The value of $$12\int_0^3 x^2 - 3x + 2 dx$$ is ______

We wish to evaluate $$12\int_0^3 |x^2 - 3x + 2| \, dx$$. Observe that $$x^2 - 3x + 2 = (x-1)(x-2)$$, which vanishes at $$x = 1$$ and $$x = 2$$, dividing the interval $$[0,3]$$ into regions where the sign of the quadratic is constant.

For $$x \in [0,1]$$ both factors $$(x-1)$$ and $$(x-2)$$ are negative, so $$x^2 - 3x + 2 > 0$$ and hence $$|x^2 - 3x + 2| = x^2 - 3x + 2$$. On $$x \in [1,2]$$ the product $$(x-1)(x-2)$$ is negative, giving $$|x^2 - 3x + 2| = -(x^2 - 3x + 2) = -x^2 + 3x - 2$$. Finally, for $$x \in [2,3]$$ both factors are nonnegative, so $$x^2 - 3x + 2 > 0$$ and $$|x^2 - 3x + 2| = x^2 - 3x + 2$$.

Let $$I_1 = \int_0^1 (x^2 - 3x + 2)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_0^1 = \frac{1}{3} - \frac{3}{2} + 2 = \frac{2-9+12}{6} = \frac{5}{6}$$.

Next, $$I_2 = \int_1^2 (-x^2 + 3x - 2)\,dx = \left[-\frac{x^3}{3} + \frac{3x^2}{2} - 2x\right]_1^2$$ $$= \left(-\frac{8}{3} + 6 - 4\right) - \left(-\frac{1}{3} + \frac{3}{2} - 2\right) = \left(-\frac{8}{3} + 2\right) - \left(-\frac{1}{3} - \frac{1}{2}\right)$$ $$= -\frac{2}{3} + \frac{1}{3} + \frac{1}{2} = -\frac{1}{3} + \frac{1}{2} = \frac{1}{6}$$.

Similarly, $$I_3 = \int_2^3 (x^2 - 3x + 2)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_2^3$$ $$= \left(9 - \frac{27}{2} + 6\right) - \left(\frac{8}{3} - 6 + 4\right) = \left(\frac{3}{2}\right) - \left(\frac{8}{3} - 2\right) = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}$$.

Combining these results gives $$12\,(I_1 + I_2 + I_3) = 12\left(\frac{5}{6} + \frac{1}{6} + \frac{5}{6}\right) = 12 \times \frac{11}{6} = 22\,.$$ The correct answer is $$\boxed{22}$$.