Question

Evaluate $\frac{{\int }_0^n\left[x\right]dx}{{\int }_0^n\left\{x\right\}dx}$, (where $\left[x\right]$ and $\left\{x\right\}$ are integral and fractional parts of $x$ and $nϵN$).

• A. $(n-1)$
• B. $n$
• C. $(n+1)$
• D. none of these

Answer 1

The correct option is A $(n-1)$

We have, ${\int }_0^n\left[x\right]dx$
$={\int }_0{1}^{0dx}+{\int }_1^{2}1dx+{\int }_2^{3}2dx+\cdots +{\int }_{n-1}^n(n-1)dx$
$=0+1(2-1)+2(3-2)+\cdots +(n-1)(n-(n-1\left)\right)$
$=1+2+3+\cdots +(n-1)=\frac{n(n-1)}{2}$ (i)
and ${\int }_0^n\left\{x\right\}dx={\int }_0^n(x-[x\left]\right)dx={\int }_0^{n}xdx-{\int }_0^n\left[x\right]dx$
$=\frac{n^2}{2}-\frac{n(n-1)}{2}$ [using equation (i)]
$=\frac{n}{2}(n-n+1)=\frac{n}{2}$ (ii)
$\therefore \frac{{\int }_0^n\left[x\right]dx}{{\int }_0^n\left\{x\right\}dx}=\frac{\frac{n(n-1)}{2}}{\frac{n}{2}}=(n-1)$