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交换积分次序:
$$
\int_0^{\frac{1}{4}} \mathrm{~d} y \int_y^{\sqrt{y}} f(x, y) \mathrm{d} x+\int_{\frac{1}{4}}^{\frac{1}{2}} \mathrm{~d} y \int_y^{\frac{1}{2}} f(x, y) \mathrm{d} x=
$$
                        
不再提醒