We find the critical points where the absolute value function changes sign to partition the interval into sub-intervals.
1. Find the roots of the cubic expression inside the absolute value function by factorizing it:
$$x^3 - 3x^2 + 2x = x(x-1)(x-2) = 0 \implies x = 0, 1, 2$$
2. Determine the sign of the polynomial expression across the integration limits $$[0, 2]$$:
* For $$x \in [0, 1]$$: The product $$x(x-1)(x-2)$$ is non-negative, so $$|x^3 - 3x^2 + 2x| = x^3 - 3x^2 + 2x$$.
* For $$x \in [1, 2]$$: The product $$x(x-1)(x-2)$$ is non-positive, so $$|x^3 - 3x^2 + 2x| = -(x^3 - 3x^2 + 2x) = -x^3 + 3x^2 - 2x$$.
3. Split the integral into two separate sequential sub-integrals:
$$\frac{I}{4} = \int_{0}^{1} (x^3 - 3x^2 + 2x) \, dx + \int_{1}^{2} (-x^3 + 3x^2 - 2x) \, dx$$
4. Find the general antiderivative formula for the polynomial terms:
$$\int (x^3 - 3x^2 + 2x) \, dx = \frac{x^4}{4} - x^3 + x^2 = F(x)$$
5. Evaluate each sub-integral by substituting its respective limits:
* Sub-integral 1: $$F(1) - F(0) = \left(\frac{1}{4} - 1 + 1\right) - 0 = \frac{1}{4}$$
* Sub-integral 2: $$[-F(2)] - [-F(1)] = \left(-4 + 8 - 4\right) - \left(-\frac{1}{4} + 1 - 1\right) = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$$
6. Sum the evaluated fractions together:
$$\frac{I}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
7. Calculate the requested target final value:
$$I = 4 \cdot \left(\frac{1}{2}\right) = 2$$