Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then the value of ${\int }_{-1}^1\left(\left|x\right|-2\left[x\right]\right)dx$ is:

• A. $3$
• B. $2$
• C. $-2$
• D. $-3$

Answer 1

The correct option is A

$3$

Explanation for the correct option.

$\begin{array}{rcl}{\int }_{-1}^1\left(\left|x\right|-2\left[x\right]\right)dx& =& {\int }_{-1}^0\left(\left|x\right|-2\left[x\right]\right)dx+\hspace{0.17em}{\int }_0^1\left(\left|x\right|-2\left[x\right]\right)dx\\ & =& {\int }_{-1}^0(-x-2(-1\left)\right)dx+\hspace{0.17em}{\int }_0^1(x-2\left(0\right))dx\\ & =& {\int }_{-1}^0(-x+2)dx+\hspace{0.17em}{\int }_0^{1}xdx\\ & =& {\left[-\frac{x^2}{2}+2x\right]}_{-1}^0+{\left[\frac{x^2}{2}\right]}_0^1\\ & =& \left[-\frac{\left(0^2\right)-{\left(-1\right)}^2}{2}+2\left(0+1\right)\right]+\left[\frac{\left(1^2-0^2\right)}{2}\right]\\ & =& \frac{1}{2}+2+\frac{1}{2}\\ & =& 3\end{array}$

Hence, option A is correct.