Question

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

Answer 1

Correct Answer: 164

To find the area \( A \), we need to evaluate the integral over the region bounded by \( y = x^2 + 2 \) and \( y = 2x + 2 \) from \( x = 0 \) to \( x = 3 \).

Step 1. Set up the area as the sum of two integrals based on the intersection point of \( y = x^2 + 2 \) and \( y = 2x + 2 \), which is \( x = 2 \):**  
  \(A = \int_{0}^{2} (x^2 + 2) \, dx + \int_{2}^{3} (2x + 2) \, dx\)

Step 2. Evaluate each integral:
  - For \( \int_{0}^{2} (x^2 + 2) \, dx \):  
    \(= \left[ \frac{x^3}{3} + 2x \right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}\)
   - For \( \int_{2}^{3} (2x + 2) \, dx \):  
    \(= \left[ x^2 + 2x \right]_2^3 = (9 + 6) - (4 + 4) = 7\)
    
Step 3. Combine the results:
  \(A = \frac{20}{3} + 7 = \frac{41}{3}\)
  
Step 4. Calculate \( 12A \):
  12A = \(12 \times \frac{41}{3}\) = 164
  
The Correct Answer is: \( 12A = 164 \)

Answer 2

Step 1: Set up the integrals to find the area.

The area between the curves is given by the integral of the difference of the functions. We can divide the area into two parts: \[ A = \int_0^2 (x^2 + 2) \, dx + \int_0^{3/2} (2x + 2) \, dx \]

Step 2: Compute 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 3: Compute the second integral.

\[ \int_0^{3/2} (2x + 2) \, dx = \left[ x^2 + 2x \right]_0^{3/2} = \left( \left(\frac{3}{2}\right)^2 + 2\left(\frac{3}{2}\right) \right) - (0 + 0) \] \[ = \left( \frac{9}{4} + 3 \right) = \frac{9}{4} + \frac{12}{4} = \frac{21}{4} \]

Step 4: Combine the results to find the total area.

Now, combine both integrals: \[ A = \frac{20}{3} + \frac{21}{4} \] To add these fractions, find the common denominator: \[ A = \frac{80}{12} + \frac{63}{12} = \frac{143}{12} \]

Step 5: Multiply by the constant factor.

Finally, multiply by the factor of 12 to get the final result: \[ 12A = 12 \times \frac{41}{3} = 41 \times 4 = 164 \]

Final Answer:

The area between the curves is: \[ \boxed{164} \]