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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