2. 考研
  3. 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

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∫y4-yf(x，y)dx等于
选项 A、∫12dy∫y2f(x，y)dx．
B、∫12dx∫14-xf(x，y)dy．
C、∫12dy∫14-yf(x，y)dx．
D、∫12dx∫x4-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∫14-yf(x，y)dx．
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考研数学三

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