Respuesta :
Answer:
[tex]T_{ave} \approx 55^{\circ}F[/tex]
Step-by-step explanation:
Given :[tex]T(t) = 54 + 11 sin (\frac{\pi t}{12})[/tex]
To Find : Find the average temperature Tave during the period from 9 AM to 9 PM.
Solution:
[tex]T(t) = 54 + 11 sin (\frac{\pi t}{12})[/tex]
Where t is hours after 9 AM
Now we are supposed to find the average temperature Tave during the period from 9 AM to 9 PM
Period : 9 a.m. to 9 p.m.
Hours between 9 am to 9 pm = 12
Interval of function: [0,12]
So, we will use mean value theorem
[tex]T_{ave}=\frac{1}{b-a}\int\limits^b_a {T(t)} \, dt[/tex]
[tex]T_{ave}=\frac{1}{12-0}\int\limits^{12}_0{ 54 + 11 sin (\frac{\pi t}{12})} \, dt[/tex]
[tex]T_{ave}=\frac{1}{12}\int\limits^{12}_0{54}\, dt+\frac{1}{12}\int\limits^{12}_0{ 11 sin (\frac{\pi t}{12})} \, dt[/tex]
[tex]T_{ave}=\frac{1}{12}\int\limits^{12}_0{54}\, dt+\frac{1}{12}\int\limits^{12}_0{ 11 sin (\frac{\pi t}{12})} \, dt[/tex]
let [tex]\frac{\pi t}{12}=u[/tex]
[tex]du =\frac{\pi}{12}dt\\\frac{12}{\pi}du=dt[/tex]
u(0)=0
u(12)=π
Now apply this substitution on right integral
[tex]T_{ave}=\frac{1}{12}[54t]^{12} _0+\frac{1}{12} \times\frac{12}{\pi}\times 11 \int\limits^{\pi}_0{ sin u} \, du[/tex]
[tex]T_{ave}=\frac{1}{12}(54 \times 12-0)+\frac{11}{\pi}(- cos u)^{\pi}_0[/tex]
[tex]T_{ave}=54+\frac{11}{\pi}(- cos {\pi}-(-cos 0))^{\pi}_0[/tex]
[tex]T_{ave}=54+\frac{11}{\pi}(- (-1)-(-1))[/tex]
[tex]T_{ave}=54+\frac{11}{\pi}(2)[/tex]
[tex]T_{ave}=54+\frac{22}{\pi}[/tex]
So, [tex]T_{ave}=54+\frac{22}{180} ^{\circ}F[/tex]
[tex]T_{ave} \approx 55^{\circ}F[/tex]
Hence the average temperature Tave during the period from 9 AM to 9 PM is [tex]55^{\circ}F[/tex]
