Answer
=$\frac{92}{3}$
Work Step by Step
$\int^4_1\int^4_0(\frac{x}{2}+\sqrt y)dxdy$
=$\int^4_1[\frac{1}{4}x^2+x\sqrt y]^4_0dy$
=$\int^4_1(4+4y^\frac{1}{2})dy$
=$[4y+\frac{8}{3}y^\frac{3}{2}]^4_1$
=$\frac{92}{3}$
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 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 874: 8
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