   Chapter 6.2, Problem 58E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Population Growth In Exercises 57 and 58, use a graphing utility to graph the growth function. Use the integration table in Appendix C to find the average value of the growth function over the interval, where N is the size of a population and t is the time in days. N = 375 1 + e 4.20 − 0.25 t ' [ 21 , 28 ]

To determine

To calculate: The average value of the provided growth function N=3751+e4.200.25t over the interval [21,28] using Appendix C. and then plot the graph of growth function over the interval [21,28] using graphic calculator

Explanation

Given Information:

The growth function is;

N=3751+e4.200.25t

Formula used:

The formula for the integral 11+eudu is:

11+eudu=uln(1+eu)+C

The Formula for average value of the function: is;

average value=1baabf(x)dx

General power differentiation Rule:

ddx[un]=nun1dudx

Calculation:

Consider u=4.200.25t.

Differentiate the considered function with respect to t using power rule of differentiation.

du=(00.25)dt=0.25dt

Consider the provided expression:

N=3751+e4.200.25t

Now, use the formula for average value to find the average value of growth function for interval [21,28].

average value=1282121283751+e4.200.25tdt=3757212811+e4.200.25tdt

Multiply and divide by 0.25.

average value=3757(0.25)2128(11+e4.200.25t)(0.25dt)=214.2852128(11+e4.200.25t)(0.25dt)

Substitute u for 4.200.25t, and du for 0.25dt.

average value=214.285212811+eudu

Use the formula stated above of integration and solve the above integral for average value as;

average value=214

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