Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 71


$25 ^{\circ} F$

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, $\overline{T}=\dfrac{25}{365}(365-0)= 25 ^{\circ} F$
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