📝 题目
6. 1.12 在下列积分中改变积分的顺序:
(1) $\displaystyle{\int }_{1}^{\mathrm{e}}\mathrm{d}x{\int }_{0}^{\ln x}f\left( {x,y}\right) \mathrm{d}y$ ; (2) $\displaystyle{\int }_{0}^{2}\mathrm{\;d}y{\int }_{{y}^{2}}^{3y}f\left( {x,y}\right) \mathrm{d}x$ ;
(3) $\displaystyle{\int }_{-1}^{1}\mathrm{\;d}x{\int }_{-\sqrt{1 - {x}^{2}}}^{1 - {x}^{2}}f\left( {x,y}\right) \mathrm{d}y$ ; (4) $\displaystyle{\int }_{1}^{2}\mathrm{\;d}x{\int }_{\sqrt{x}}^{2}f\left( {x,y}\right) \mathrm{d}y$ .
💡 答案与解析
6. 1.12 (1) $\displaystyle{\int }_{0}^{1}\mathrm{\;d}y{\int }_{{\mathrm{e}}^{y}}^{\mathrm{e}}f\left( {x,y}\right) \mathrm{d}x$ ;
(2) $\displaystyle{\int }_{0}^{4}\mathrm{\;d}x{\int }_{\frac{x}{3}}^{\sqrt{x}}f\left( {x,y}\right) \mathrm{d}y + {\int }_{4}^{6}\mathrm{\;d}x{\int }_{\frac{x}{3}}^{2}f\left( {x,y}\right) \mathrm{d}y$ ；
(3) $\displaystyle{\int }_{-1}^{0}\mathrm{\;d}y{\int }_{-\sqrt{1 - {y}^{2}}}^{\sqrt{1 - {y}^{2}}}f\left( {x,y}\right) \mathrm{d}x$ ;
(4) $\displaystyle{\int }_{0}^{\sqrt{2}}\mathrm{\;d}y{\int }_{0}^{{y}^{2}}f\left( {x,y}\right) \mathrm{d}x + {\int }_{\sqrt{2}}^{2}\mathrm{\;d}y{\int }_{0}^{2}f\left( {x,y}\right) \mathrm{d}x$ .