$$\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx} $$

$$ = \int\limits_1^2 {\left| {2x - \left( {3x - \left\{ {3x} \right\}} \right)} \right|dx} $$

$$ = \int\limits_1^2 {\left| {\left\{ {3x} \right\} - x} \right|dx} $$

We know, 0 $$ \le $$ $$\left\{ {3x} \right\} < 1$$ and x > 1

$$\therefore \left\{ {3x} \right\} - x < 0$$

So, $$\left| {\left\{ {3x} \right\} - 2} \right| = - \left[ {\left\{ {3x} \right\} - x} \right]$$

$$ = \int\limits_1^2 {\left( {x - \left\{ {3x} \right\}} \right)dx} $$

$$ = \left[ {{{{x^2}} \over 2}} \right]_1^2 - \int\limits_{3 \times {1 \over 3}}^{6 \times {1 \over 3}} {\left\{ {3x} \right\}dx} $$   [Period of {x} = 1, so period of {3x} = $${1 \over 3}$$]

$$ = \left( {2 - {1 \over 2}} \right) - \left( {6 - 3} \right)\int\limits_0^{{1 \over 3}} {3xdx} $$

$$ = {3 \over 2} - 3 \times 3\left[ {{{{x^2}} \over 2}} \right]_0^{{1 \over 3}}$$

$$ = {3 \over 2} - {9 \over 2}\left[ {{1 \over 9} - 0} \right]$$

$$ = {3 \over 2} - {1 \over 2}$$ = 1