Question

cálcule a integral dupla

Answer 1

$\int\limits^2_1 \int\limits^{3y}_y {(x+y)} \, dx \, dy \\\\ integrando\,em\,x\\\\ \left[\ \frac{x^2}{2} \right] \limits^{3y}_y = \frac{(3y)^2}{2}+y(3y) - \frac{y^2}{2}+y.y\\ \frac{9y^2+6y^2}{2} - \frac{y^2 + 2y^2}{2} = \frac{15y^2 - 3y^2}{2} = \\ \frac{12y^2}{2} = 6y^2\\\\ agora, \,integrando \,em\, y\\\\ \int\limits^2_1 {6y^2} \, dy = \left[ 6 \frac{y^3}{3} \right]\limits^2_1 = \\ \left[2y^3\right]\limits^2_1 = 2(2)^3 - 2(1)^3 = 16 - 2 = 14$