# The volume of inhaled air in the lungs at time t.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 5.5, Problem 65E
To determine

Expert Solution

## Answer to Problem 65E

The volume of inhaled air in the lungs is 54π[1cos(2πt5)]liters_.

### Explanation of Solution

Given:

The function is f(t)=12sin(2πt5).

Calculation:

The volume is given in terms of integral function as,

V(t)=0tf(t)dt

Substitute [12sin(2πt5)] for f(t).

V(t)=0tf(t)dt=0t[12sin(2πt5)]dt (1)

The region lies between t=0 and t=t.

Consider u=2πt5 (2)

Differentiate both sides of the equation (2).

du=2π5dtdt=52πdu

Calculate the lower limit value of u using equation (2).

Substitute 0 for t in equation (2).

u=2π(0)5=0

Calculate the upper limit value of u using equation (2).

Substitute t for t in equation (2).

u=2π(t)5=2πt5

Apply lower and upper limits for u in equation (1).

Substitute u for (2πt5) and (52πdu) for dt in equation (1).

0t[12sin(2πt5)]dt=02πt512sinu(52πdu)=12×52π02πt5sinudu=54π02πt5sinudu (3)

Integrate equation (3).

54π02πt5sinudu=54π[cosu]02πt5=54π[cos(2πt5)cos0]=54π[cos(2πt5)1]=54π[1cos(2πt5)]

Hence, the volume of inhaled air in the lungs is 54π[1cos(2πt5)]liters_.

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