y = 2 − 1 2 x , x − a x i s y = 0 x = 1 x = 2 {\displaystyle {\begin{aligned}y=2-{\frac {1}{2}}x,x-axis\\[1ex]y=0\\[1ex]x=1\\[1ex]x=2\\[1ex]\end{aligned}}}
π ∫ 1 2 [ ( 2 − 1 2 ) 2 ] d x = π ∫ 1 2 [ ( 4 − 2 x + 1 4 x 2 ) ] d x = 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 _{1}^{2}\left[(2-{\frac {1}{2}})^{2}\right]dx&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx\\[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}y=2-{\frac {1}{2}}x,x-axis\\[1ex]y=0\\[1ex]x=1\\[1ex]x=2\\[1ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ad570f5891fb3315e2cff706966f174d13d9edcd)
![{\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}})^{2}\right]dx&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx\\[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/b8e32d2520786845b34612093f81a845b80af9c1)