Question

lf $f\left(x\right)=x-\left[x\right]$, where $\left[x\right]$ is the integral part of $x$, then ${\int }_{-1}^{1}f\left(x\right)dx=$

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

Answer 1

The correct option is A $1$

$f\left(x\right)=x-\left[x\right]=\left\{x\right\}$ (where $\left\{\right\}$ denotes fractional part function)
$\therefore {\int }_{-1}^1(x-[x\left]\right)dx={\int }_{-1}^0(x+1)dx+{\int }_0^{1}xdx$

$={|\frac{x^2}{2}+x|}_{-1}^0+{\left|\frac{x^2}{2}\right|}_0^1$
$=[0-(\frac{1}{2}-1)]+\left[\frac{1}{2}\right]$
$=1$

Hence, $\therefore {\int }_{-1}^{1}f\left(x\right)dx=1$