Mean value of x^2 on the interval [0,3]

Text

Compute the integral mean value of $f(x)=x^2$ on the interval $[0,3]$, find the point $c\in(0,3)$ predicted by the mean value theorem for integrals, and verify that $f(c)=\mu$.

Solution

Mean value. $\mu=\frac{1}{3-0}\int_0^3 x^2\,dx=\frac13\left[\frac{x^3}{3}\right]_0^3=\frac13\cdot 9=3.$

Point $c$. Imposing $f(c)=\mu$: $c^2=3 \implies c=\sqrt3\approx 1.7321\in(0,3).$

Check. $f(\sqrt3)=(\sqrt3)^2=3=\mu$. ✓

$\boxed{\mu=3,\qquad c=\sqrt3\approx 1.7321}$