$\begin{aligned} & \int_1^2|2 x-[3 x]| d x=\int_1^{\frac{4}{3}}|2 x-3| d x+\int_{\frac{4}{3}}^{\frac{5}{3}}|2 x-4| d x+\int_{\frac{5}{3}}^2|2 x-5| d x \\ & =\int_1^{\frac{4}{3}}(3-2 x) d x+\int_{\frac{4}{3}}^{\frac{5}{3}}(4-2 x) d x+\int_{\frac{5}{3}}^2(5-2 x) d x \\ & =\left[3 x-x^2\right]_1^{4 / 3}+\left[4 x-x^2\right]_{4 / 3}^{5 / 3}+\left[5 x-x^2\right]_{5 / 3}^2 \\ & =3[x]_1^{4 / 3}+[4 x]_{4 / 3}^{5 / 3}+5\left[x^2\right]_{5 / 3}^2-\left[x^2\right]_1^2 \\ & =(3+4+5) \times \frac{1}{3}-\left(2^2-1^2\right) \\ & =4-3=1\end{aligned}$