Answer
$$e-1$$
Work Step by Step
$$\iint_{R} dA=\int_0^{1} \int_0^{3y^2} e^{y^3} dx dy \\= \int_0^{1} 3y^2e^{y^3} dy $$
Suppose $ y^3 =u $ and $3y^2 dy =du$
Now, $$\int_0^{1} e^{u} du=[e^u]_0^{1}\\=e^1-e^0\\=e-1$$
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 14.2 - Double Integrals over General Regions - Exercises - Page 768: 52
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