Question

The value of ${\int }_0^{\left[x\right]}(x-[x\left]\right)dx$, (where $[.]$ denotes the greatest integer function) is equal to

• A. $\left[x\right]$
• B. $\frac{\left[x\right]}{2}$
• C. $x\left[x\right]$
• D. none of these

Answer 1

Answer: (B)

$\frac{\left[x\right]}{2}$

Let $I={\int }_0^{\left[x\right]}(x-\left[x\right])dx$
$\because x-\left[x\right]$ is periodic function with period 1
$I=\left[x\right]{\int }_0^1(x-\left[x\right])dx$
$=2\left[x\right]{\left(\frac{x^2}{2}\right)}_0^1-\left[x\right]{\int }_0^1\left[x\right]dx=\frac{\left[x\right]}{2}$