The expression $\frac{\int\limits_0^n[x] d x}{\int\limits_0^n\{x\} d x}$, where $[x]$ and $\{x\}$ are integral and fractional parts of $x$ and $n \in N$ is equal to
Answer & explanation
Correct answer: option 4
We have,
$I_1=\int\limits_0^n[x] d x=\int\limits_0^1 0 d x+\int\limits_1^2 1 . d x+\int\limits_2^3 2 d x+...+\int\limits_{n-1}^n(n-1) d x$
$\Rightarrow I_1=1(2-1)+2(3-2)+3(4-3)+...+(n-1)(n-(n-1))$
$\Rightarrow I_1=1+2+3+...+(n-1)=\frac{n(n-1)}{2}$
and,
$I_2=\int\limits_0^n\{x\} d x=\int\limits_0^n(x-[x]) d x=\frac{n}{2}$ $\left[\begin{array}{r}\text { Using graph of } \\ f(x)=x-[x]\end{array}\right]$
∴ $\frac{I_1}{I_2}=n-1$