Let [ t ] denote the greatest integer ≤ t . Then the value of 8 · ∫ - 1 2 1 ( [ 2 x ] + | x…

Let $I={\int }_{-\frac{1}{2}}^1\left(\right[2x]+|x\left|\right)dx$

$\left[2x\right]$ will be discontinuous at $x = 0, \frac{1}{2} ; x \in \left(\frac{-1}{2}, 1\right)$

We know that $\left|x\right| = x, x \ge 0 ; -x , x <0.$

$={\int }_{-\frac{1}{2}}^0(-1-x)dx+{\int }_0^{1/2}(0+x)dx+{\int }_{\frac{1}{2}}^1(1+x)dx$

We know that $\int x^{n}dx = x^{\frac{n+1}{n+1}} + c$

$={\left(-x-\frac{x^2}{2}\right)}_{-\frac{1}{2}}^0+{\left(\frac{x^2}{2}\right)}_0^{1/2}+{\left(x+\frac{x^2}{2}\right)}_{\frac{1}{2}}^1$

$=\left(0\right)-\left(+\frac{1}{2}-\frac{1}{8}\right)+\frac{1}{8}+\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{8}\right)=\frac{5}{8}$

$\Rightarrow 8{\int }_{-\frac{1}{2}}^1\left(\right[2x]+|x\left|\right)dx=5$.