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设f(x，y)是连续函数，则累次积分∫₀¹dy∫_1-y^1+y2f(x，y)dx=()
A、∫₀¹dx∫_1-x¹f(x，y)dy
B、∫₀¹dx∫₁^1-xf(x，y)dy+∫₁²dx∫₁^√x-1f(x，y)dy
C、∫₁²dx∫₁^√x-1f(x，y)dy+∫₀¹dx∫_x¹(x，y)dy
D、∫₁²dx∫_x-1^x2+1f(x，y)dy
【正确答案】：B