Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 409: 41


$$\int^{\sqrt 3/2}_{0}\frac{dr}{\sqrt{1-r^2}}=\frac{\pi}{3}$$

Work Step by Step

$$A=\int^{\sqrt 3/2}_{0}\frac{dr}{\sqrt{1-r^2}}$$ According to Table 1, we have $$\int \frac{1}{\sqrt{1-x^2}}dx=\sin^{-1}x+C$$ Therefore, $$A=\sin^{-1}r\Bigg]^{\sqrt 3/2}_{0}$$ $$A=(\sin^{-1}\frac{\sqrt 3}{2}-\sin^{-1}0)$$ $$A=\frac{\pi}{3}-0$$ $$A=\frac{\pi}{3}$$
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.