the integration24

$\begin{aligned} I & = \int_{1.1}^{2.5} \frac {\lfloor 2x \rfloor}{\lceil x-1 \rceil} dx \\ & = \int_{1.1}^{1.5} \frac {\lfloor 2x \rfloor}{\lceil x-1 \rceil} dx + \int_{1.5}^2 \frac {\lfloor 2x \rfloor}{\lceil x-1 \rceil} dx + \int_2^{2.5} \frac {\lfloor 2x \rfloor}{\lceil x-1 \rceil} dx \\ & = \int_{1.1}^{1.5} 2\ dx + \int_{1.5}^2 3\ dx + \int_2^{2.5} \frac 42\ dx \\ & = 2x \ \bigg|_{1.1}^{1.5} + 3x \ \bigg|_{1.5}^2 + 2x \ \bigg|_2^{2.5} \\ & = 0.8 + 1.5 + 1 \\ & = \boxed{3.3} \end{aligned}$

The result of integration is $2\times (1.5-1.1)+3\times (2-1.5)+2\times (2.5-2)=\boxed {3.3}$ .