Integral and inequality

[Papapetros Vaggelis]

Let \(\displaystyle{f:\left[a,b\right]\longrightarrow \mathbb{R}}\) be a continuous function . If for every

\(\displaystyle{x\in\left[a,b\right)}\) there exists \(\displaystyle{y\in\left(x,b\right)}\) such that



\(\displaystyle{\int_{x}^{y}f(t)\,\mathrm{d}t>0}\), then prove that \(\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x>0}\) .

[Tolaso J Kos]

Let \(\displaystyle{f:\left[a,b\right]\longrightarrow \mathbb{R}}\) be a continuous function . If for every

\(\displaystyle{x\in\left[a,b\right)}\) there exists \(\displaystyle{y\in\left(x,b\right)}\) such that



\(\displaystyle{\int_{x}^{y}f(t)\,\mathrm{d}t>0}\), then prove that \(\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x>0}\) .

Hi there.

We consider the function \( \displaystyle F(y)=\int_a^y f(t)\, {\rm d}t, \;\; a\leq y \leq b \). Apparently \( F \) is continuous since is differentiable. Hence the problem now using \(F\) is rephrased to the following:

Let \(F \) be a continuous on \([a, b]\) such that \(F(a)=0\) and for every \(x \in [a,b)\) there exists a \(y \in (x, b)\) such that \(F(y)>F(x)\). Prove that \(F(b)>0\).

Since \(F\) is continuous on \([a, b]\) it clearly attains a maximum value. From the hypothesis for \(x=a\) there exists a \(y\) such that \(F(y)>0\). Hence the maximum is positive for sure. We will prove that the maximum is attained at \(b\). If not, then the maximum is attained at a point \(a<x<b\). From the hypothesis there exists a \(y'\) such that \(F(y')>F(x)\), a contradiction since \(F(x)\) was the maximum of \(F\) and the exercise is complete.