The value of 12 ∫ _ 0 ^ 3 | x ^ 2 - 3x + 2|dx is ______

$I = {∫}_{0}^{3}\left|{x}^{2} - 3x + 2\right|dx$

$= {∫}_{0}^{3}\left|\left(x - 1\right)\left(x - 2\right)\right|dx$

$= {∫}_{0}^{1}\left({x}^{2} - 3x + 2\right)dx - {∫}_{1}^{2}\left({x}^{2} - 3x + 2\right)dx + {∫}_{2}^{3}\left({x}^{2} - 3x + 2\right)dx$

$= {\left[\dfrac{{x}^{3}}{3} - 3\left(\dfrac{{x}^{2}}{2}\right) + 2x\right]}_{0}^{1} - {\left[\dfrac{{x}^{3}}{3} - 3\left(\dfrac{{x}^{2}}{2}\right) + 2x\right]}_{1}^{2} + {\left[\dfrac{{x}^{3}}{3} - 3\left(\dfrac{{x}^{2}}{2}\right) + 2x\right]}_{2}^{3}$

$= \left(\dfrac{1}{3} - \dfrac{3}{2} + 2\right) - \left[\left(\dfrac{8}{3} - 6 + 4\right) - \left(\dfrac{1}{3} - \dfrac{3}{2} + 2\right)\right] + \left[\left(9 - \dfrac{27}{2} + 6\right) - \left(\dfrac{8}{3} - 6 + 4\right)\right]$

$= \left(\dfrac{2}{3} - 3 + 4\right) - \left(\dfrac{16}{3} - 12 + 8\right) + \left(9 - \dfrac{27}{2} + 6\right)$

$= 20 - \dfrac{109}{6} = \dfrac{11}{6}$

So,

$12I = 12 \times \dfrac{11}{6} = 22$