Question

The value of ${\int }_{-1}^2|\left[x\right]-\{x\left\}\right|dx$ where $\left[x\right]$ the greatest integer less then or equal to $x$ and $\left\{x\right\}$ is the fractional part of $x$ is

• A. $7/2$
• B. $5/2$
• C. $1/2$
• D. $3/2$

Answer 1

The correct option is B $5/2$

For any $xϵR,x\left[x\right]+\left\{x\right\}$ so
$\left[x\right]-\left\{x\right\}=2\left[x\right]-x$ Thus
${\int }_{-1}^2\left|\right[x]-\left\{x\right\}|dx=$
$={\int }_{-1}^0\left|2\right[x]-\left\{x\right\}|dx+{\int }_0^1\left|2\right[x]-x|dx+{\int }_1^2\left|2\right[x]-x|dx$
$={\int }_{-1}^0|2+x|dx+{\int }_0^1\left|x\right|dx+{\int }_1^2|2-x|dx$
$={\int }_{-1}^0|2+x|dx+{\int }_0^{1}xdx+{\int }_1^2(2-x)dx$
$=-(-2+\frac{1}{2})+\frac{1}{2}+2-\frac{3}{2}=\frac{5}{2}$