Question

Let $I_1={\int }_0^n\left[x\right]dx$ and $I_2={\int }_0^n\left\{x\right\}dx$, where $\left[x\right]$ and $\left\{x\right\}$ are integral and fractional parts of $x$ and $nϵN-\left\{1\right\}$. Then $I_1/I_2$ is equal to

• A. $\frac{1}{n-1}$
• B. $\frac{1}{n}$
• C. $n$
• D. $n-1$

Answer 1

Answer: (D)

$n-1$

$I_1={\int }_0^n\left[x\right]dx={\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx+....+{\int }_{n-1}^n(n-1)dx$

$=0+(2-1)+2(3-2)+3(4-3)+......+(n-1)(n-(n-1\left)\right)$

$=0+1+2+3+.....+(n-1)$

$=\frac{n(n-1)}{2}$,
$I_2={\int }_0^n\left\{x\right\}dx$

$={\int }_0^{n}xdx-{\int }_0^n\left[x\right]dx$

$={\int }_0^{n}xdx-I_1$

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

$\therefore I_1/I_2=n-1$