Use the table of integrals to find the average value of the growth function $N = \frac { 330 } { 1 + e ^ { 5.7 - 0.25 t } }$ over the interval $[ 22,27 ]$ , where N the size of a population and t is the time in days. Round your answer to three decimal places.
A)200.507
B)758.790
C)198.507
D)391.543
E)321.407

Question

Quizplus · Accepted Answer

The average value of a function $f(t)$ over the interval $[a, b]$ is given by $\frac{1}{b-a} \int_{a}^{b} f(t) dt$. For the given growth function $N(t) = \frac{330}{1 + e^{5.7 - 0.25t}}$, we need to find the average value over the interval $[22, 27]$. This involves integrating the function over the interval and then dividing by the length of the interval, which is $27 - 22 = 5$. The exact integration process involves using the table of integrals and possibly substitution to handle the exponential part, but the specific steps depend on the table provided and are not detailed here. After performing the integration and division, the result is approximately $198.507$, which corresponds to choice C.

Use the Table of Integrals to Find the Average Value N=3301+e5.7−0.25tN = \frac { 330 } { 1 + e ^ { 5.7 - 0.25 t } }N=1+e5.7−0.25t330​

Use the Table of Integrals to Find the Average Value $N = \frac { 330 } { 1 + e ^ { 5.7 - 0.25 t } }$