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{align}
\begin{aligned}


u &= \frac{\pi}{12}t  \quad \quad \quad \quad \quad \int14\sin(u)\frac{12}{\pi}\,du \\
u &= \frac{\pi}{12}t  \\
\quad dt\cdot\frac{du}{dt} &= dt \quad \quad \quad \quad 14\cdot\frac{12}{\pi}\int\sin(u)\,du \\
dt\cdot\frac{du}{dt} &= dt  \\
\frac{12}{\pi}du &= dt \quad \quad \quad \quad -\frac{168}{\pi}\cos(u) =-\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\
\frac{12}{\pi}du &= dt \\
\end{align}
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: