Let the area of the region `{(x, y): 0 le x le 3`, `0 le y le min{x^2+ 2, 2x + 2}}`be A. Then 12A is equal to _____.

Answer

To solve the problem, we need to find the area \( A \) of the region defined by the inequalities \( 0 \leq x \leq 3 \) and \( 0 \leq y \leq \min(x^2 + 2, 2x + 2) \).

### Step 1: Identify the curves
We have two functions:
1. \( y = x^2 + 2 \) (a parabola)
2. \( y = 2x + 2 \) (a line)

### Step 2: Find the intersection points
To find the area, we first need to determine where these two curves intersect. We set them equal to each other:
\[ x^2 + 2 = 2x + 2 \]
Subtracting \( 2 \) from both sides gives:
\[ x^2 = 2x \]
Rearranging this, we have:
\[ x^2 - 2x = 0 \]
Factoring out \( x \):
\[ x(x - 2) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{and} \quad x = 2 \]

### Step 3: Determine the area under each curve
Next, we need to find the area under each curve from \( x = 0 \) to \( x = 3 \). The region can be split into two parts:
1. From \( x = 0 \) to \( x = 2 \), the upper curve is \( y = x^2 + 2 \).
2. From \( x = 2 \) to \( x = 3 \), the upper curve is \( y = 2x + 2 \).

### Step 4: Calculate the area
We calculate the area \( A \) as follows:
\[ A = \int_0^2 (x^2 + 2) \, dx + \int_2^3 (2x + 2) \, dx \]

Calculating the first integral:
\[ \int_0^2 (x^2 + 2) \, dx = \left[ \frac{x^3}{3} + 2x \right]_0^2 = \left( \frac{2^3}{3} + 2 \cdot 2 \right) - \left( 0 + 0 \right) = \frac{8}{3} + 4 = \frac{8}{3} + \frac{12}{3} = \frac{20}{3} \]

Calculating the second integral:
\[ \int_2^3 (2x + 2) \, dx = \left[ x^2 + 2x \right]_2^3 = \left( 3^2 + 2 \cdot 3 \right) - \left( 2^2 + 2 \cdot 2 \right) = (9 + 6) - (4 + 4) = 15 - 8 = 7 \]

### Step 5: Combine the areas
Now, we combine the two areas:
\[ A = \frac{20}{3} + 7 = \frac{20}{3} + \frac{21}{3} = \frac{41}{3} \]

### Step 6: Calculate \( 12A \)
Finally, we calculate \( 12A \):
\[ 12A = 12 \times \frac{41}{3} = 4 \times 41 = 164 \]

Thus, the final answer is:
\[ \boxed{164} \]