Question

∬(𝑦 + 𝑥𝑦

−2)𝑑𝐴

𝑅

, 𝑅 = {(𝑥, 𝑦)|0 ≤ 𝑥 ≤ 2, 1 ≤ 𝑦 ≤ 2}

Answer 1

∬(𝑦 + 𝑥𝑦 −2)𝑑𝐴 = $\int_{1}^{2}\int_{0}^{2}(y+xy-2)dxdy\\$

$=\int_{1}^{2}[\int_{0}^{2}(y+xy-2)dy]dx\\ =\int_{1}^{2}[y^2/2+xy^2/2-2y]_{0}^{2}dx\\ =\int_{1}^{2} [2+2x-4-0]dx\\ =\int_{1}^{2}(2x-2)dx\\ =[2x^2/2-2x]_{1}^{2}\\ =[x^2-2x]_{1}^{2}\\ =4-4+1-2\\ =-1$