Does the Definite Integral Equal Zero for a Continuous Function?

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    Continuity Integrals
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Discussion Overview

The discussion revolves around the implications of a continuous function having a definite integral that is bounded by the square of the interval length. Participants explore whether this condition necessitates that the function equals zero for all real numbers.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant posits that if the definite integral of a continuous function f from m to n is less than or equal to (n-m)², then f must be zero for all x in R.
  • Another participant questions the assumption that f(x) would necessarily be zero, suggesting that the reasoning does not hold.
  • Some participants propose that if f is non-zero at a point a, then there exists a neighborhood around a where f remains positive, which could lead to a contradiction regarding the integral's bounds.
  • There is mention of using the Mean Value Theorem and the Fundamental Theorem of Calculus to explore the problem, though participants express difficulty in applying these theorems effectively.
  • A participant suggests that continuity implies that if f(x) is greater than some positive K in a neighborhood, then the integral over that interval would exceed a certain value, hinting at a contradiction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the condition implies that f(x) must be zero for all x in R. Multiple competing views are presented, with some participants supporting the idea and others challenging it.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the implications of continuity and the properties of definite integrals that are not fully explored in the discussion.

Who May Find This Useful

Readers interested in mathematical analysis, particularly in the properties of continuous functions and definite integrals, may find this discussion relevant.

mike1988
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Let f : R to R be a continuous function, and suppose that definite integral from m to n |∫(m to n)f(x)dx|≤(n-m)^2 for every closed bounded interval [m, n] in R. Then is it the case that f(x) = 0 for all x in R?

I tried using fundamental theorem of calculus but got stuck, since I only got that F'(x)=f(x)≤ 0.

Any help/suggestion would be appreciated.
 
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Why do you think f(x) would be 0? Doesn't make much sense.
 
If f is non-zero at a point a then there exists some b, K > 0 s.t. f(x) > K for all x in [a-b,a+b]. Can you finish the proof?
 
xAxis said:
Why do you think f(x) would be 0? Doesn't make much sense.

At first I thought it was wrong, so was trying to find a counterexample. But I saw this question in a book asks to show that f(x)=0 for all x in R, which means it should be true. I was using Mean Value theorem and fundamental theorem of calc but I am not really heading anywhere with those.

Thanks and any further help will be appreciated!
 
deluks917 said:
If f is non-zero at a point a then there exists some b, K > 0 s.t. f(x) > K for all x in [a-b,a+b]. Can you finish the proof?

I guess I don't really get what this means (are you using the sing-change property of continuity?). I am really stuck!

Thanks
 
mike1988 said:
I guess I don't really get what this means (are you using the sing-change property of continuity?). I am really stuck!

Thanks


Using continuity of f is that the condition is fulfilled, so [tex]\,f(x)>K \,\,\forall\,x\in [a-b,a+b]\Longrightarrow \int_{a-b}^{a+b}f(x)dx>K\int_{a-b}^{a+b}dx = 2Kb[/tex] Can you see now how to finish with a contradiction?

DonAntonio
 
note also if 0<b<c then f(x) >K on [a-c,a+c].
 

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