Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 27


a)2$\pi$ b)$\pi$

Work Step by Step

a) $\int_{-2}^2 \sqrt{4-x^2}dx$ =$\frac{1}{2}[x(2)^2]$ =$2\pi$ b) $\int_0^2\sqrt{4-x^2}dx$ =$\frac{1}{4}[x(2)^2]$ =$\pi$
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