设积分区域D是由坐标轴及直线x+y=1所围成,求二重积分∫∫D(2x+3y)dxdy.
∵D:{0≤y≤1-x 0≤x≤1 ∴∫∫D(2x+3y)dxdy=∫01dx∫01-x(2x+3y)dy =∫01[2xy+3/2y2]∣01-xdx =∫01[2xy-2x2+3/2(1-x)2]dx =[x2-(3/2)x3-1/2(1-x)3]∣01 =5/6.
设积分区域D是由坐标轴及直线x+y=1所围成,求二重积分∫∫D(2x+3y)dxdy.
∵D:{0≤y≤1-x 0≤x≤1 ∴∫∫D(2x+3y)dxdy=∫01dx∫01-x(2x+3y)dy =∫01[2xy+3/2y2]∣01-xdx =∫01[2xy-2x2+3/2(1-x)2]dx =[x2-(3/2)x3-1/2(1-x)3]∣01 =5/6.