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: \( 12A = 164 \)