Answer:
The average temperature during the period of 9 am to 9 pm is 58.276 ° F
Step-by-step explanation:
From the given information, the time given as 9 pm implies that t = 12, since there is a period of 12 hours between 9 am to 9 pm.
Hence,
Let the average temperature be represented with [tex]T_{avg}[/tex]
Then:
[tex]T_{avg} = \dfrac{1}{12-0}\int^{12}_{0} T(t) dt[/tex]
[tex]T_{avg} = \dfrac{1}{12}\int^{12}_{0} (50+13 \ sin ( \dfrac{\pi t}{12}))dt[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 50t + 13 \begin {pmatrix} -\dfrac{cos \dfrac{\pi t}{12} }{\dfrac{\pi}{12}} \end {pmatrix} \end {bmatrix}^{12}_{0}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 50t - \dfrac{13(12)}{\pi} \ cos \begin {pmatrix} \dfrac{\pi t}{12} \end {pmatrix} \end {bmatrix}^{12}_{0}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 50(12) - \dfrac{13(12)}{\pi} \ cos \begin {pmatrix} \dfrac{\pi (12)}{12}\end {pmatrix} - 50 (0) + \dfrac{13(12)}{\pi} cos \begin{pmatrix}\dfrac{\pi (0)}{12} \end {pmatrix} \end {bmatrix}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 50(12) - \dfrac{156}{\pi} \ cos \pi - 50 (0) + \dfrac{156}{\pi} cos (0) \end {bmatrix}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 50(12) - \dfrac{156}{\pi} (-1)- 50 (0) + \dfrac{156}{\pi} (1) \end {bmatrix}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 600 + \dfrac{156}{\pi} + \dfrac{156}{\pi} \end {bmatrix}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 600 + \dfrac{312}{\pi} \end {bmatrix}[/tex]
[tex]T_{avg} = \dfrac{1}{12} \begin {bmatrix} 699.3126845 \end {bmatrix}[/tex]
[tex]\mathbf{T_{avg} = 58.276}[/tex]
Therefore, the average temperature during the period of 9 am to 9 pm is 58.276 ° F
