Let f: [ 0, ∞ ) → [ 0, ∞ ) be defined as f (x) = ∫ _ 0 ^ x [y]dy...

$f: [ 0, \infty ) \to [ 0, \infty )$, $f\left(x\right) = {∫}_{0}^{x}\left[y\right]dy$

Let $x = n + f, f \in ( 0, 1 )$

$⇒ \left[x\right] = n, \left\{x\right\} = f$.

So

$f\left(x\right) = {∫}_{0}^{1}\left[y\right]dy + {∫}_{1}^{2}\left[y\right]dy + {∫}_{2}^{3}\left[y\right]dy + \ldots + {∫}_{n - 1}^{n}\left[y\right]dy + {∫}_{n}^{n + f}\left[y\right]dy$

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

$f\left(x\right) = 0\left[1 - 0\right] + 1\left[2 - 1\right] + 2\left[3 - 2\right] + \ldots + \left(n - 1\right)\left[n - \left(n - 1\right)\right] + n\left[n + f - n\right]$

$f\left(x\right) = 0 + 1 + 2 + \ldots + \left(n - 1\right) + nf$

$⇒ f\left(x\right) = \dfrac{n ( n - 1 )}{2} + nf$

$= \dfrac{[ x ] ( [ x ] - 1 )}{2} + \left[x\right]\left\{x\right\}$

Note $\lim_{x \to {n}^{+}}f\left(x\right) = \dfrac{n\left(n - 1\right)}{2}$

$\lim_{x \to {n}^{-}}f\left(x\right) = \dfrac{( n - 1 ) ( n - 2 )}{2} + \left(n - 1\right)$

$= \dfrac{n ( n - 1 )}{2}$

$f\left(x\right) = \dfrac{n\left(n - 1\right)}{2}\left(n \in {N}_{0}\right)$

so $f ( x )$ is continuous $\forall x \ge 0$ and differentiable except at integer.'