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

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

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

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