Question

let $I_1=\underset{0}{\overset{n}{\int }}\left[x\right]dx$ and $I_2=\underset{0}{\overset{n}{\int }}\left\{x\right\}dx,$ where $\left[x\right]$ and $\left\{x\right\}$ are integral and fractional parts of $x$ and $n\in N-\left\{1\right\}$. Then $\frac{I_1}{I_2}$ is equal to

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

Answer 1

The correct option is D $n-1$

$I_1=\underset{0}{\overset{n}{\int }}\left[x\right]dx$ and $I_2=\underset{0}{\overset{n}{\int }}\left\{x\right\}dx,$
$I_1=\underset{0}{\overset{n}{\int }}\left[x\right]dx=\underset{0}{\overset{1}{\int }}0dx+\underset{1}{\overset{2}{\int }}1dx+\cdots +\underset{n-1}{\overset{n}{\int }}\left(n-1\right)dx=0+1+2+\cdots +(n-1)=\frac{n(n-1)}{2}$
$I_2=\underset{0}{\overset{n}{\int }}\left\{x\right\}dx=\underset{0}{\overset{n}{\int }}xdx-I_1=\frac{n^2}{2}-I_1=\frac{n}{2}\Rightarrow \frac{I_1}{I_2}=n-1$