #3
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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:
  • #4
HallsofIvy
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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?
 
  • #5
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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?
I see what you mean, but why would it be harder for any subinterval?

(sorry, I am just incredibly confused. :confused:)
 
  • #6
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I see what you mean, but why would it be harder for any subinterval?

(sorry, I am just incredibly confused. :confused:)
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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