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)dxd

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)dxd

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)=4xy+1.

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