The Mean Value Theorem for Integrals says that if $f(t)$ is continuous on [ $a$ , $b$ ], then there exists a number m in [ $a$ , $b$ ] such that $f(m)=f_{a v e}=\frac{1}{a-b} \int_{a}^{b} f(t) d t .$

Question

Quizplus · Accepted Answer

The correct formula for the average value of a function is $f_{ave} = \frac{1}{b-a} \int_{a}^{b} f(t) dt$, not $\frac{1}{a-b}$.

The Mean Value Theorem for Integrals Says That If f(t)f(t)f(t) Is Continuous on [

The Mean Value Theorem for Integrals Says That If $f(t)$ Is Continuous on [