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

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

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function T(t) = 45 + 20 sin t 12 Find the average temperature Tave during the period from 9 AM to 9 PM. Step 1 Let 9:00 AM correspond to t = 0 hours. Then, 9:00 PM corresponds to t = 12 12 hours. Step 2 1 12 12 We must find T ave (45 + 20 sin dt. 12 12 Step 3 Using properties of the integral, we know the following. r12 r12 r12 45 + 20 sin sin dt = 45 dt + 20 By evaluating the first integral, we get 12 112 45 dt =|45t 45t = 540 540 Step 4 12 t dt can be done with the substitution u = 12 t and 12 The second integral 20 sin du = 12 dt. 12
Transcribed Image Text:This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function T(t) = 45 + 20 sin t 12 Find the average temperature Tave during the period from 9 AM to 9 PM. Step 1 Let 9:00 AM correspond to t = 0 hours. Then, 9:00 PM corresponds to t = 12 12 hours. Step 2 1 12 12 We must find T ave (45 + 20 sin dt. 12 12 Step 3 Using properties of the integral, we know the following. r12 r12 r12 45 + 20 sin sin dt = 45 dt + 20 By evaluating the first integral, we get 12 112 45 dt =|45t 45t = 540 540 Step 4 12 t dt can be done with the substitution u = 12 t and 12 The second integral 20 sin du = 12 dt. 12
Step 4 "12 nt The second integral 20 t sin dt can be done with the substitution u = and 12 du = 12 dt. 12 Step 5 With this substitution, the integration limits change from t = 0 to u = ol and fromt = 12 to u = Tt Step 6 •12 Now, 20 sin t dt = 20 sin(u) du. Jo Submit Skip (you cannot come back).
Transcribed Image Text:Step 4 "12 nt The second integral 20 t sin dt can be done with the substitution u = and 12 du = 12 dt. 12 Step 5 With this substitution, the integration limits change from t = 0 to u = ol and fromt = 12 to u = Tt Step 6 •12 Now, 20 sin t dt = 20 sin(u) du. Jo Submit Skip (you cannot come back).
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