Question

The area of the region : R = {(x, y) : 5x² ≤ y ≤ 2x² + 9} is :

• A. 6√3
• B. 9√3
• C. 11√3
• D. 12√3

Hint:

Always exploit symmetry! Since both $5x^2$ and $2x^2+9$ are even functions, the area from $-\sqrt{3}$ to $\sqrt{3}$ is simply twice the area from $0$ to $\sqrt{3}$.

Answer 1

The Correct Option is D

Step 1: Find intersection points: $5x^2 = 2x^2 + 9 \Rightarrow 3x^2 = 9 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}$.
Step 2: Area $= \int_{-\sqrt{3}}^{\sqrt{3}} [(2x^2 + 9) - 5x^2] dx = \int_{-\sqrt{3}}^{\sqrt{3}} [9 - 3x^2] dx$.
Step 3: $2 \int_0^{\sqrt{3}} (9 - 3x^2) dx = 2 [9x - x^3]_0^{\sqrt{3}}$.
Step 4: $2 [9\sqrt{3} - 3\sqrt{3}] = 2 [6\sqrt{3}] = 12\sqrt{3}$.