Let I = $$\int\limits_0^3 {|{x^2} - 3x + 2|dx} $$

$$ = \int\limits_0^3 {|(x - 1)(x - 2)|dx} $$

$$ = \int\limits_0^1 { + ({x^2} - 3x + 2)dx + \int\limits_1^2 { - ({x^2} - 3x + 2)dx + \int\limits_2^3 {({x^2} - 3x + 2)dx} } } $$

$$ = \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_0^1 - \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_1^2 + \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_2^3$$

$$ = \left( {{1 \over 3} - {3 \over 2} + 2} \right) - \left[ {\left( {{8 \over 3} - 6 + 4} \right) - \left( {{1 \over 3} - {3 \over 2} + 2} \right)} \right] + \left[ {\left( {{{27} \over 3} - {{27} \over 2} + 6} \right) - \left( {{8 \over 3} - 6 + 4} \right)} \right]$$

$$ = {5 \over 6} - \left( {{2 \over 3} - {5 \over 6}} \right) + \left( {{3 \over 2} - {2 \over 3}} \right)$$

$$ = {5 \over 6} + {5 \over 6} - {2 \over 3} - {2 \over 3} + {3 \over 2}$$

$$ = {{10} \over 6} - {4 \over 3} + {3 \over 2}$$

$$ = {{10 - 8 + 9} \over 6} = {{11} \over 6}$$

$$ \therefore $$ 12I = $$12 \times {{11} \over 6}$$ = 22