## Calculus (3rd Edition)

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