Single Variable Calculus, Chapter 6, 6.5, Section 6.5, Problem 2

Determine the average value of the function $f(x) = \sin 4x $ on the interval $[- \pi, \pi]$


$
\begin{equation}
\begin{aligned}

f_{ave} =& \frac{1}{b - a} \int^b_a f(x) dx
\\
\\
f_{ave} =& \frac{1}{\pi - (- \pi)} \int^8_1 \sin 4x dx
\\
\\
\text{Let } u =& 4x, \text{ then}
\\
\\
du =& 4dx

\end{aligned}
\end{equation}
$


Also, make sure that the upper and lower limits are now in terms of $u$.


$
\begin{equation}
\begin{aligned}

f_{ave} =& \frac{1}{2 \pi} \left(\frac{1}{4} \right) \int^{4(8)}_{4(1)} \sin u du
\\
\\
f_{ave} =& \frac{1}{8 \pi} \int^{32}_4 \sin u du
\\
\\
f_{ave} =& \frac{1}{8 \pi} [- \cos u]^{32}_4
\\
\\
f_{ave} =& \frac{1}{8 \pi} [- \cos (32) - (- \cos (4))]
\\
\\
f_{ave} =& -0.059

\end{aligned}
\end{equation}
$