Question:

Step 1: Analyze the integrand and determine its sign. First, we analyze the quadratic expression inside the absolute value, $x^2 - 3x + 2$. To remove the absolute value, we need to find the roots of this quadratic equation, as these points determine where the expression changes sign. $$x^2 - 3x + 2 = 0$$ Factoring the quadratic, we get: $$(x-1)(x-2) = 0$$ The roots are $x=1$ and $x=2$. These roots divide the number line into intervals where the expression $x^2 - 3x + 2$ has a consistent sign. Step 2: Determine the sign of the integrand over the integration interval and split the integral. The integration interval is $[0, 3]$. We examine the sign of $x^2 - 3x + 2$ in sub-intervals defined by the roots $x=1$ and $x=2$ within $[0, 3]$. * For $x \in [0, 1)$: $(x-1)$ is negative, $(x-2)$ is negative. So, $(x-1)(x-2) > 0$. * For $x \in (1, 2)$: $(x-1)$ is positive, $(x-2)$ is negative. So, $(x-1)(x-2) < 0$. * For $x \in (2, 3]$: $(x-1)$ is positive, $(x-2)$ is positive. So, $(x-1)(x-2) > 0$. Based on this, the absolute value function can be written as: $$|x^2 - 3x + 2| = \begin{cases} x^2 - 3x + 2 & \text{if } x \in [0, 1] \cup [2, 3] \\ -(x^2 - 3x + 2) & \text{if } x \in (1, 2) \end{cases}$$ Now, we can split the definite integral into three parts: $$I = \int_0^3 |x^2 - 3x + 2|\,dx = \int_0^1 (x^2 - 3x + 2)\,dx + \int_1^2 -(x^2 - 3x + 2)\,dx + \int_2^3 (x^2 - 3x + 2)\,dx$$ Step 3: Evaluate each definite integral. First, find the antiderivative of $x^2 - 3x + 2$: $$\int (x^2 - 3x + 2)\,dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x + C$$ Let's evaluate each part: For the first integral: $$\int_0^1 (x^2 - 3x + 2)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_0^1$$ $$= \left(\frac{1^3}{3} - \frac{3(1^2)}{2} + 2(1)\right) - \left(\frac{0^3}{3} - \frac{3(0^2)}{2} + 2(0)\right)$$ $$= \left(\frac{1}{3} - \frac{3}{2} + 2\right) - 0 = \frac{2 - 9 + 12}{6} = \frac{5}{6}$$ For the second integral: $$\int_1^2 -(x^2 - 3x + 2)\,dx = -\left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_1^2$$ $$= -\left[\left(\frac{2^3}{3} - \frac{3(2^2)}{2} + 2(2)\right) - \left(\frac{1^3}{3} - \frac{3(1^2)}{2} + 2(1)\right)\right]$$ $$= -\left[\left(\frac{8}{3} - 6 + 4\right) - \left(\frac{1}{3} - \frac{3}{2} + 2\right)\right]$$ $$= -\left[\left(\frac{8}{3} - 2\right) - \left(\frac{2 - 9 + 12}{6}\right)\right]$$ $$= -\left[\left(\frac{8 - 6}{3}\right) - \frac{5}{6}\right] = -\left[\frac{2}{3} - \frac{5}{6}\right]$$ $$= -\left[\frac{4 - 5}{6}\right] = -\left[-\frac{1}{6}\right] = \frac{1}{6}$$ For the third integral: $$\int_2^3 (x^2 - 3x + 2)\,dx = \left[\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right]_2^3$$ $$= \left(\frac{3^3}{3} - \frac{3(3^2)}{2} + 2(3)\right) - \left(\frac{2^3}{3} - \frac{3(2^2)}{2} + 2(2)\right)$$ $$= \left(9 - \frac{27}{2} + 6\right) - \left(\frac{8}{3} - 6 + 4\right)$$ $$= \left(15 - \frac{27}{2}\right) - \left(\frac{8}{3} - 2\right)$$ $$= \left(\frac{30 - 27}{2}\right) - \left(\frac{8 - 6}{3}\right)$$ $$= \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}$$ Step 4: Sum the results and multiply by the constant factor. Now, we add the values of the three integrals: $$I = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{5+1+5}{6} = \frac{11}{6}$$ The problem asks for the value of $12\displaystyle\int_0^3|x^2-3x+2|\,dx$. We multiply our result $I$ by 12: $$12 \times I = 12 \times \frac{11}{6} = 2 \times 11 = 22$$ The final answer is $\boxed{22}$.