6.5 Average Value of a Function/17: Difference between revisions
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<math> | <math> | ||
\frac{1}{12}\int_{0}^{12} 50 + 14\sin(\frac{\pi}{12}t)\,dt =\frac{1}{12}[50t-\frac{168}{\pi}\cos(\frac{\pi}{12}t)]\bigg|_{12}^{0}=\frac{1}{12}[(50)(12)-\frac{168}{\pi}\cos(\pi))(0-\frac{168}{\pi}\cos(0)]=\frac{1}{12}[600-\frac{168}{\pi}(-1)+\frac{168}{\pi}(1)] =\frac{1}{12}[600+\frac{168}{\pi}+\frac{168}{\pi}] | \frac{1}{12}\int_{0}^{12} 50 + \underbrace{14\sin(\frac{\pi}{12}t)}\,dt =\frac{1}{12}[50t-\frac{168}{\pi}\cos(\frac{\pi}{12}t)]\bigg|_{12}^{0}=\frac{1}{12}[(50)(12)-\frac{168}{\pi}\cos(\pi))(0-\frac{168}{\pi}\cos(0)]=\frac{1}{12}[600-\frac{168}{\pi}(-1)+\frac{168}{\pi}(1)] =\frac{1}{12}[600+\frac{168}{\pi}+\frac{168}{\pi}] | ||
</math> | </math> | ||
<math> | <math> | ||
\begin{ | \begin{aligned} | ||
u &= \frac{\pi}{12}t \ | u &= \frac{\pi}{12}t \\ | ||
\ | dt\cdot\frac{du}{dt} &= dt \\ | ||
\frac{12}{\pi}du &= dt \\ | |||
\end{ | integrate for 14\sin(u)\frac{12}{\pi} \\ | ||
\int14\sin(u)\frac{12}{\pi}\,du \\ | |||
14\cdot\frac{12}{\pi}\int\sin(u)\,du \\ | |||
-\frac{168}{\pi}\cos(u) \\ | |||
-\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\ | |||
\end{aligned} | |||
</math> | </math> |
Revision as of 20:22, 29 November 2022
1. Use the Average Value from a to b: