Question

If ${\int }_0\left[x\right]xdx={\int }_{0}x\left[x\right]dx$. where $[.]$ denotes greatest integer function of $x$

then${\int }_1^{2}xdx$ equals?

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $1$
• D. None of the above

Answer 1

Answer: (B)

$\frac{3}{2}$

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

${\int }_0^x\left[x\right]dx={\int }_0^n\left[x\right]dx+{\int }_n^x\left[x\right]dx$

Let $\left[x\right]=n$
$=\frac{n(n-1)}{2}+n{\int }_n^{x}1dx=\frac{n(n-1)}{2}+n(x-n)$

$=nx+\frac{n(n-1)}{2}-n^2=nx+\frac{n^2}{2}-\frac{n}{2}-n^2$

$=nx-\frac{n^2}{2}-\frac{n}{2}=nx-\frac{n(n+1)}{2}=x\left[x\right]-\frac{\left[x\right]\left(\right[x]+1)}{2}$ ...(2)

From $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \frac{\left[x\right]}{2}=x\left[x\right]-\frac{\left[x\right]\left(\right[x]+1)}{2}=\frac{1}{2}=x-\frac{\left[x\right]}{2}-\frac{1}{2}$

$\Rightarrow x=1+\frac{\left[x\right]}{2}\Rightarrow {\int }_1^2(1+\frac{\left[x\right]}{2})dx=1+\frac{1}{2}=\frac{3}{2}$