Question

${\int }_{10}^{100}(x-\left[x\right])dx$ is equal to

Answer 1

$I={\int }_{10}^{100}(x-\left[x\right])dx={\int }_{10}^{100}\left\{x\right\}dx$

where $x$ is periodic function with period 1

now using ${\int }_{aT}^{bT}f\left(x\right)dx=(b-a){\int }_0^{T}f\left(x\right)dx$

$I=90{\int }_0^1(x-\left[x\right])dx$

$=90[{\int }_0^{1}xdx-{\int }_0^1\left[x\right]dx]$
$=90[{\left[\frac{x^2}{2}\right]}_0^1-0]=\frac{90}{2}=45$