Question

The area of the region $S = \{(x, y) : 3x^2 \leq 4y \leq 6x + 24\}$ is _________.

Hint:

The area between a parabola $y = ax^2 + bx + c$ and a chord intersecting at $x_1$ and $x_2$ can also be computed as $\frac{|a|}{6}(x_2 - x_1)^3$. For this problem: $\frac{3/4}{6}(4 - (-2))^3 = \frac{1}{8}(6)^3 = 27$.

Answer 1

Correct Answer: 27

Step 1: Understanding the Concept:
The region is bounded by the parabola $y \geq \frac{3x^2}{4}$ and the straight line $y \leq \frac{6x + 24}{4} = \frac{3x}{2} + 6$. We find the intersection points and integrate the difference of the functions.
Step 2: Detailed Explanation:
Find intersection points:
\[ \frac{3x^2}{4} = \frac{6x + 24}{4} \implies 3x^2 - 6x - 24 = 0 \implies x^2 - 2x - 8 = 0 \]
\[ (x - 4)(x + 2) = 0 \implies x = -2, 4 \]
Area $= \int_{-2}^4 \left[ \left( \frac{3x}{2} + 6 \right) - \frac{3x^2}{4} \right] dx$:
\[ \text{Area} = \left[ \frac{3x^2}{4} + 6x - \frac{x^3}{4} \right]_{-2}^4 \]
Upper limit $(x = 4)$: $\frac{3(16)}{4} + 6(4) - \frac{64}{4} = 12 + 24 - 16 = 20$.
Lower limit $(x = -2)$: $\frac{3(4)}{4} + 6(-2) - \frac{-8}{4} = 3 - 12 + 2 = -7$.
Total Area $= 20 - (-7) = 27$.
Step 3: Final Answer:
The area of the region is 27.