Answer
$24$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{1}^{2}[(2x) \int_{0}^{4} y dy ]dx=\int_{1}^{2}[(2x) (\dfrac{(4)^2}{0}-0) dx$
This implies that
$\int_{1}^{2} 16 x dx= 8(2^2-1^2)$
Thus, we have, $32-8=24$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 759: 1
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