Answer
$\frac{9}{2}$
Work Step by Step
$\int^{1}_{-2} \int^{-y^2}_{y-2}dx dy $
=$\int^{1}_{-2} (-y^2-y+2)dy$
=$[\frac{-y^3}{3}-\frac{y^2}{2}+2y]^1_{-2}$
=$(\frac{-1}{3}-\frac{1}{2}+2)-(\frac{8}{3}-2-4)$
$
=\frac{9}{2}$
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 15: Multiple Integrals - Section 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 3
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
