Key Concepts and Formulas

• Definite Integral for Area: The area $A$ bounded by a curve $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$ is given by $A = \int_{a}^{b} |f(x)| \, dx$.
• Absolute Value Definition: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Fundamental Theorem of Calculus: $\int_{a}^{b} F'(x) \, dx = F(b) - F(a)$.

Step 1: Simplify the function $y=x|x-3|$ in the interval $[-1, 2]$

We need to rewrite the absolute value function. Since we are integrating from $x=-1$ to $x=2$, $x$ will always be less than $3$. Therefore $x-3 < 0$ in this interval, so $|x-3| = -(x-3) = 3-x$. Substituting this into the function gives: $y = x|x-3| = x(3-x) = 3x - x^2$ Thus, we will be integrating $f(x) = 3x - x^2$ from $x=-1$ to $x=2$.

Step 2: Set up the area integral and analyze the sign of $f(x) = 3x - x^2$ in the interval $[-1, 2]$

The area is given by: $A = \int_{-1}^{2} |3x - x^2| \, dx$ We need to determine where $3x - x^2$ is positive and negative in the interval $[-1, 2]$. We find the roots by setting $3x - x^2 = 0$, which gives $x(3-x) = 0$, so $x=0$ or $x=3$. Since $3$ is outside our interval $[-1, 2]$, we only need to consider the root $x=0$.

• For $x \in [-1, 0)$, let's test $x=-0.5$. Then $f(-0.5) = 3(-0.5) - (-0.5)^2 = -1.5 - 0.25 = -1.75 < 0$. Therefore, $f(x) < 0$ on $[-1, 0)$.
• For $x \in (0, 2]$, let's test $x=1$. Then $f(1) = 3(1) - (1)^2 = 3 - 1 = 2 > 0$. Therefore, $f(x) > 0$ on $(0, 2]$.

Therefore, we have $|3x - x^2| = \begin{cases} x^2 - 3x & \text{if } -1 \le x < 0 \\ 3x - x^2 & \text{if } 0 \le x \le 2 \end{cases}$

Step 3: Split the integral and evaluate

Now, we split the integral at $x=0$: $A = \int_{-1}^{0} (x^2 - 3x) \, dx + \int_{0}^{2} (3x - x^2) \, dx$

First integral: $\int_{-1}^{0} (x^2 - 3x) \, dx = \left[\frac{x^3}{3} - \frac{3x^2}{2}\right]_{-1}^{0} = \left(\frac{0^3}{3} - \frac{3(0)^2}{2}\right) - \left(\frac{(-1)^3}{3} - \frac{3(-1)^2}{2}\right) = 0 - \left(-\frac{1}{3} - \frac{3}{2}\right) = \frac{1}{3} + \frac{3}{2} = \frac{2}{6} + \frac{9}{6} = \frac{11}{6}$

Second integral: $\int_{0}^{2} (3x - x^2) \, dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_{0}^{2} = \left(\frac{3(2)^2}{2} - \frac{(2)^3}{3}\right) - \left(\frac{3(0)^2}{2} - \frac{(0)^3}{3}\right) = \left(\frac{12}{2} - \frac{8}{3}\right) - 0 = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$

Step 4: Calculate the total area

$A = \frac{11}{6} + \frac{10}{3} = \frac{11}{6} + \frac{20}{6} = \frac{31}{6}$

Step 5: Calculate $12A$

$12A = 12 \left(\frac{31}{6}\right) = 2(31) = 62$

Common Mistakes & Tips:

• Remember to use the absolute value when calculating area, and to split the integral where the function changes sign.
• Be careful with signs when evaluating the definite integrals.
• Double-check your arithmetic when dealing with fractions.

Summary:

We simplified the function, found the intervals where the function was positive and negative, split the integral accordingly, and evaluated each integral. Finally, we calculated the value of $12A$.

The final answer is $\boxed{62}$.