若函数f(x,y)在矩形区域D：0

题目

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)=( )

答案

note that ∫∫f(x,y)dxdy is a constant,let it be c,thenxy*c^2=f(x,y)-1f(x,y)=xy*c^2+1,take the integral,we get∫∫f(x,y)dxdy=(c^2)/4+1but as assumed,it equals c.solve:(c^2)/4+1=c we get c=2thus,f(x,y)...