2. 考研
  3. 设函数f(x，y)连续，则∫12dx∫x2f(x，y)dy+∫12dy∫y4一yf(x，y) dx=

设函数f(x，y)连续，则∫12dx∫x2f(x，y)dy+∫12dy∫y4一yf(x，y) dx=

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问题 设函数f(x，y)连续，则∫12dx∫x2f(x，y)dy+∫12dy∫y4一yf(x，y) dx=
选项 A、∫12dx∫14一xf(x，y)dy．
B、∫12dx∫x4一xf(x，y)dy．
C、∫12dy∫14一yf(x，y)dx．
D、∫12dy∫y2f (x，y)dx．
答案C
解析 原式=∫12dy∫14一yf(x，y)dx，故应选(C)．
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考研数学二

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