R + f ( y ) = 1 R = 1 − f ( y ) r = 1 2 x 2 + 3 ( x − 1 ) ( x − 2 ) = 2 x 2 + 3 ( x 2 − 3 x + 2 ) = 2 x 2 + 3 x 2 − 9 x + 6 = 5 x 2 − 9 x + 6 {\displaystyle R+f(y)=1R=1-f(y)r=1{\begin{aligned}2x^{2}+3(x-1)(x-2)&=2x^{2}+3(x^{2}-3x+2)\\&=2x^{2}+3x^{2}-9x+6\\&=5x^{2}-9x+6\end{aligned}}}
π ∫ 0 1 [ ( 1 ) 2 − ( 1 − y 2 ) 2 ] d y = π ∫ 0 1 [ ( 1 − ( 1 − 2 y 2 + y 4 ) ] d y = π ∫ 0 1 [ ( 2 y 2 − y 4 ) ] d y = π [ 2 y 3 3 − y 5 5 ] | 0 1 = π [ 2 3 − 1 5 ] = π [ 10 15 − 3 15 ] = 7 π 15 {\displaystyle {\begin{aligned}\pi \int _{0}^{1}\left[(1)^{2}-(1-y^{2})^{2}\right]dy&=\pi \int _{0}^{1}\left[(1-(1-2y^{2}+y^{4})\right]dy=\pi \int _{0}^{1}\left[(2y^{2}-y^{4})\right]dy\\[2ex]&=\pi \left[{\frac {2y^{3}}{3}}-{\frac {y^{5}}{5}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {2}{3}}-{\frac {1}{5}}\right]=\pi \left[{\frac {10}{15}}-{\frac {3}{15}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\pi \int _{0}^{1}\left[(1)^{2}-(1-y^{2})^{2}\right]dy&=\pi \int _{0}^{1}\left[(1-(1-2y^{2}+y^{4})\right]dy=\pi \int _{0}^{1}\left[(2y^{2}-y^{4})\right]dy\\[2ex]&=\pi \left[{\frac {2y^{3}}{3}}-{\frac {y^{5}}{5}}\right]{\Bigg |}_{0}^{1}\\[2ex]&=\pi \left[{\frac {2}{3}}-{\frac {1}{5}}\right]=\pi \left[{\frac {10}{15}}-{\frac {3}{15}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a3c37d9a43456ede6764d9feefdd9789505aea31)