Let [t] denote the greatest integer less than or equal to t.
Then the value of $\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx}$ is ______.

$\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx}$ <br><br> $$ = \int\limits_1^2 {\left| {2x - \left( {3x - \left\{ {3x} \right\}} \right)} \right|dx} $$<br><br> $= \int\limits_1^2 {\left| {\left\{ {3x} \right\} - x} \right|dx}$<br><br> We know, 0 $\le$ $\left\{ {3x} \right\} &lt; 1$ and x &gt; 1<br><br> $\therefore \left\{ {3x} \right\} - x &lt; 0$<br><br> So, $$\left| {\left\{ {3x} \right\} - 2} \right| = - \left[ {\left\{ {3x} \right\} - x} \right]$$<br><br> $= \int\limits_1^2 {\left( {x - \left\{ {3x} \right\}} \right)dx}$<br><br> $$ = \left[ {{{{x^2}} \over 2}} \right]_1^2 - \int\limits_{3 \times {1 \over 3}}^{6 \times {1 \over 3}} {\left\{ {3x} \right\}dx} $$&nbsp;&nbsp; [Period of {x} = 1, so period of {3x} = ${1 \over 3}$]<br><br> $$ = \left( {2 - {1 \over 2}} \right) - \left( {6 - 3} \right)\int\limits_0^{{1 \over 3}} {3xdx} $$ <br><br> $= {3 \over 2} - 3 \times 3\left[ {{{{x^2}} \over 2}} \right]_0^{{1 \over 3}}$<br><br> $= {3 \over 2} - {9 \over 2}\left[ {{1 \over 9} - 0} \right]$<br><br> $= {3 \over 2} - {1 \over 2}$ = 1