Key Concepts and Formulas

• Area under a curve: If $y = f(x)$ is a non-negative function on the interval $[a, b]$, the area under the curve is given by $A = \int_a^b f(x) \, dx$.
• Finding intersection points: To find where two curves $f(x)$ and $g(x)$ intersect, solve the equation $f(x) = g(x)$.
• Properties of definite integrals: If $a < c < b$, 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: Understanding the Problem and Identifying the Functions

We are given the region defined by $0 \leq x \leq 3$ and $0 \leq y \leq \min\{x^2+2, 2x+2\}$. This means we need to find the area under the curve $y = \min\{x^2+2, 2x+2\}$ from $x=0$ to $x=3$. The functions are $f_1(x) = x^2 + 2$ and $f_2(x) = 2x + 2$. We need to determine which function is the minimum in different intervals.

Step 2: Finding the Intersection Points of the Functions

To find where the two functions intersect, we set them equal to each other: $x^2 + 2 = 2x + 2$ Subtracting $2x+2$ from both sides gives: $x^2 - 2x = 0$ Factoring out $x$, we get: $x(x - 2) = 0$ The solutions are $x = 0$ and $x = 2$. These are the points where the two functions intersect. Since both values are within the given interval $[0, 3]$, we will need to consider both.

Step 3: Determining the Minimum Function in Each Interval

We have two intervals to consider: $[0, 2]$ and $[2, 3]$.

• Interval $[0, 2]$: Let's choose a test point, say $x = 1$. Then $f_1(1) = 1^2 + 2 = 3$ and $f_2(1) = 2(1) + 2 = 4$. Since $3 < 4$, $f_1(x) = x^2 + 2$ is the smaller function in this interval.
• Interval $[2, 3]$: Let's choose a test point, say $x = 2.5$. Then $f_1(2.5) = (2.5)^2 + 2 = 6.25 + 2 = 8.25$ and $f_2(2.5) = 2(2.5) + 2 = 5 + 2 = 7$. Since $7 < 8.25$, $f_2(x) = 2x + 2$ is the smaller function in this interval.

Therefore, the minimum function is: $\min\{x^2+2, 2x+2\} = \begin{cases} x^2+2 & \text{if } 0 \leq x \leq 2 \\ 2x+2 & \text{if } 2 < x \leq 3 \end{cases}$

Step 4: Setting Up the Definite Integrals

The area $A$ is given by the sum of two integrals: $A = \int_0^2 (x^2 + 2) \, dx + \int_2^3 (2x + 2) \, dx$

Step 5: Evaluating the First Integral

$\int_0^2 (x^2 + 2) \, dx = \left[ \frac{x^3}{3} + 2x \right]_0^2 = \left( \frac{2^3}{3} + 2(2) \right) - \left( \frac{0^3}{3} + 2(0) \right) = \frac{8}{3} + 4 = \frac{8}{3} + \frac{12}{3} = \frac{20}{3}$

Step 6: Evaluating the Second Integral

$\int_2^3 (2x + 2) \, dx = \left[ x^2 + 2x \right]_2^3 = (3^2 + 2(3)) - (2^2 + 2(2)) = (9 + 6) - (4 + 4) = 15 - 8 = 7$

Step 7: Calculating the Total Area A

$A = \frac{20}{3} + 7 = \frac{20}{3} + \frac{21}{3} = \frac{41}{3}$

Step 8: Calculating 12A

$12A = 12 \left( \frac{41}{3} \right) = 4 \times 41 = 164$

Common Mistakes & Tips

• Always sketch the graphs to visualize the region and verify the intersection points.
• Be careful with the arithmetic when evaluating the definite integrals.
• Remember to split the integral if the function defining the upper or lower boundary changes.

Summary

We found the intersection points of the two functions, determined which function was the minimum in each interval, set up the definite integrals, evaluated them, and found the total area $A$. Finally, we calculated $12A$.

The final answer is \boxed{164}.