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Chapter 4: Problem 63

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.

Short Answer

Expert verified
The average volume of air \(V_{avg}\) in the lungs during the respiratory cycle can be calculated by evaluating the integral and dividing by the length of the interval, i.e., \(\frac{1}{5} ([0.1729/2 \cdot 5^2 + 0.1522/3 \cdot 5^3 - 0.0374/4 \cdot 5^4])\). Further calculation details are provided in Step 6.

Step by step solution

01

Understand the problem

The problem gives a function \(V(t) = 0.1729 t + 0.1522 t^2 - 0.0374 t^3\) that models the volume of air in the lungs at any time \(t\) during a 5-second respiratory cycle. The task is to find the average volume of air in the lungs during this cycle.
02

Find the average value of the function

The formula for the average value of a function \(f(t)\) over the interval \([a, b]\) is given by \(\frac{1}{b-a} \int_{a}^{b} f(t) dt\). Applying this to our function \(V(t)\), the average volume \(V_{avg}\) is given by \(\frac{1}{5-0} \int_{0}^{5} (0.1729 t + 0.1522 t^2 - 0.0374 t^3) dt\).
03

Calculate the integral

To find the average volume, we need to calculate the definite integral \(\int_{0}^{5} (0.1729 t + 0.1522 t^2 - 0.0374 t^3) dt\). Using the power rule for integrals, the result is \([0.1729/2 \cdot t^2 + 0.1522/3 \cdot t^3 - 0.0374/4 \cdot t^4] \Bigg|_0^5\).
04

Evaluate the definite integral

Next, evaluate this expression at the upper and lower limits of integration (5 and 0). The result is \([0.1729/2 \cdot 5^2 + 0.1522/3 \cdot 5^3 - 0.0374/4 \cdot 5^4] - [0.1729/2 \cdot 0^2 + 0.1522/3 \cdot 0^3 - 0.0374/4 \cdot 0^4]\).
05

Calculate the average

The result from Step 4 represents the integral part of the average volume formula. Divide it by the length of the interval, which is 5, to find the average volume. Thus, \(V_{avg} = \frac{1}{5} ([0.1729/2 \cdot 5^2 + 0.1522/3 \cdot 5^3 - 0.0374/4 \cdot 5^4])\).
06

Carry out the calculations

Perform the calculations to attain the average volume of air in the lungs during a respiratory cycle.

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