September 1991 #2

Problem Statement: Suppose that f is a continuous real valued function on (0,1] such that f(x)\geq 0 for every x\in (0,1] and f(1/2)>0. Show \int_{0}^{1}f(x)dx>0.

Proof: Since f is continuous on (0,1] we know f is Riemann Integrable. We are given that f(1/2)=m>0 and since f is continuous we know that for every \varepsilon>0 there exists a \delta>0 such that |f(x)-m|<\varepsilon for every |x-1/2|<\delta. This implies that 0<f(x)<\varepsilon+m for every |x-1/2|<\delta. This implies that \int_{1/2-\delta}^{1/2+\delta}f(x)dx>0. Thus we may conclude that

\int_{0}^{1}f(x)dx=\int_{0}^{1/2-\delta}f(x)dx + \int_{1/2-\delta}^{1/2+\delta}f(x)dx +\int_{1/2+\delta}^{1}f(x)dx>0


Reflection: We know we can even consider the integral since f is continuous on the interval. It is the fact the f is continuous that really gives us the meat of this proof. Since it’s continuous we know there’s some interval of positive length where f(x)>0 since f(1/2)>0.

This entry was posted in Analysis, Continuity, Riemann Integrable. Bookmark the permalink.

3 Responses to September 1991 #2

  1. Sir Wadesworth says:

    I would maybe show that if a continuous function is >0 on an interval of positive length then the integral is positive. I’m not sure if that is something that can be used without proof.

    Or, I would just do it explicitly for this problem. So instead of doing 0<f(x)<e+m, I would show that 0<k0.

    But you definitely have the main idea of the proof. I just have no idea what they take points off for, and if they would dock you because they actually wanted you to show the integral was >0 explicitly.

    • Sir Wadesworth says:

      Hrm, part of my message was chopped off. 2nd paragraph should say,

      0< k< f (x) therefore, the integral of f on the interval from 1/2-d to 1/2d is 2dk.

      • Sir Wadesworth says:

        Man, there should be an edit button for comments, lol…

        so the integral is greater than (not equal to necessarily) 2dk.

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