Answer
$\mathbf{a}.\text{ }158.3{}^\circ ,\text{ }\mathbf{b}\text{. }125{}^\circ ,\text{ }\mathbf{c}.\text{ }133.3{}^\circ $
Work Step by Step
\[\begin{align}
& T\left( x \right)=160-0.05{{x}^{2}} \\
& \text{The average value for the interval }\left[ a,b \right]\text{ is given by } \\
& {{f}_{avg}}=\frac{1}{b-a}\int_{a}^{b}{\left( 160-0.05{{x}^{2}} \right)}dx \\
& {{f}_{avg}}=\frac{1}{b-a}\left[ 160x-\frac{0.05{{x}^{3}}}{3} \right]_{a}^{b} \\
& {{f}_{avg}}=\frac{1}{b-a}\left[ \left( 160\left( b \right)-\frac{0.05{{b}^{3}}}{3} \right)-\left( 160\left( a \right)-\frac{0.05{{a}^{3}}}{3} \right) \right] \\
& {{f}_{avg}}=\frac{1}{b-a}\left( 160b-\frac{{{b}^{3}}}{60}-160a+\frac{{{a}^{3}}}{60} \right)\text{ }\left( \mathbf{1} \right) \\
& \\
& \mathbf{a}.\text{ }\left[ 0,10 \right]\to a=0,\text{ }b=10 \\
& \text{Substituting into }\left( \mathbf{1} \right) \\
& {{f}_{avg}}=\frac{1}{10-0}\left( 160\left( 10 \right)-\frac{{{\left( 10 \right)}^{3}}}{60}-160\left( 0 \right)+\frac{{{\left( 0 \right)}^{3}}}{60} \right)\text{ } \\
& {{f}_{avg}}=\frac{1}{10}\left( \frac{4750}{3} \right) \\
& {{f}_{avg}}=\frac{475}{3} \\
& {{f}_{avg}}=158.3{}^\circ \\
& \\
& \mathbf{b}.\text{ }\left[ 10,40 \right]\to a=10,\text{ }b=40 \\
& \text{Substituting into }\left( \mathbf{1} \right) \\
& {{f}_{avg}}=\frac{1}{40-10}\left( 160\left( 40 \right)-\frac{{{\left( 40 \right)}^{3}}}{60}-160\left( 10 \right)+\frac{{{\left( 10 \right)}^{3}}}{60} \right)\text{ } \\
& {{f}_{avg}}=\frac{1}{30}\left( 3750 \right) \\
& {{f}_{avg}}=125 \\
& {{f}_{avg}}=125{}^\circ \\
& \\
& \mathbf{c}.\text{ }\left[ 0,40 \right]\to a=10,\text{ }b=40 \\
& \text{Substituting into }\left( \mathbf{1} \right) \\
& {{f}_{avg}}=\frac{1}{40-0}\left( 160\left( 40 \right)-\frac{{{\left( 40 \right)}^{3}}}{60}-160\left( 0 \right)+\frac{{{\left( 0 \right)}^{3}}}{60} \right)\text{ } \\
& {{f}_{avg}}=\frac{1}{40}\left( \frac{16000}{3} \right) \\
& {{f}_{avg}}=\frac{400}{3} \\
& {{f}_{avg}}=133.3{}^\circ \\
\end{align}\]
