## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 5 - Practice Exercises - Page 344: 119

#### 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 $[m,n]$ is: $\dfrac{1}{n-m}\int_m^n f'(x) dx=\dfrac{1}{n-m}[f'(x)]_m^n$ or, $=\dfrac{f(n)-f(m)}{n-m}$ This represents an average change of $f(x)$ on $[m,n]$ Here, $f(x)=37 \sin [(2\pi/365) (x-101) )+25 dx$ We are given the interval: $[0,365]$ Then, $T_{avg}=\dfrac{1}{n-m}\int_m^n f'(x) dx=\dfrac{1}{365-0}\int_{0}^{365} 37 \sin (2\pi/365 (x-101) )+25 dx=\dfrac{37}{365}\int_{0}^{365} \sin (2\pi/365 (x-101) ) dx+\dfrac{25}{365}\int_{0}^{365} dx$ Here, the integral of $\sin(2\pi/365) (x-101)$ is $0$ over the interval because it has a period of $365$ . or, $T_{avg}=(0)+\dfrac{25}{365}(365-0)= 25 ^{\circ} F$

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