6.5 Average Value of a Function/17

In a certain city the temperature (in Fahrenheit) t hours after a 9 AM 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) }

Find the average temperature during the period 9 AM to 9 PM

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

a=0 (start at 9 AM) b=12 (From 9 AM to 9 PM)

${\frac {1}{12-0}}\int _{0}^{12}50+14\sin \left({\frac {\pi }{12}}t\right)={\frac {1}{12}}\int _{0}^{12}50+\underbrace {14\sin \left({\frac {\pi }{12}}t\right)} _{\begin{aligned}u&={\frac {\pi }{12}}t\\dt\cdot {\frac {du}{dt}}&=dt\\{\frac {12}{\pi }}du&=dt\\integratefor\,14\sin(u){\frac {12}{\pi }}\\\int 14\sin(u){\frac {12}{\pi }}\,du14\cdot {\frac {12}{\pi }}\int \sin(u)\,du\\-{\frac {168}{\pi }}\cos(u)\\-{\frac {168}{\pi }}\cos({\frac {\pi }{12}}t)\end{aligned}}\,dt={\frac {1}{12}}[50t-{\frac {168}{\pi }}\cos({\frac {\pi }{12}}t)]{\bigg |}_{12}^{0}={\frac {1}{12}}[(50)(12)-{\frac {168}{\pi }}\cos(\pi ))(0-{\frac {168}{\pi }}\cos(0)]$

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} }