∫ − 1 2 ( x − 2 | x | ) d x = ∫ − 1 0 ( x − 2 ( − x ) ) d x + ∫ 0 2 ( x − 2 ( x ) ) d x = ( 1 2 x 2 + x 2 ) | − 1 0 + ( 1 2 x 2 − x 2 ) | 0 2 = 0 − ( 1 2 ( − 1 ) 2 + ( − 1 ) 2 ) + ( 1 2 ( 2 ) 2 − ( 2 ) 2 ) − 0 = ( 1 2 + 1 ) + ( 1 2 4 ) − 4 ) {\displaystyle {\begin{aligned}\int \limits _{-1}^{2}(x-2|x|)dx=\int \limits _{-1}^{0}(x-2(-x))dx+\int \limits _{0}^{2}(x-2(x))dx\\[2ex]&=\left({\frac {1}{2}}{x^{2}}+x^{2}\right){\bigg |}_{-1}^{0}+\left({\frac {1}{2}}{x^{2}}-x^{2}\right){\bigg |}_{0}^{2}\\[2ex]&=0-\left({\frac {1}{2}}(-1)^{2}+(-1)^{2}\right)+\left({\frac {1}{2}}(2)^{2}-(2)^{2}\right)-0\\[2ex]&=\left({\frac {1}{2}}+1\right)+\left({\frac {1}{2}}4)-4\right)\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}\int \limits _{-1}^{2}(x-2|x|)dx=\int \limits _{-1}^{0}(x-2(-x))dx+\int \limits _{0}^{2}(x-2(x))dx\\[2ex]&=\left({\frac {1}{2}}{x^{2}}+x^{2}\right){\bigg |}_{-1}^{0}+\left({\frac {1}{2}}{x^{2}}-x^{2}\right){\bigg |}_{0}^{2}\\[2ex]&=0-\left({\frac {1}{2}}(-1)^{2}+(-1)^{2}\right)+\left({\frac {1}{2}}(2)^{2}-(2)^{2}\right)-0\\[2ex]&=\left({\frac {1}{2}}+1\right)+\left({\frac {1}{2}}4)-4\right)\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/605b390a2b2c0c4624e0f08a7640e2a753e694e8)