Answer
$\dfrac{32}{3}$
Work Step by Step
The area $(A)$ can be calculated as:
$A=\int_{-2}^2 [(2-x^2)-(-2)] \ dx \\= \int_{-2}^2 (4-x^2) \ dx \\=[4x-\dfrac{x^3}{3}]_{-2}^2 \\=\dfrac{16}{3}+\dfrac{16}{3} \\=\dfrac{32}{3}$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 1
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