In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function Te) 41 + 14 sin 12 Find the average temperature Tave during the period from 9 AM to 9 PM Step 1 Let 9:00 AM correspond to t-O hours. Then, 9:00 PM corresponds to t 12 t-1212 hours Step 2 We must find Tave 41 +14 sin 12 o Step 3 Using properties of the integral, we know the following. 112( (쯤))-...(12 12s n )dt 41+14 sin 41 ft + 14 By evaluating the first integral, we get 1241-1 12 41 de = | 41t 41r 492 Step 4 12 )dt can be done with the substitution u-12 and The second integral 14 dt can be done with the substitution12 and du dt 12 12
Step 5 -B+ å½and fromt-12 to With this substitution, the integration limits change from t-0 to u Step 6 Now, 14sin sin(u) du. Step 7 This can be evaluated as -cos(u) | 336 Step 8 Therefore, T660+3306x)- 55+336 Tave 12 ΧãF, which to the nearest degree is approximately 162 Submit Skip (you cannot come back)
Show transcribed image text In a certain city the temperature (in °F) t hours after 9 AM was modeled by the function Te) 41 + 14 sin 12 Find the average temperature Tave during the period from 9 AM to 9 PM Step 1 Let 9:00 AM correspond to t-O hours. Then, 9:00 PM corresponds to t 12 t-1212 hours Step 2 We must find Tave 41 +14 sin 12 o Step 3 Using properties of the integral, we know the following. 112( (쯤))-...(12 12s n )dt 41+14 sin 41 ft + 14 By evaluating the first integral, we get 1241-1 12 41 de = | 41t 41r 492 Step 4 12 )dt can be done with the substitution u-12 and The second integral 14 dt can be done with the substitution12 and du dt 12 12
Step 5 -B+ å½and fromt-12 to With this substitution, the integration limits change from t-0 to u Step 6 Now, 14sin sin(u) du. Step 7 This can be evaluated as -cos(u) | 336 Step 8 Therefore, T660+3306x)- 55+336 Tave 12 ΧãF, which to the nearest degree is approximately 162 Submit Skip (you cannot come back)