Question

Let $f(x)=\max \{x+|x|,x-[x]\}$, where $[x]$ denotes the greatest integer $\le x$. Then, the value of $\underset{-3}{\overset{3}{\int }}f(x)dx$ is

• A. $0$
• B. $51/2$
• C. $21/2$
• D. $1$

Answer 1

Answer: (C)

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