∫ − 1 0 d x ∫ 0 1 − x x d y + ∫ 0 1 d x ∫ x − 1 1 − x 2 x d y = ∫ − 1 0 x 1 − x d x + ∫ 0 1 x ( 1 − x 2 − ( x − 1 ) ) d x = = ∫ − 1 0 x 1 − x d x + ∫ 0 1 x 1 − x 2 d x + ∫ 0 1 ( x − x 2 ) d x = A ∫ ( x − x 2 ) d x = x 2 2 − x 3 3 + C ∫ 0 1 ( x − x 2 ) d x = ( x 2 2 − x 3 3 ) | 0 1 = 1 2 − 1 3 = 1 6 ∫ x 1 − x 2 d x = ∫ 1 2 1 − x 2 d ( x 2 ) = ? t = 1 − x 2 , x 2 = 1 − t 2 , d ( x 2 ) = − 2 t d t ∫ 1 2 1 − x 2 d ( x 2 ) = ∫ 1 2 t ( − 2 t ) d t = − t 3 3 + C = − 1 − x 2 3 3 + C ∫ 0 1 x 1 − x 2 d x = − 1 − x 2 3 3 | 0 1 = 1 3 ∫ x 1 − x d x = ? t = 1 − x , x = 1 − t 2 , d x = − 2 t d t ∫ x 1 − x d x = ∫ ( 1 − t 2 ) t ( − 2 t ) d t = − 2 ∫ ( t 2 − t 4 ) = − 2 ( t 3 3 − t 5 5 ) + C = 2 ( 1 − x 5 5 − 1 − x 3 3 ) + C ∫ − 1 0 x 1 − x d x = 2 ( 1 − x 5 5 − 1 − x 3 3 ) | − 1 0 = 2 ( 1 5 − 1 3 − 2 5 5 + 2 3 3 ) A = 7 30 − 4 2 15 {\displaystyle {\begin{aligned}&\int \limits _{-1}^{0}dx\int \limits _{0}^{\sqrt {1-x}}x\,dy+\int \limits _{0}^{1}dx\int \limits _{x-1}^{\sqrt {1-x^{2}}}x\,dy=\int \limits _{-1}^{0}x{\sqrt {1-x}}\,dx+\int \limits _{0}^{1}x\left({\sqrt {1-x^{2}}}-(x-1)\right)\,dx=\\&=\int \limits _{-1}^{0}x{\sqrt {1-x}}\,dx+\int \limits _{0}^{1}x{\sqrt {1-x^{2}}}\,dx+\int \limits _{0}^{1}(x-x^{2})\,dx=A\\&\\&\int (x-x^{2})\,dx={\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}+C\\&\int \limits _{0}^{1}(x-x^{2})\,dx=\left({\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}\right){\Bigr |}_{0}^{1}={\frac {1}{2}}-{\frac {1}{3}}={\frac {1}{6}}\\&\\&\int x{\sqrt {1-x^{2}}}\,dx=\int {\frac {1}{2}}{\sqrt {1-x^{2}}}\,d(x^{2})=?\\&t={\sqrt {1-x^{2}}},\quad x^{2}=1-t^{2},\quad d(x^{2})=-2tdt\\&\int {\frac {1}{2}}{\sqrt {1-x^{2}}}\,d(x^{2})=\int {\frac {1}{2}}t(-2t)\,dt=-{\frac {t^{3}}{3}}+C=-{\frac {{\sqrt {1-x^{2}}}^{\;3}}{3}}+C\\&\int \limits _{0}^{1}x{\sqrt {1-x^{2}}}\,dx=\left.-{\frac {{\sqrt {1-x^{2}}}^{\;3}}{3}}\right|_{0}^{1}={\frac {1}{3}}\\&\\&\int x{\sqrt {1-x}}\,dx=?\\&t={\sqrt {1-x}},\quad x=1-t^{2},\quad dx=-2tdt\\&\int x{\sqrt {1-x}}\,dx=\int (1-t^{2})t(-2t)dt=-2\int (t^{2}-t^{4})=-2\left({\frac {t^{3}}{3}}-{\frac {t^{5}}{5}}\right)+C=2\left({\frac {{\sqrt {1-x}}^{\;5}}{5}}-{\frac {{\sqrt {1-x}}^{\;3}}{3}}\right)+C\\&\int \limits _{-1}^{0}x{\sqrt {1-x}}\,dx=\left.2\left({\frac {{\sqrt {1-x}}^{\;5}}{5}}-{\frac {{\sqrt {1-x}}^{\;3}}{3}}\right)\right|_{-1}^{0}=2\left({\frac {1}{5}}-{\frac {1}{3}}-{\frac {{\sqrt {2}}^{5}}{5}}+{\frac {{\sqrt {2}}^{3}}{3}}\right)\\&\\&A={\frac {7}{30}}-{\frac {4{\sqrt {2}}}{15}}\end{aligned}}}
∫ − 1 0 d y ∫ 0 1 + y x d x + ∫ 0 1 d y ∫ − 1 1 − y 2 x d x + ∫ 1 2 d y ∫ − 1 1 − y 2 x d x = B ∫ x d x = x 2 2 + C ∫ − 1 0 d y ∫ 0 1 + y x d x = ∫ − 1 0 ( 1 + y ) 2 2 d ( 1 + y ) = ( 1 + y ) 3 6 | − 1 0 = 1 6 ∫ 0 1 d y ∫ − 1 1 − y 2 x d x = ∫ 0 1 ( 1 − y 2 2 − 1 2 ) d y = ∫ 0 1 − y 2 2 d y = y 3 6 | 0 1 = − 1 6 ∫ 1 2 d y ∫ − 1 1 − y 2 x d x = ∫ 1 2 d y ( ( 1 − y 2 ) 2 2 − 1 2 ) = ∫ 1 2 ( y 4 2 − y 2 ) d y = ( y 5 10 − y 3 3 ) | 0 2 = 2 5 10 − 2 3 3 − 1 10 + 1 3 B = 7 30 − 4 2 15 {\displaystyle {\begin{aligned}&\int \limits _{-1}^{0}dy\int \limits _{0}^{1+y}x\,dx+\int \limits _{0}^{1}dy\int \limits _{-1}^{\sqrt {1-y^{2}}}x\,dx+\int \limits _{1}^{\sqrt {2}}dy\int \limits _{-1}^{1-y^{2}}x\,dx=B\\&\\&\int x\,dx={\frac {x^{2}}{2}}+C\\&\\&\int \limits _{-1}^{0}dy\int \limits _{0}^{1+y}x\,dx=\int \limits _{-1}^{0}{\frac {(1+y)^{2}}{2}}\,d(1+y)=\left.{\frac {(1+y)^{3}}{6}}\right|_{-1}^{0}={\frac {1}{6}}\\&\\&\int \limits _{0}^{1}dy\int \limits _{-1}^{\sqrt {1-y^{2}}}x\,dx=\int \limits _{0}^{1}\left({\frac {1-y^{2}}{2}}-{\frac {1}{2}}\right)\,dy=\int \limits _{0}^{1}-{\frac {y^{2}}{2}}\,dy=\left.{\frac {y^{3}}{6}}\right|_{0}^{1}=-{\frac {1}{6}}\\&\\&\int \limits _{1}^{\sqrt {2}}dy\int \limits _{-1}^{1-y^{2}}x\,dx=\int \limits _{1}^{\sqrt {2}}dy\left({\frac {(1-y^{2})^{2}}{2}}-{\frac {1}{2}}\right)=\int \limits _{1}^{\sqrt {2}}\left({\frac {y^{4}}{2}}-y^{2}\right)\,dy=\left.\left({\frac {y^{5}}{10}}-{\frac {y^{3}}{3}}\right)\right|_{0}^{\sqrt {2}}={\frac {{\sqrt {2}}^{5}}{10}}-{\frac {{\sqrt {2}}^{3}}{3}}-{\frac {1}{10}}+{\frac {1}{3}}\\&\\&B={\frac {7}{30}}-{\frac {4{\sqrt {2}}}{15}}\end{aligned}}}
