University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.3 - Area by Double Integration - Exercises - Page 772: 3

Answer

$\dfrac{9}{2}$

Work Step by Step

Our aim is to integrate the integral as follows: $ Area =\iint_D dA $ $\int^{1}_{-2} \int^{-y^2}_{y-2} \space dx \space dy= \int^{1}_{-2} (-y^2-(y-2) \space dy $ or, $=[\dfrac{-y^3}{3}-\dfrac{y^2}{2}+2y]^1_{-2}$ or, $=(\dfrac{-1}{3}-\dfrac{1}{2}+2)-(\dfrac{8}{3}-\dfrac{4}{2}-4)$ or, $=\dfrac{9}{2}$
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