∫[0,1] dx∫[-x^2,1] f(x,y)dy交换积分次序
答案是∫[-1,0] ∫[√-y,1] f(x,y)dx吗？它围成的图形为x轴以下？
人气：355 ℃ 时间：2020-04-18 22:10:35
解答
∫[0,1] dx∫[-x^2,1] f(x,y)dy
=∫[-1,0] dy∫[(-y)^(1/2),1] f(x,y)dx+∫[0,1] dy∫[0,1] f(x,y)dx
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