Answer
$$\dfrac{3 \ln 2}{2}$$
Work Step by Step
$$\iint_{R} dA= \int_{1}^{2} \int_{x}^{2x} xy^{-1} dy dx \\= \int_{1}^{2}[x \ln (y) ]_x^{2x} \\=\int_{1}^{2} [x \ln 2x- x \times \ln x] dx \\=\int_1^{2} \ln (2 x) dx \\=\dfrac{3 \ln 2}{2}$$
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 767: 25
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