Answer
\[{{f}_{avg}}=\frac{58}{3}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{x}^{3}}-5{{x}^{2}}+30,\text{ on the interval }0\le x\le 4 \\
& \text{The average value is given by} \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{4-0}\int_{0}^{4}{\left( {{x}^{3}}-5{{x}^{2}}+30 \right)dx} \\
& {{f}_{avg}}=\frac{1}{4}\int_{0}^{4}{\left( {{x}^{3}}-5{{x}^{2}}+30 \right)dx} \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{1}{4}\left[ \frac{1}{4}{{x}^{4}}-\frac{5}{3}{{x}^{3}}+30x \right]_{0}^{4} \\
& {{f}_{avg}}=\frac{1}{4}\left[ \frac{1}{4}{{\left( 4 \right)}^{4}}-\frac{5}{3}{{\left( 4 \right)}^{3}}+30\left( 4 \right) \right]-\frac{1}{4}\left[ 0 \right] \\
& {{f}_{avg}}=\frac{58}{3} \\
\end{align}\]
