Mean value theorem for integrals

In summary, the conversation discusses a question asking for a 2*pi periodic function that is integrable on [-pi, pi] but fails to satisfy the mean value theorem for integrals. The function g(x) = sin(1/x) is provided as a potential solution, but the questioner is unsure of how to show that it fails to satisfy the mean value theorem. The expert suggests looking for a function that is not continuous at one or two points in [-pi, pi], and explains that it would be harder to find a function that fails to satisfy the mean value theorem on any subinterval.
  • #1
titaniumx3
53
0
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=0Now 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!
 
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  • #2
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
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
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
HallsofIvy said:
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
titaniumx3 said:
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.
 

What is the Mean Value Theorem for Integrals?

The Mean Value Theorem for Integrals states that for a continuous function f(x) on the interval [a,b], there exists a value c in [a,b] such that the average value of f(x) on the interval [a,b] is equal to f(c). In other words, there exists a point where the function takes on its average value over a given interval.

How is the Mean Value Theorem for Integrals different from the Mean Value Theorem for Derivatives?

The Mean Value Theorem for Integrals is used to find a specific value c where the function takes on its average value over a given interval, while the Mean Value Theorem for Derivatives is used to find a specific value c where the slope of the tangent line at point c is equal to the average rate of change over a given interval.

How can the Mean Value Theorem for Integrals be applied in real-world situations?

The Mean Value Theorem for Integrals is commonly used in physics and engineering to find average values of quantities such as velocity, acceleration, and force over a certain time period. It can also be used in economics to find the average value of a function representing a company's revenue over a given time period.

What are the conditions that must be met for the Mean Value Theorem for Integrals to hold?

The function must be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Additionally, the average value of the function over the interval [a,b] must exist.

Can the Mean Value Theorem for Integrals be extended to higher dimensions?

Yes, there is a multivariable version of the Mean Value Theorem for Integrals, which states that for a continuous function f(x,y) on a closed and bounded region R, there exists a point (c,d) in R such that the average value of f(x,y) over R is equal to f(c,d). This concept can also be extended to higher dimensions with more variables.

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