Floor Integration

$x=\left\lfloor x \right\rfloor +\left\{ x \right\}$

$\int _{ 1 }^{ 4 }{ (x-\left\lfloor x \right\rfloor ) } dx\quad =\quad \int _{ 1 }^{ 4 }{ \left\{ x \right\} } dx$

and $\left\{ x \right\}$ is a Periodic function with period 1.

Hence $\int _{ 1 }^{ 4 }{ \left\{ x \right\} } dx\quad =\quad (4-1)\int _{ 0 }^{ 1 }{ \left\{ x \right\} } dx$

$=3*\left( \frac { 1 }{ 2 } *1*1 \right)$ $=1.5$ y={x}

Observe that for $n \leq x < n+1$ we have $\left \lfloor x \right \rfloor = n$ and therefore

$\int_1^4 (x - \left \lfloor x \right \rfloor) dx = \sum_{n=1}^3 \int_n^{n+1} (x - n)dx = \sum_{n=1}^3 \left( \frac{x^2}{2} - nx \right)_n^{n+1} = \sum_{n=1}^3 \frac{1}{2} = \frac{3}{2}.$