Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x$, then the value of ${\int }_0^2\left[|x-2|+\left[x\right]\right]dx$ is:

• A. $2$
• B. $3$
• C. $1$
• D. $4$
• E. $\frac{3}{2}$

Answer 1

The correct option is B

$3$

Explanation for correct option.

$\begin{array}{rcl}{\int }_0^2\left[|x-2|+\left[x\right]\right]dx& =& {\int }_0^2\left|x-2\right|dx+{\int }_0^2\left[x\right]dx\\ & =& {\int }_0^2\left(2-x\right)dx+{\int }_0^1\left[x\right]dx+{\int }_1^2\left[x\right]dx\left[\mathrm{if}x<2,then\left|x-2\right|=-\left(x-2\right)\right]\\ & =& {\left[2x-\frac{x^2}{2}\right]}_0^2+{\int }_0^{1}0dx+{\int }_1^{2}dx\left[\text{the value of}\left[x\right]\mathrm{between}0\mathrm{to}1\mathrm{is}0,\mathrm{and}\mathrm{between}1to2\mathrm{is}1\right]\\ & =& \left[2(2-0)-\frac{\left(2^2-0^2\right)}{2}\right]+0+{\left[x\right]}_1^2\\ & =& 2+1\\ & =& 3\end{array}$

Hence, option B is correct.