In a certain city the temperature (in \text{F}^{\circ}) t hours after a 9AM was modeled by the function
T ( t ) = 50 + 14 sin ( π 12 t ) {\displaystyle T(t)=50+14\sin({\frac {\pi }{12}}t)} 1. Use the Average Value from a to b:
f avg = 1 b − a ∫ a b f ( x ) d x {\displaystyle f_{\text{avg}}={\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx}
1 12 ∫ 0 12 50 + 14 sin ( π 12 t ) ⏟ u = π 12 t d t ⋅ d u d t = d t 12 π d u = d t i n t e g r a t e f o r 14 sin ( u ) 12 π ∫ 14 sin ( u ) 12 π d u 14 ⋅ 12 π ∫ sin ( u ) d u − 168 π cos ( u ) − 168 π cos ( π 12 t ) d t = 1 12 [ 50 t − 168 π cos ( π 12 t ) ] | 12 0 = 1 12 [ ( 50 ) ( 12 ) − 168 π cos ( π ) ) ( 0 − 168 π cos ( 0 ) ] {\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\\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)]}
= 1 12 [ 600 − 168 π ( − 1 ) + 168 π ( 1 ) ] = 1 12 [ 600 + 168 π + 168 π ] = 1 12 [ 600 + 336 π ] = 50 + 336 12 π = 50 + 28 π = 59 F ∘ {\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 }}
1. Use the Average Value from a to b:
![{\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\\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)]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/2238f7f30c839cd4232dcf387eb7c1ccd1f6bf45)
![{\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 }}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9e7814e98f24cc5e565981b7c7557e93b699146f)