Question

If $f\left(x\right)=x-\left[x\right]$, for every real number $x$, where $\left[x\right]$ is the integral part of $x$. Then, ${\int }_{-1}^{1}f\left(x\right)dx$ is equal to

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

Answer 1

The correct option is A $1$

$f\left(x\right)=x-\left[x\right],-1\le x<0$

$\Rightarrow f\left(x\right)=x+1$

When $0\le x<1$

$\Rightarrow f\left(x\right)=x$
${\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx$

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

$=0-[\frac{(-1{)}^2}{2}-1]+\frac{1}{2}=1$

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