Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of $$\int_1^2 |2x - [3x]| \; dx$$ is ___________.

We have to evaluate the definite integral $$\displaystyle \int_{1}^{2}\, \left|\,2x-[3x]\,\right|\,dx,$$

where $$[t]$$ denotes the greatest integer less than or equal to $$t$$ (the “floor” function).

The integrand involves $$[3x]$$, so we first find the sub-intervals of $$x$$ in $$[1,2]$$ on which $$[3x]$$ remains constant. We note that

$$3x\in[3,6]\quad\text{when}\quad x\in[1,2].$$

Because $$[3x]$$ changes only at integer values of $$3x$$, we examine the points $$3,4,5,6$$:

$$\begin{aligned} 3\le 3x < 4 &\Longrightarrow& 1\le x < \dfrac{4}{3},\\ 4\le 3x < 5 &\Longrightarrow& \dfrac{4}{3}\le x < \dfrac{5}{3},\\ 5\le 3x < 6 &\Longrightarrow& \dfrac{5}{3}\le x < 2. \end{aligned}$$

Thus $$[3x]$$ takes the constant values $$3,4,5$$ respectively on the intervals

$$\bigl[1,\tfrac43\bigr),$$ $$\bigl[\tfrac43,\tfrac53\bigr),$$ $$\bigl[\tfrac53,2\bigr).$$

On each of these intervals we now write the expression inside the absolute value:

$$\left|\,2x-[3x]\,\right|=\begin{cases} |2x-3|, & 1\le x<\dfrac43,\\[4pt] |2x-4|, & \dfrac43\le x<\dfrac53,\\[4pt] |2x-5|, & \dfrac53\le x<2\quad(\text{and at }x=2\text{ a single point of zero measure}).\\ \end{cases}$$

We next determine the sign of each linear expression to remove the absolute value:

$$\begin{aligned} &\text{For }x\in[1,\tfrac43):\quad 2x\in[2,\tfrac83)\subset(2,3),\text{ so }2x-3<0,\\ &\phantom{\text{For }}\Rightarrow |2x-3| = 3-2x.\\[4pt] &\text{For }x\in[\tfrac43,\tfrac53):\quad 2x\in[\tfrac83,\tfrac{10}3)\subset(2,4),\text{ so }2x-4<0,\\ &\phantom{\text{For }}\Rightarrow |2x-4| = 4-2x.\\[4pt] &\text{For }x\in[\tfrac53,2):\quad 2x\in[\tfrac{10}3,4)\subset(3,5),\text{ so }2x-5<0,\\ &\phantom{\text{For }}\Rightarrow |2x-5| = 5-2x. \end{aligned}$$

The entire integral therefore breaks up into three plain integrals:

$$\int_{1}^{2}\left|\,2x-[3x]\,\right|dx =\int_{1}^{\tfrac43}(3-2x)\,dx+\int_{\tfrac43}^{\tfrac53}(4-2x)\,dx+\int_{\tfrac53}^{2}(5-2x)\,dx.$$

We now evaluate each part, stating first the antiderivative formula

$$\int (a-2x)\,dx = ax - x^2.$$

First integral:

$$\begin{aligned} \int_{1}^{\tfrac43}(3-2x)\,dx &=\Bigl[3x-x^{2}\Bigr]_{1}^{\tfrac43}\\ &= \left(3\cdot\dfrac43-\left(\dfrac43\right)^{2}\right)-\left(3\cdot1-1^{2}\right)\\ &= \left(4-\dfrac{16}{9}\right)-(3-1)\\ &= \dfrac{20}{9}-2\\ &= \dfrac{2}{9}. \end{aligned}$$

Second integral:

$$\begin{aligned} \int_{\tfrac43}^{\tfrac53}(4-2x)\,dx &=\Bigl[4x-x^{2}\Bigr]_{\tfrac43}^{\tfrac53}\\ &= \left(4\cdot\dfrac53-\left(\dfrac53\right)^{2}\right)-\left(4\cdot\dfrac43-\left(\dfrac43\right)^{2}\right)\\ &= \left(\dfrac{20}{3}-\dfrac{25}{9}\right)-\left(\dfrac{16}{3}-\dfrac{16}{9}\right)\\ &= \left(\dfrac{60}{9}-\dfrac{25}{9}\right)-\left(\dfrac{48}{9}-\dfrac{16}{9}\right)\\ &= \dfrac{35}{9}-\dfrac{32}{9}\\ &= \dfrac{3}{9}\\ &= \dfrac{1}{3}. \end{aligned}$$

Third integral:

$$\begin{aligned} \int_{\tfrac53}^{2}(5-2x)\,dx &=\Bigl[5x-x^{2}\Bigr]_{\tfrac53}^{2}\\ &= \left(5\cdot2-2^{2}\right)-\left(5\cdot\dfrac53-\left(\dfrac53\right)^{2}\right)\\ &= (10-4)-\left(\dfrac{25}{3}-\dfrac{25}{9}\right)\\ &= 6-\left(\dfrac{75}{9}-\dfrac{25}{9}\right)\\ &= 6-\dfrac{50}{9}\\ &= \dfrac{54}{9}-\dfrac{50}{9}\\ &= \dfrac{4}{9}. \end{aligned}$$

Adding all three values:

$$\int_{1}^{2}\left|\,2x-[3x]\,\right|dx = \dfrac{2}{9}+\dfrac{1}{3}+\dfrac{4}{9} = \dfrac{2}{9}+\dfrac{3}{9}+\dfrac{4}{9} = \dfrac{9}{9} = 1.$$

So, the answer is $$1$$.