Find the number(s) $u$ guaranteed by the Mean Value Theorem for Integrals for $\int_{2}^{6}x^{2}dx.$

Solution The Mean Value Theorem for Integrals states there is a number $u,$ $2\leq u\leq 6,$ for which $$\int_{2}^{6}x^{2}dx=f(u) (6-2) =4u^{2}\qquad\qquad {\color{#0066A7}{\hbox{\(f(u)=u^2\)}}}$$

We integrate to obtain $$\int_{2}^{6}x^{2}dx=\left[\dfrac{x^{3}}{3}\right] _{2}^{6}=\dfrac{1}{3}(216-8) =\dfrac{208}{3}$$

Then $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l@{\hskip-2pc}} \dfrac{208}{3} &= 4u^{2} \\ u^{2} &= \dfrac{52}{3} & {\color{#0066A7}{2\leq u\leq 6}} \\ u &= \sqrt{\dfrac{52}{3}}\approx 4.163 & {\color{#0066A7}{\hbox{Disregard the negative solution since \(u>0\).}}} \end{array} \end{eqnarray*}$$