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. $51/2$
• C. $21/2$
• D. $1$

Answer 1

The correct option is C $21/2$

Given : $f\left(x\right)=max\{x+\left|x\right|,x-\left|x\right|\}$

${\int }_{-3}^{3}f\left(x\right)dx=?$

We know $f\left(x\right)=max\{2x:x20,0;(x<0)x-[x[=\{x\left\}\right\}$

From the curve we can see that

$f\left(x\right)=\{2xx>0\left\{x\right\}x<0\}$

Then,

${\int }_{-3}^{3}f\left(x\right)dx={\int }_{-3}^0\left\{x\right\}dx+{\int }_0^{3}2xdx={\int }_{-3}^0[x-(x\left)\right]dx+2{\int }_0^{3}xdx={\int }_{-3}^{-2}[x-(x\left)\right]dx+{\int }_{-2}^{-1}[x-(x\left)\right]dx+{\int }_{-1}^0[x-(x\left)\right]dx+2{\int }_0^{3}xdx={\int }_{-3}^{-2}xdx+{\int }_{-2}^{-1}xdx+{\int }_{-1}^{0}xdx+2{\int }_0^{3}xdx-[{\int }_{-3}^{-2}\left[x\right]dx+{\int }_{-2}^{-1}\left[x\right]dx+{\int }_{-1}^0\left[x\right]dx]={\int }_{-3}^0[x-(x\left)\right]dx+2{\int }_0^{3}xdx-[{\int }_{-3}^{-2}(-3)dx+{\int }_{-2}^{-1}-2dx+{\int }_{-1}^0-1dx]={\left[\frac{x^2}{2}\right]}_{-3}^0+2{\left[\frac{x^2}{2}\right]}_0^3+[{3\left[x\right]}_{-3}^{-2}+{2\left[x\right]}_{-2}^{-1}+{\left[x\right]}_{-1}^0]=0-\frac{9}{2}+(9-0)+(3+2+1)=15-\frac{9}{2}=\frac{21}{2}$

Hence the correct answer is $\frac{21}{2}$