Answer
\[{{f}_{avg}}=2000\]
Work Step by Step
\[\begin{align}
& \text{Let }y=30x\left( 20-x \right)\text{ on the interval }\left[ 0,20 \right] \\
& \text{The average value is given by} \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{f\left( x \right)dx} \\
& \text{Therefore,} \\
& {{f}_{avg}}=\frac{1}{20-0}\int_{0}^{20}{30x\left( 20-x \right)dx} \\
& {{f}_{avg}}=\frac{1}{20}\int_{0}^{20}{\left( 600x-30{{x}^{2}} \right)dx} \\
& \text{Integrating} \\
& {{f}_{avg}}=\frac{1}{20}\left[ 300{{x}^{2}}-10{{x}^{3}} \right]_{0}^{20} \\
& {{f}_{avg}}=\frac{1}{20}\left[ 300{{\left( 20 \right)}^{2}}-10{{\left( 20 \right)}^{3}} \right]-\frac{1}{20}\left[ 300{{\left( 0 \right)}^{2}}-10{{\left( 0 \right)}^{3}} \right] \\
& {{f}_{avg}}=2000 \\
\end{align}\]
