Question

Integrate The function
${\int }_{-1}^1{x}^2+x-\left[x\right]dx$

Answer 1

$\begin{array}{c}{\int }_{-1}^1\left[x^2+x-\left[x\right]\right]dx={\int }_{-1}^0\left[x^2+x-\left(-1\right)\right]dx+{\int }_0^1\left[x^2+x-0\right]dx\\ ={\int }_{-1}^0\left[x^2+x+1\right]dx+{\int }_0^1\left[x^2+x\right]dx\\ ={\int }_{-1}^0\left[x^2+x\right]dx+{\int }_{-1}^{0}1dx+{\int }_0^{\alpha }0dx+{\int }_{\alpha }^{2}1dx\\ ={\int }_{-1}^0-1dx+1\left(1\right)+0+1\left(2\alpha \right)\\ =-1\left(1\right)+1+2-\alpha \\ =2-\alpha \\ =2-\left(\frac{\sqrt{3}-1}{2}\right)\\ =\frac{4-\sqrt{3+1}}{2}\end{array}$