Question

If $f\left(x\right)=x-\left[x\right]$, for every $x$ where $\left[x\right]$ is the greatest integer less than or equal to $x$ then ${\int }_{-1}^{1}f\left(x\right)dx$

• A. $1$
• B. $2$
• C. $3$
• D. $0$
• E. $\frac{1}{2}$

Answer 1

The correct option is A

$1$

Explanation for the correct option.

Find the value of the integral.

From $-1\le x<0$, the value of $\left[x\right]$ is $-1$ and in the interval $0\le x<1$, the value of $\left[x\right]$ is $0$. So the given function $f\left(x\right)=x-\left[x\right]$ can be written as:

$f\left(x\right)=\left\{\begin{array}{l}x-\left(-1\right)-1\le x<0\\ x-00\le x<1\end{array}\right.$

Now the integral can be split into two and its value can be found as:

$\begin{array}{rcl}{\int }_{-1}^{1}f\left(x\right)dx& =& {\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx\\ & =& {\int }_{-1}^0\left(x+1\right)dx+{\int }_0^1\left(x\right)dx\\ & =& {\left[\frac{x^2}{2}+x\right]}_{-1}^0+{\left[\frac{x^2}{2}\right]}_0^1\\ & =& 0+0-\frac{{\left(-1\right)}^2}{2}-\left(-1\right)+\frac{1^2}{2}-0\\ & =& -\frac{1}{2}+1+\frac{1}{2}\\ & =& 1\end{array}$

So the value of the integral ${\int }_{-1}^{1}f\left(x\right)dx$ is $1$.

Hence, the correct option is A.