Question

Let $f\left(x\right)=max\{x+\left|x\right|,x-\left|x\right|\}$, where $\left|x\right|$ denotes the greatest integer $\le x$. Then, the value of ${\int }_{-3}^{3}f\left(x\right)dx$ is

• A. $0$
• B. $\frac{51}{2}$
• C. $\frac{21}{2}$
• D. $1$

Answer 1

The correct option is C $\frac{21}{2}$

Given, $f\left(x\right)=max\{x+\left|x\right|,x-\left[x\right]\}$
$=\{2x,x\ge 0x-\left[x\right],x\le 0\}$
$\therefore {\int }_{-3}^{3}f\left(x\right)dx={\int }_{-3}^{0}x-\left[x\right]dx+{\int }_0^{3}2xdx$
$=3{\int }_{-1}^0(1+x)dx+2{\int }_0^{3}xdx$
[$\because x-\left[x\right]$ is a periodic function at $x=1$]
$=3{[x+\frac{x^2}{2}]}_{-1}^0+2{\left[\frac{x^2}{2}\right]}_0^3$
$=3[0-0-(-1+\frac{1}{2})]+3^2-0$
$=3\left[\frac{1}{2}\right]+9=\frac{3}{2}+9$
$=\frac{21}{2}$