6.5 Average Value of a Function/17

In a certain city the temperature (in \text{F}^{\circ}) t hours after a 9AM was modeled by the function

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 T(t)=50+14\sin(\frac{\pi}{12}t) } 1. Use the Average Value from a to b:

$f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx$

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 + \underbrace{14\sin\left(\frac{\pi}{12}t\right)}_{ \begin{aligned} u &= \frac{\pi}{12}t\\dt\cdot\frac{du}{dt} &= dt\\ \frac{12}{\pi}du &= dt \\ 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}} \,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)] }

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}[600-\frac{168}{\pi}(-1)+\frac{168}{\pi}(1)] =\frac{1}{12}[600+\frac{168}{\pi}+\frac{168}{\pi}]=\frac{1}{12}[600+\frac{336}{\pi}]= 50+\frac{336}{12\pi}=50+\frac{28}{\pi}= 59\text{F}^{\circ} }