Question

求复合函数的二重积分，写出详细解答过程，谢谢大神了

Answer 1

(0,x)(0,y)∫∫2e^-(2x+y)dxdy=(0,y)∫dy(0,x)∫2e^-(2x+y)dx
先求: (0,x)∫2e^-(2x+y)dx=(0,x)∫-e^-(2x+y)d-(2x+y)=(0,x)|-e^-(2x+y)=--e^-(2x+y)+e^-y
再求;(0,y)∫-e^-(2x+y)dy+(0,y)∫e^-ydy
=(0,y)∫-e^-(2x+y)d-(2x+y)+(0,y)∫e^-yd(-y)=(0,y)|e^-(2x+y)+(0,y)|e^(-y)=e^-(2x+y)-e^-2x+e^(-y)-1

追问

错了结果是：(1-e^(-2x))(1-e^(-y))...请再算一下吧，，谢谢了

追答

我这个神太小了还是等等看有那位大神来协助吧!