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Question Number 57329 by naka3546 last updated on 02/Apr/19

$$\underset{-1}{\int }\stackrel{2}{}\mid x\mid \lfloor x\rfloor dx=?$$

Commented by turbo msup by abdo last updated on 02/Apr/19

$$I={\int }_{-1}^0\mid x\mid \left[x\right]dx+{\int }_0^1\mid x\mid \left[x\right]dx/+{\int }_1^2\mid x\mid \left[x\right]dx$$$$={\int }_{-1}^0-x(-1)dx+0+{\int }_1^{2}xdx$$$$={\left[\frac{x^2}{2}\right]}_{-1}^0+{\left[\frac{x^2}{2}\right]}_1^2=-\frac{1}{2}+2-\frac{1}{2}$$$$\Rightarrow I=1.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Apr/19

$${\int }_{-1}^0\mid x\mid \lfloor x\rfloor dx+{\int }_0^1\mid x\mid \lfloor x\rfloor dx+{\int }_1^2\mid x\mid \lfloor x\rfloor dx$$$$now..$$$$x\lfloor x\rfloor \mid x\mid \mid x\mid \lfloor x\rfloor$$$$0>x⩾-1-1-xx$$$$1>x⩾00x0$$$$2>x⩾11xx$$$${\int }_{-1}^{0}xdx+{\int }_0^{1}0\times dx+{\int }_1^{2}xdx$$$$\frac{1}{2}\mid x^2{\mid }_{-1}^0+0+\frac{1}{2}\mid x^2{\mid }_1^2$$$$\frac{1}{2}(0-1)+\frac{1}{2}(4-1)$$$$\frac{3-1}{2}=1answer$$

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