Answer
2
Work Step by Step
$\int^{2}_{0} \int^{2-x}_{0} dy dx$
=$\int^{2}_{0}(2-x)dx$
=$[2x-\frac{x^2}{2}]^2_0$
=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: 1
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