若函数f(x,y)在矩形区域D：0<=x<=1,0<=y<=1上连续,且xy（∫∫f(x,y)dxdy）^2=f(x,y)-1,则f(x,y)=( )
网友回答
note that ∫∫f(x,y)dxdy is a constant,let it be c,then
xy*c^2=f(x,y)-1
f(x,y)=xy*c^2+1,
take the integral,we get
∫∫f(x,y)dxdy=(c^2)/4+1
but as assumed,it equals c.
solve:
(c^2)/4+1=c we get c=2
thus,f(x,y)=4xy+1.