Mean Value Theorem for Integrals

A continuous function actually ATTAINS its average value somewhere on the interval.

The Mean Value Theorem for Integrals guarantees that if $f$ is continuous on $[a,b]$ , there exists at least one point $c\in[a,b]$ where $f(c)$ equals $f$ 's average value on that interval: $f(c)=\dfrac{1}{b-a}\int_a^b f(x)\,dx$ . It strengthens "the average value is a number computed from an integral" to "the function actually reaches that number somewhere."

Examples

For $f(x)=x^2$ on $[0,3]$ with average value $3$ , the MVT guarantees some $c\in[0,3]$ with $c^2=3$ , i.e. $c=\sqrt{3}$ .