Answer
$=\frac{32}{3}.$
Work Step by Step
We have the integral
$$
\int_{0}^{2} \int_{3}^{5} y(x-y) d x d y= \int_{0}^{2} \left( \frac{yx^2}{2}-y^2x \right)_3^5d y\\
= \int_{0}^{2} 8y-2y^2d y=4y^2-\frac{2}{3}y^3|_0^2=\frac{32}{3}.
$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 16 - Multiple Integration - Chapter Review Exercises - Page 907: 5
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