Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 31



Work Step by Step

$\int^{\frac{1}{2}}_{0}\frac{4}{\sqrt{1-x^2}}dx$ =4$sin^{-1}x|^{\frac{1}{2}}_0$ =4$sin_{-1}(\frac{1}{2})-4sin^{-1}0$ =4($\frac{\pi}{6}-4(0)$) =$\frac{2\pi}{3}$
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