Key Concepts and Formulas

• Absolute Value Function: The definition of the absolute value function is crucial for integrating expressions involving absolute values: $|u| = \begin{cases} u & \text{if } u \ge 0 \\ -u & \text{if } u < 0 \end{cases}$
• Definite Integration of Absolute Value Functions: To evaluate $\int_a^b |f(x)| dx$, we first find the roots of $f(x)$ within the interval $[a, b]$. These roots divide the interval into subintervals. Within each subinterval, $f(x)$ has a constant sign. We then rewrite the integral as a sum of integrals over these subintervals, removing the absolute value by multiplying $f(x)$ by $-1$ where $f(x) < 0$ and leaving $f(x)$ as is where $f(x) \ge 0$.
• Properties of Definite Integrals: If $c$ is a point between $a$ and $b$, then $\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$.

Step-by-Step Solution

Step 1: Analyze the integrand and find its roots. We need to evaluate $12\int\limits_0^3 {\left| {{x^2} - 3x + 2} \right|dx}$. Let $f(x) = x^2 - 3x + 2$. To remove the absolute value, we first find the roots of $f(x) = 0$. $x^2 - 3x + 2 = 0$ Factoring the quadratic equation, we get: $(x - 1)(x - 2) = 0$ The roots are $x = 1$ and $x = 2$.

Step 2: Determine the sign of the integrand in the given interval. The roots $x=1$ and $x=2$ divide the interval of integration $[0, 3]$ into three subintervals: $[0, 1]$, $[1, 2]$, and $[2, 3]$. We need to determine the sign of $x^2 - 3x + 2$ in each of these subintervals.

• Interval [0, 1]: Choose a test value, e.g., $x = 0.5$. $f(0.5) = (0.5)^2 - 3(0.5) + 2 = 0.25 - 1.5 + 2 = 0.75 > 0$. So, $|x^2 - 3x + 2| = x^2 - 3x + 2$ for $x \in [0, 1]$.

• Interval [1, 2]: Choose a test value, e.g., $x = 1.5$. $f(1.5) = (1.5)^2 - 3(1.5) + 2 = 2.25 - 4.5 + 2 = -0.25 < 0$. So, $|x^2 - 3x + 2| = -(x^2 - 3x + 2) = -x^2 + 3x - 2$ for $x \in [1, 2]$.

• Interval [2, 3]: Choose a test value, e.g., $x = 2.5$. $f(2.5) = (2.5)^2 - 3(2.5) + 2 = 6.25 - 7.5 + 2 = 0.75 > 0$. So, $|x^2 - 3x + 2| = x^2 - 3x + 2$ for $x \in [2, 3]$.

Step 3: Split the integral into sub-integrals based on the sign changes. Using the property of definite integrals, we can rewrite the integral as: $\int_0^3 |x^2 - 3x + 2| dx = \int_0^1 (x^2 - 3x + 2) dx + \int_1^2 (-x^2 + 3x - 2) dx + \int_2^3 (x^2 - 3x + 2) dx$

Step 4: Evaluate each definite integral. We will find the antiderivative of $x^2 - 3x + 2$, which is $\frac{x^3}{3} - \frac{3x^2}{2} + 2x$.

• First integral: $\int_0^1 (x^2 - 3x + 2) dx = \left[ \frac{x^3}{3} - \frac{3x^2}{2} + 2x \right]_0^1$ $= \left( \frac{1^3}{3} - \frac{3(1)^2}{2} + 2(1) \right) - \left( \frac{0^3}{3} - \frac{3(0)^2}{2} + 2(0) \right)$ $= \left( \frac{1}{3} - \frac{3}{2} + 2 \right) - 0 = \frac{2 - 9 + 12}{6} = \frac{5}{6}$

• Second integral: $\int_1^2 (-x^2 + 3x - 2) dx = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_1^2$ $= \left( -\frac{2^3}{3} + \frac{3(2)^2}{2} - 2(2) \right) - \left( -\frac{1^3}{3} + \frac{3(1)^2}{2} - 2(1) \right)$ $= \left( -\frac{8}{3} + 6 - 4 \right) - \left( -\frac{1}{3} + \frac{3}{2} - 2 \right)$ $= \left( -\frac{8}{3} + 2 \right) - \left( -\frac{1}{3} + \frac{3}{2} - 2 \right)$ $= \left( \frac{-8 + 6}{3} \right) - \left( \frac{-2 + 9 - 12}{6} \right)$ $= -\frac{2}{3} - \left( -\frac{5}{6} \right) = -\frac{2}{3} + \frac{5}{6} = \frac{-4 + 5}{6} = \frac{1}{6}$

• 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 5: Sum the results of the sub-integrals. $\int_0^3 |x^2 - 3x + 2| dx = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{5 + 1 + 5}{6} = \frac{11}{6}$

Step 6: Multiply by the constant factor. The original question asks for the value of $12\int\limits_0^3 {\left| {{x^2} - 3x + 2} \right|dx}$. $12 \times \frac{11}{6} = 2 \times 11 = 22$

Common Mistakes & Tips

• Incorrectly identifying roots: Always double-check the factorization or the quadratic formula when finding roots. Even a small error in roots can lead to incorrect intervals.
• Sign errors in sub-integrals: When negating the quadratic in the interval where it's negative, be careful with the signs of each term. A common mistake is to forget to negate all terms.
• Calculation errors: Definite integration involves arithmetic with fractions. It's advisable to perform these calculations carefully, perhaps by finding a common denominator for all terms at once.

Summary

To evaluate the definite integral of an absolute value function, we first find the roots of the expression inside the absolute value. These roots divide the integration interval into subintervals. We then determine the sign of the expression in each subinterval and remove the absolute value accordingly. The integral is then split into a sum of integrals over these subintervals. After evaluating each sub-integral, we sum their results and multiply by any constant factor present in the original problem. In this case, the roots of $x^2 - 3x + 2$ are 1 and 2, dividing the interval $[0, 3]$ into $[0, 1]$, $[1, 2]$, and $[2, 3]$. The integral was split and evaluated over these intervals, yielding a total value of $\frac{11}{6}$ for the integral. Multiplying by 12 gives the final answer.

The final answer is \boxed{22}.