Question

During a certain24 ​-hour ​period, the temperature at time t​ (measured in hours from the start of the​ period) was T(t)=50+6t-1/3t^2
degrees. What was the average temperature during that​ period?

Answer 1

Answer:

The value of average temperature during that period=$58^{\circ}$

Step-by-step explanation:

We are given that

Temperature at time t  is given by

$T(t)=50+6t-\frac{1}{3}t^2$

We have to find the average temperature during a certain  24-hour period.

We know that

Average value of function

$f_{av}=\frac{1}{b-a}\int_{a}^{b}f(x)dx$

Using the formula

Average temperature during the period

$=\frac{1}{24-0}\int_{0}^{24}(50+6t-\frac{1}{3}t^2)dt$

$=\frac{1}{24}[50t+3t^2-\frac{1}{9}t^3]^{24}_{0}$

$=\frac{1}{24}(50\times 24+3(24)^2-\frac{1}{9}(24)^3)$

$=58^{\circ}$

Hence, the value of average temperature during that period=$58^{\circ}$