Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 847: 37

Answer

$6 \ln (3)$

Work Step by Step

Let us consider $I=\iint_{R} \dfrac{x}{y} \ dA$ We can write the given integral with limits to obtain: $I=\int_1^3 \int_{-2}^{4} \dfrac{x}{y} \ dx \ dy\\=\int_1^3 \dfrac{1}{2} [\dfrac{x^2}{y}]_{-2}^4\\=\dfrac{1}{2} \int_1^3 [\dfrac{(4)^2}{y}- \dfrac{(-2)^2}{y}] \ dy \\=\int_1^3 \dfrac{6}{y} \ dy \\=6 [\ln |y|]_1^3 \\=6 [\ln |3|-\ln |1|] \\=6 \ln (3)$
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