   Chapter 4.4, Problem 60E

Chapter
Section
Textbook Problem

Respiratory Cycle The volume V in liters, of air in the lungs during a five-second respiratory cycle is approximated by the model V = 0.1729 t + 0.1522 t 2 − 0.0374 t 3 , where t is the time in seconds. Approximate the average volume of air in the lungs during one cycle.

To determine

To calculate: Average volume of air in lungs during 5-second respiratory cycle where, the function is V(x)=0.1729t+0.1522t20.0374t3.

Explanation

Given:

The function, V(x)=0.1729t+0.1522t20.0374t3 on the interval [0,5].

Formula used:

If f is integrable on a closed interval [a,b] then the average value of f on the interval is

1baabf(x)dx.

The integration can be broken to parts as shown below.

(f(x)+g(x))dx=f(x)dx+g(x)dx.

Calculation:

On considering the given function V(x) that models the air in lungs during a five-=second respiratory cycle.

Average value of V(x)=0.1729t+0.1522t20.0374t3 is calculated as,

Average=15(0)050.1729t+0.1522t20.0374t3dt=15(050.1729tdt+050

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