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