Answer
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)
$\int_0^2\sqrt{4-x^2}dx$
=$\frac{1}{4}[x(2)^2]$
=$\pi$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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