∫ arctan(4t) d t {\displaystyle \int {\text{arctan(4t) }}dt}
π ∫ 1 2 [ ( 2 − 1 2 x ) 2 ] d x = 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 12 ) ] = π [ 4 + 8 12 − 3 − 1 12 ] = π [ 1 + 7 12 ] = π [ 12 12 + 7 12 ] = π [ 19 12 ] = 19 π 12 {\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[\left(2-{\frac {1}{2}}x\right)^{2}\right]dx&=\pi \int _{1}^{2}\left[\left(4-2x+{\frac {1}{4}}x^{2}\right)\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[\left(4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}\right)-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[\left(8-4+{\frac {8}{12}}\right)-\left(4-1+{\frac {1}{12}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{12}}\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[\left(2-{\frac {1}{2}}x\right)^{2}\right]dx&=\pi \int _{1}^{2}\left[\left(4-2x+{\frac {1}{4}}x^{2}\right)\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[\left(4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}\right)-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[\left(8-4+{\frac {8}{12}}\right)-\left(4-1+{\frac {1}{12}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{12}}\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/b626b7619f18339f05d8e4a2e4256b6d9eed3a0a)