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}
$