If a function on average is greater than another over the same interval, does this always imply that the integral of the bigger function is greater?

Question

I am trying to compare two integrals over the same continuous interval and would like to know if I could use the fact that one is on average greater than the other over a specified interval in order to at least prove that they aren’t equal.

Answer 1

The average of $f(x)$ on $(a,b)$ is defined as $\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)\,dx$, so if $\displaystyle \frac{1}{b-a}\int_{a}^{b}f(x)\,dx>\frac{1}{b-a}\int_{a}^{b}g(x)\,dx$, then multiplying by $b-a$ yields $\displaystyle \int_a^b f(x)\,dx>\int_a^bg(x)\,dx$