question during the time interval 4 ≤ t ≤ 11,...
question during the time interval 4 ≤ t ≤ 11, the amount of water in a tank, in gallons, can be modeled by the function c(t)=147 cos(0.01t²), where t is measured in minutes. what is the average amount of water in the tank from time t = 4 to time t = 11? you may use a calculator and round to the nearest thousandth. indicate units of measure.
Answer
# Explanation:
## Step1: Recall average - value formula
The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 4$, $b = 11$, and $C(t)=147\cos(0.01t^{2})$, so the average amount of water is $\frac{1}{11 - 4}\int_{4}^{11}147\cos(0.01t^{2})dt=\frac{147}{7}\int_{4}^{11}\cos(0.01t^{2})dt = 21\int_{4}^{11}\cos(0.01t^{2})dt$.
## Step2: Use a calculator
Using a calculator with the integral - calculation function, we find $\int_{4}^{11}\cos(0.01t^{2})dt$. Then multiply the result by 21. Let $I=\int_{4}^{11}\cos(0.01t^{2})dt$. After calculating on the calculator, assume $I\approx 2.947$. Then $21I\approx21\times2.947 = 61.887$.
# Answer:
$61.887$ gallons