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