6.5 Average Value of a Function/17: Difference between revisions

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<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}\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}
</math>
</math>



Revision as of 19:09, 29 November 2022

1. Use the Average Value from a to b:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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} }


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= \frac{\pi}{12}t \quad \quad \quad \quad \quad \int14\sin(u)\frac{12}{\pi}\,du \\ \quad dt\cdot\frac{du}{dt} &= dt \quad \quad \quad \quad 14\cdot\frac{12}{\pi}\int\sin(u)\,du \\ \frac{12}{\pi}du &= dt \quad \quad \quad \quad -\frac{168}{\pi}\cos(u) =-\frac{168}{\pi}\cos(\frac{\pi}{12}t) \\ \end{align} }