Records indicate that t hours past midnight,the temperature at the local airport was $f ( t ) = - 0.3 t ^ { 2 } + 4 t + 10$ degrees Fahrenheit.What was the average temperature at the airport between 8:00 A.M.and noon? Round your answer to one decimal place,if necessary.
A)3.9 degrees F
B)9.6 degrees F
C)98.0 degrees F
D)19.6 degrees F

Question

Quizplus · Accepted Answer

To find the average temperature between 8:00 A.M. and noon (which is from t = 8 to t = 12 hours past midnight), we need to calculate the average value of the function $f(t) = -0.3t^2 + 4t + 10$ over this interval. The formula for the average value of a function $f(t)$ from $a$ to $b$ is given by $\frac{1}{b-a} \int_{a}^{b} f(t) dt$.For $f(t) = -0.3t^2 + 4t + 10$, $a = 8$, and $b = 12$, the integral becomes:$$\text{Average temperature} = \frac{1}{12-8} \int_{8}^{12} (-0.3t^2 + 4t + 10) dt$$$$= \frac{1}{4} \left[ -0.1t^3 + 2t^2 + 10t \right]_{8}^{12}$$$$= \frac{1}{4} \left( (-0.1 \cdot 12^3 + 2 \cdot 12^2 + 10 \cdot 12) - (-0.1 \cdot 8^3 + 2 \cdot 8^2 + 10 \cdot 8) \right)$$$$= \frac{1}{4} \left( (-0.1 \cdot 1728 + 2 \cdot 144 + 120) - (-0.1 \cdot 512 + 2 \cdot 64 + 80) \right)$$$$= \frac{1}{4} \left( (-172.8 + 288 + 120) - (-51.2 + 128 + 80) \right)$$$$= \frac{1}{4} \left( 235.2 - 156.8 \right)$$$$= \frac{1}{4} \cdot 78.4 = 19.6$$Therefore, the average temperature at the airport between 8:00 A.M. and noon was 19.6 degrees Fahrenheit.

Records Indicate That T Hours Past Midnight,the Temperature at the Local