## 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 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

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

$\Box$

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 $f(1/2)>0$.

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### 3 Responses to September 1991 #2

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.