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

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: $12A = 164$