设D是xoy平面上由直线y=1,2x-y+3=0与2x-y-3=0所围成的区域,求∫∫（2x-y）dxdy.在D域内.题目是高等数学二重积分的计算：∫∫（2x-y)dxdy,D是由y=1,2x-y+3=0,x+y-3=0围成的区域

Question

设D是xoy平面上由直线y=1,2x-y+3=0与2x-y-3=0所围成的区域,求∫∫（2x-y）dxdy.
在D域内.
题目是高等数学二重积分的计算：∫∫（2x-y)dxdy,D是由y=1,2x-y+3=0,x+y-3=0围成的区域

Answer 1

先积y,
∫∫ (2x-y) dxdy
=∫[0→1] dx∫[3-x→2x+3] (2x-y) dy
=∫[0→1] [2xy-(1/2)y²] |[3-x→2x+3]dx
=∫[0→1] [2x(2x+3)-(1/2)(2x+3)²-2x(3-x)+(1/2)(3-x)²] dx
=∫[0→1] [4x²+6x-(1/2)(2x+3)²-6x+2x²+(1/2)(3-x)²] dx
=[(4/3)x³+3x²-(1/12)(2x+3)³-3x²+(2/3)x³-(1/6)(3-x)³] |[0→1]
=4/3 + 3 - 125/12 - 3 + 2/3 - 4/3 + 27/12 + 27/6
=-3
若有不懂请追问,如果解决问题请点下面的“选为满意答案”.