Question

The volume V, in liters, of air in the lungs during a five-second respiratory cycle is approximated by the model V + 0.1729t + 0.1522t² - 0.0374t³, where t is the time in seconds. Approximate the average volume of air in the lungs during one cycle.

Answer 1

To find the average volume of air in the lungs during one cycle, calculate the definite integral of V(t) over the interval t = 0 to t = 5 seconds.

Step-by-step explanation:

The average volume of air in the lungs during one cycle can be approximated by finding the average value of the function V(t) over the interval t = 0 to t = 5 seconds. To do this, we need to find the definite integral of V(t) over the given interval:

∫[0,5] V(t) dt

After calculating the integral, divide the result by the length of the interval (5 seconds) to find the average volume of air in the lungs during one cycle.

Answer 2

The average volume of air in the lungs during one cycle is 0.53176 liters.

Step-by-step explanation:

Given : The volume V, in liters, of air in the lungs during a five-second respiratory cycle is approximated by the model
`V=0.1729t+0.1522t^2- 0.0374t^3`, where t is the time in seconds.

To find : The average volume of air in the lungs during one cycle ?

Solution :

The volume V of air in the lungs during a five-second respiratory cycle is
`V=0.1729t+0.1522t^2- 0.0374t^3`

Then the average volume of air in the lungs during one cycle [0,5] is

`V_a=(1)/(b-a)\int\limits^a_b {V(t)} \, dt`

`V_a=(1)/(5-0)\int\limits^0_5 {0.1729t+0.1522t^2- 0.0374t^3} \, dt`

`V_a=(1)/(5)[(0.1729t^2)/(2)+(0.1522t^3)/(3)- (0.0374t^4)/(4)}]^5_0`

`V_a=(1)/(5)[0.08645t^2+0.05073t^3- 0.00935t^4]^5_0`

`V_a=(1)/(5)[0.08645(5)^2+0.05073(5)^3- 0.00935(5)^4-0.08645(0)^2+0.05073(0)^3- 0.00935(0)^4]`

`V_a=(1)/(5)[2.1613+6.3413-5.8438]`

`V_a=(1)/(5)[2.6588]`

`V_a=0.53176\ l`

The average volume of air in the lungs during one cycle is 0.53176 liters.