Answer
$\int^{1}_{0}$ $\int^{x}_{x^{2}}$ dydx
Work Step by Step
$\int^{1}_{0}$ $\int^{\sqrt y}_{y}$dxdy = $\int^{1}_{0}$ $\int^{x}_{x^{2}}$dydx
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 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 882: 35
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