Answer to Question #276440 in Calculus for Air

Question #276440

∬(𝑦 + π‘₯𝑦



βˆ’2)𝑑𝐴



𝑅



, 𝑅 = {(π‘₯, 𝑦)|0 ≀ π‘₯ ≀ 2, 1 ≀ 𝑦 ≀ 2}

1
Expert's answer
2021-12-07T13:35:02-0500

∬(𝑦 + π‘₯𝑦 βˆ’2)𝑑𝐴 = "\\int_{1}^{2}\\int_{0}^{2}(y+xy-2)dxdy\\\\"

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



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