Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} y=2-\frac{1}{2}x, y=0, x=1, x=2 }
π ∫ 1 2 [ ( 2 − 1 2 x ) 2 ] d y = π ∫ 1 2 [ ( 4 − 2 x + 1 4 x 2 ) ] d x = π [ 4 x − x 2 + 1 12 x 3 ] | 1 2 = π [ 4 ( 2 ) − ( 2 ) 2 + 1 12 ( 2 ) 3 − ( 4 ( 1 ) − ( 1 ) 2 + 1 12 ( 1 ) 3 ) ] = π [ 8 − 4 + 8 12 − ( 4 − 1 + 1 2 ) ] = π [ 4 + 8 12 − 3 − 1 3 ] = π [ 1 + 7 12 ] = π [ 12 12 + 7 12 ] = π [ 19 12 ] = 19 π 12 {\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}}x)^{2}\right]dy&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[8-4+{\frac {8}{12}}-\left(4-1+{\frac {1}{2}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{3}}\right]=\pi \left[1+{\frac {7}{12}}\right]\\[2ex]&=\pi \left[{\frac {12}{12}}+{\frac {7}{12}}\right]=\pi \left[{\frac {19}{12}}\right]\\[2ex]&={\frac {19\pi }{12}}\end{aligned}}}
![{\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}}x)^{2}\right]dy&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[8-4+{\frac {8}{12}}-\left(4-1+{\frac {1}{2}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{3}}\right]=\pi \left[1+{\frac {7}{12}}\right]\\[2ex]&=\pi \left[{\frac {12}{12}}+{\frac {7}{12}}\right]=\pi \left[{\frac {19}{12}}\right]\\[2ex]&={\frac {19\pi }{12}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/69509bfb3754e2ad7a1ac8b47f87b3a656adaa58)