1. We need to evaluate $I=\int_1^2 \left|2x-[3x]\right|\,dx.$

2. Since the floor function $[3x]$ changes value when $3x$ crosses an integer, we split the interval $[1,2]$ at points where $3x$ is an integer: $x=1,\;\frac43,\;\frac53,\;2.$

So we consider:

• $1\le x<\frac43$
• $\frac43\le x<\frac53$
• $\frac53\le x<2$
• and the endpoint $x=2$ does not affect the integral.

3. Find $[3x]$ on each interval.

• For $1\le x<\frac43$, we have $3\le 3x<4$, so $[3x]=3.$ Then $|2x-[3x]|=|2x-3|.$ Since $1\le x<\frac43$, we have $2\le 2x<\frac83<3$, so $2x-3<0$. Thus $|2x-3|=3-2x.$

• For $\frac43\le x<\frac53$, we have $4\le 3x<5$, so $[3x]=4.$ Then $|2x-[3x]|=|2x-4|.$ Here $\frac83\le 2x<\frac{10}3<4$, so $2x-4<0$. Thus $|2x-4|=4-2x.$

• For $\frac53\le x<2$, we have $5\le 3x<6$, so $[3x]=5.$ Then $|2x-[3x]|=|2x-5|.$ Here $\frac{10}3\le 2x<4<5$, so $2x-5<0$. Thus $|2x-5|=5-2x.$

Hence $I=\int_1^{4/3}(3-2x)\,dx+\int_{4/3}^{5/3}(4-2x)\,dx+\int_{5/3}^{2}(5-2x)\,dx.$

4. Evaluate each part.

First: $\int_1^{4/3}(3-2x)dx=[3x-x^2]_1^{4/3}$ $=\left(4-\frac{16}{9}\right)-(3-1)=\frac{20}{9}-2=\frac{2}{9}.$

Second: $\int_{4/3}^{5/3}(4-2x)dx=[4x-x^2]_{4/3}^{5/3}$ $=\left(\frac{20}{3}-\frac{25}{9}\right)-\left(\frac{16}{3}-\frac{16}{9}\right)$ $=\frac{35}{9}-\frac{32}{9}=\frac{1}{3}.$

Third: $\int_{5/3}^{2}(5-2x)dx=[5x-x^2]_{5/3}^{2}$ $=(10-4)-\left(\frac{25}{3}-\frac{25}{9}\right)$ $=6-\frac{50}{9}=\frac{4}{9}.$

5. Add them: $I=\frac{2}{9}+\frac{1}{3}+\frac{4}{9}$ $=\frac{2}{9}+\frac{3}{9}+\frac{4}{9}=\frac{9}{9}=1.$

Therefore, the value of the integral is $\boxed{1}.$