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In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function nt T(t) = 45 + 20 sin 12 Find the average temperature Tave during the period from 9 AM to 9 PM.

In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function nt T(t) = 45 + 20 sin 12 Find the average temperature Tave during the period from 9 AM to 9 PM.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 54E
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In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function 45 + 20 sin 12 T(t) Find the average temperature Tave during the period from 9 AM to 9 PM. Step 1 Let 9:00 AM correspond tot = 0 hours. Then, 9:00 PM corresponds to t = |12 12 hours. Step 2 1 12 12 nt 45 + 20 sin 12 We must find Tave dt. 12 12 | Step 3 Using properties of the integral, we know the following. L"(45 + 20 an) er - {"45 at + 20/" )a '12 '12 sin() Jat = 45 + 20 45 dt + 20 sin dt By evaluating the first integral, we get 12 12 45 dt = |45t 45t or = 540 540 Step 4 12 Tt nt and 12 The second integral 20 sin dt can be done with the substitution u = du = 12 dt. 12
Transcribed Image Text:In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function 45 + 20 sin 12 T(t) Find the average temperature Tave during the period from 9 AM to 9 PM. Step 1 Let 9:00 AM correspond tot = 0 hours. Then, 9:00 PM corresponds to t = |12 12 hours. Step 2 1 12 12 nt 45 + 20 sin 12 We must find Tave dt. 12 12 | Step 3 Using properties of the integral, we know the following. L"(45 + 20 an) er - {"45 at + 20/" )a '12 '12 sin() Jat = 45 + 20 45 dt + 20 sin dt By evaluating the first integral, we get 12 12 45 dt = |45t 45t or = 540 540 Step 4 12 Tt nt and 12 The second integral 20 sin dt can be done with the substitution u = du = 12 dt. 12
Step 5 With this substitution, the integration limits change from t = 0 to u = 0 0 and fromt = 12 to u = T Step 6 12 12 12 t sin 12 Now, 20 dt = 20 sin(u) du. Step 7 This can be evaluated as 240 240 cos(u) 2 %D 480 152.78874 Step 8 1 Therefore, Tave = °F, which to the nearest degree is 12 approximately °F. Submit Skip (you cannot come back) Need Help? Read It
Transcribed Image Text:Step 5 With this substitution, the integration limits change from t = 0 to u = 0 0 and fromt = 12 to u = T Step 6 12 12 12 t sin 12 Now, 20 dt = 20 sin(u) du. Step 7 This can be evaluated as 240 240 cos(u) 2 %D 480 152.78874 Step 8 1 Therefore, Tave = °F, which to the nearest degree is 12 approximately °F. Submit Skip (you cannot come back) Need Help? Read It
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