设函数f(x，y)连续，则累次积分∫12dx∫x2f(x，y)dy+∫12dy∫y4-yf(x，y)dx等于

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问题设函数f(x，y)连续，则累次积分∫₁²dx∫_x²f(x，y)dy+∫₁²dy∫_y^4-yf(x，y)dx等于

选项 A、∫₁²dy∫_y²f(x，y)dx．
B、∫₁²dx∫₁^4-xf(x，y)dy．
C、∫₁²dy∫₁^4-yf(x，y)dx．
D、∫₁²dx∫_x^4-xf(x，y)dy．

答案C

解析由题设所给累次积分可画出积分区域如图所示，故有
I=∫₁²dy∫₁^4-yf(x，y)dx．

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考研数学三

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