Answer
$25 ^{\circ} F$
Work Step by Step
The fundamental theorem of calculus states that the average value of any function $f(x) $ on the interval $[p,q]$ is:
$\dfrac{1}{q-p}\int_p^q f'(x) dx=\dfrac{1}{q-p}[f'(x)]_p^q=\dfrac{f(q)-f(p)}{q-p}$
This represents an average change of $f(x)$ on $[q,p]$
Consider: $f(x)=37 \sin [(2\pi/365) (x-101) )+25 dx$
Need to find the integral for the intervals $[0,365]$
Then, $\overline{T}=\dfrac{1}{(365-0)}\int_{0}^{365} (37 \sin (\dfrac{(2\pi)}{365} (x-101) )+25) dx$
$\implies \overline{T}=\dfrac{37}{365}\int_{0}^{365} \sin (\dfrac{(2\pi)}{365} (x-101) ) dx+\dfrac{25}{365}\int_{0}^{365} dx$
The term $\dfrac{2\pi}{365}(x-101 )$ shows a period of $365$, that is, its value is $0$.
Hence, $\overline{T}=\dfrac{25}{365}(365-0)= 25 ^{\circ} F$
