∫ − 1 2 ( x − 2 | x | ) d x = ∫ − 1 0 ( x − 2 ( − x ) ) d x + ∫ 0 2 ( x − 2 ( x ) ) d x = ∫ − 1 0 3 x d x − ∫ 0 2 x d x = ( 3 x 2 2 ) | − 1 0 − ( x 2 2 ) | 0 2 = [ 3 ( 0 ) 2 2 − 3 ( − 1 ) 2 2 ] − [ ( 2 ) 2 2 − ( 0 ) 2 2 ] = [ − 3 2 ] − [ 4 2 ] = − 7 2 {\displaystyle {\begin{aligned}\int _{-1}^{2}(x-2|x|)dx&=\int _{-1}^{0}(x-2(-x))dx+\int _{0}^{2}(x-2(x))dx=\int _{-1}^{0}3x\,dx-\int _{0}^{2}x\,dx\\[2ex]&=\left({\frac {3x^{2}}{2}}\right){\bigg |}_{-1}^{0}-\left({\frac {x^{2}}{2}}\right){\bigg |}_{0}^{2}\\[2ex]&=\left[{\frac {3(0)^{2}}{2}}-{\frac {3(-1)^{2}}{2}}\right]-\left[{\frac {(2)^{2}}{2}}-{\frac {(0)^{2}}{2}}\right]\\[2ex]&=\left[-{\frac {3}{2}}\right]-\left[{\frac {4}{2}}\right]\\[2ex]&=-{\frac {7}{2}}\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{-1}^{2}(x-2|x|)dx&=\int _{-1}^{0}(x-2(-x))dx+\int _{0}^{2}(x-2(x))dx=\int _{-1}^{0}3x\,dx-\int _{0}^{2}x\,dx\\[2ex]&=\left({\frac {3x^{2}}{2}}\right){\bigg |}_{-1}^{0}-\left({\frac {x^{2}}{2}}\right){\bigg |}_{0}^{2}\\[2ex]&=\left[{\frac {3(0)^{2}}{2}}-{\frac {3(-1)^{2}}{2}}\right]-\left[{\frac {(2)^{2}}{2}}-{\frac {(0)^{2}}{2}}\right]\\[2ex]&=\left[-{\frac {3}{2}}\right]-\left[{\frac {4}{2}}\right]\\[2ex]&=-{\frac {7}{2}}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/af98d9b8724f27d71606df5207f112a545a67752)