Question

关于二重积分的，求高手解答啊

Answer 1

先得出∫e^(-2x)e^(-y)dxdy的的二重积分为：
[-1/2e^(-2x) +1/2e^0]*[ -e^(-y)+e^(0)] 其中x趋于无穷，y趋于无穷
=1/2(e^(-2x)+1)(e^(-y)-1)x趋于无穷，y趋于无穷
=1/2(0+1)(0-1)=-1/2
c*(-1/2)=1
c=-2
第二个：∫dx∫(0,x) 2e^-(2x+y)dy
=∫2e^(-2x)dx∫(0,x)e^(-y)dy
=∫2e^(-2x)[-e^(-x)+e^(-0)]dx
=-∫2e^(-3x)+∫e^(-2x)dx
=[2/3e^(-3x)-1/2e^(-2x)] (0,无穷）
=（0-0）-（2/3*1-1/2）=1/2-2/3=-1/6

Answer 2

Answer 3

1. C∫(0,+∞)∫(0,+∞)e^[-(2x+y)] dxdy
=C∫(0,+∞) -(1/2)e^[-(2x+y)]|[x=(0,+∞)] dy
=C∫(0,+∞) -(1/2) [e^(-∞)-e^(-y)] dy
=C∫(0,+∞) (1/2)e^(-y) dy
=-(C/2)e^(-y)|(0,+∞)
=-(C/2)[e^(-∞)-e^0]
=C/2=1
∴C=2

2. ∫(0,+∞)dx∫(0,x)2e^[-(2x+y)] dy
=∫(0,+∞) [-2e^-(2x+y)]|[y=(0,x)]dx
=∫(0,+∞) [-2e^(-3x)+2e^(-2x)]dx
=[(2/3)e^(-3x)-e^(-2x)]|(0,+∞)
=[(2/3)e^(-∞)-e^(-∞)]-[(2/3)e^0-e^0]
=0+1/3
=1/3

Answer 4