The expression ∫ _ 0 ^ n [x] dx ∫ _ 0 ^ n \ x\ dx , where [x] and \...

We have, ${∫}_{0}^{n}\left[x\right]dx$

$= {∫}_{0}^{1}0dx + {∫}_{1}^{2}1dx + {∫}_{2}^{3}2dx + \ldots + {∫}_{n - 1}^{n}\left(n - 1\right)dx$

$= 1\left(2 - 1\right) + 2\left(3 - 2\right) + 3\left(4 - 3\right) + \ldots \ldots + \left(n - 1\right)\left\{n - \left(n - 1\right)\right\}$

$= 1 + 2 + 3 + \ldots. + \left(n - 1\right) = \dfrac{n\left(n - 1\right)}{2}$

and ${∫}_{0}^{n}\left\{x\right\}dx = {∫}_{0}^{n}\left(x - \left[x\right]\right)dx = \dfrac{n}{2}$

$∴ \dfrac{{∫}_{0}^{n}\left[x\right]dx}{{∫}_{0}^{n}\left\{x\right\}dx} = n - 1$