6.5 Average Value of a Function/2: Difference between revisions

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<math>
<math>
f(x) = \sin{(4x)}\text{,}\quad [-\pi, \pi] \\[2ex]
f(x) = \sin{(4x)}\text{,}\quad [-\pi, \pi]
</math>
</math>


</math>
 
<math>
\begin{align}
\begin{align}


f_{avg} &= \frac{1}{\pi-(-\pi)}\int_{-\pi}^{\pi}\sin{(4x)}\,dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sin{(4x)}\,dx \\[2ex]
f_{avg} &= \frac{1}{\pi-(-\pi)}\int_{-\pi}^{\pi}\sin{(4x)}\,dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}\sin{(4x)}\,dx \\[2ex]


&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\frac{1}{4}\,du = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]
&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]


&= -\frac{1}{8\pi}\cos(u)\bigg|_{-4\pi}^{4\pi} \\[2ex]
&= -\frac{1}{8\pi}\cos(u)\bigg|_{-4\pi}^{4\pi} \\[2ex]


&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right]
&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] = \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]
 
&= 0


\end{align}
\end{align}
</math>
</math>


U-Sub notes:<br>
<math>
<math>
\begin{align}
\begin{align}

Latest revision as of 18:10, 7 September 2022



U-Sub notes:


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