Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 427: 89

Answer

The volume of inhaled air in the lungs at time $~t~$ is $~\frac{5}{4\pi}~ (1-cos~\frac{2\pi~t}{5})~liters$

Work Step by Step

We can find an expression for the volume of inhaled air in the lungs at time $t$: $\int_{0}^{t}f(t)~dt$ $= \int_{0}^{t}\frac{1}{2}sin(\frac{2\pi~t}{5})~dt$ Let $u = \frac{2\pi~t}{5}$ $\frac{du}{dt} = \frac{2\pi}{5}$ $dt = \frac{5~du}{2\pi}$ $\int_{0}^{2\pi~t/5}\frac{1}{2}sin~u~\frac{5~du}{2\pi}$ $=\int_{0}^{2\pi~t/5}\frac{5}{4\pi}sin~u~du$ $=\frac{5}{4\pi}(-cos~u)~\vert_{0}^{2\pi~t/5}$ $=\frac{5}{4\pi}\cdot [(-cos~\frac{2\pi~t}{5})-(-cos~0)]$ $=\frac{5}{4\pi}~ (1-cos~\frac{2\pi~t}{5})$ The volume of inhaled air in the lungs at time $~t~$ is $~\frac{5}{4\pi}~ (1-cos~\frac{2\pi~t}{5})~liters$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.