${\int }_{-3/2}^{10} \left\{2\mathrm{x}\right\}\mathrm{dx}$ where {.} denotes the fractional part of  x, is equal to

$\mathrm{f}\left(\mathrm{x}\right)=\left\{2\mathrm{x}\right\}$ is a periodic function with period  $\frac{1}{2}$

Let

$\begin{array}{l}\mathrm{I}={\int }_{-3/2}^{10} \left\{2\mathrm{x}\right\}\mathrm{dx}={\int }_{-3(1/2)}^{20(1/2)} \left\{2\mathrm{x}\right\}\mathrm{dx}\\ =23{\int }_0^{1/2} 2\mathrm{xdx}\\ \left[\text{ as }\left\{2\mathrm{x}\right\}=2\mathrm{x}-\left[2\mathrm{x}\right]\text{ and when }\in \left[0,\frac{1}{2}\right),\left[2\mathrm{x}\right]=0\right]\\ =23{\left[{\mathrm{x}}^2\right]}_0^{1/2}=\frac{23}{4}\end{array}$