Definite integral and inequality

Let $f$ be a continuous function on $[a, b]$. If for every $x \in [a, b)$ there exists a $y \in (x, b)$ such that $\displaystyle \int_x^y f(t)\, {\rm d}t>0$ then prove that $\displaystyle \int_a^b f(x)\, {\rm d}x>0$.

Solution



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.