Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 4

Determine the average value of the function $f(x) = x^2 \sqrt{1 + x^3}$ on the interval $[0,2]$.


$\begin{equation} \begin{aligned} f_{ave} =& \frac{1}{b - a} \int^b_a f(x) dx \\ \\ f_{ave} =& \frac{1}{2 - 0} \int^2_0 x^2 \sqrt{1 + x^3} dx \\ \\ \text{Let } u =& 1 + x^3 \\ \\ du =& 3x^2 dx \end{aligned} \end{equation}$


Make sure that your upper and lower limits are also in terms of $u$.


$\begin{equation} \begin{aligned} f_{ave} =& \frac{1}{2} \left( \frac{1}{3} \right) \int^{1 + (2)^3}_{1 + (0)^3} u^{\frac{1}{2}} du \\ \\ f_{ave} =& \frac{1}{6} \int^9_1 u^{\frac{1}{2}} du \\ \\ f_{ave} =& \frac{1}{6} \left[ \frac{u^{\frac{3}{2}}}{\displaystyle \frac{3}{2}} \right]^9_1 \\ \\ f_{ave} =& \frac{2}{18} \left[ 9^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] \\ \\ f_{ave} =& \frac{26}{9} \end{aligned} \end{equation}$