<p>The area (in square units) of the region bounded by the curves \(y + 2x^2 = 0\) and \(y + 3x^2 = 1\), is equal to</p>
Step-by-Step Solution
Key Concept: Find intersection points of the two curves by solving them simultaneously, then integrate the difference of upper and lower curves. The curve that lies above changes depending on the x-interval.
<p><strong>Step 1:</strong> Find intersection points by setting the curves equal:</p><p>From y + 2x² = 0, we get y = -2x²</p><p>From y + 3x² = 1, we get y = 1 - 3x²</p><p>Setting them equal: -2x² = 1 - 3x²</p><p>x² = 1 → x = ±1</p><p><strong>Step 2:</strong> Determine which curve is above the other in [-1, 1]:</p><p>At x = 0: y₁ = 0 and y₂ = 1, so y = 1 - 3x² is above y = -2x²</p><p><strong>Step 3:</strong> Set up the integral for area between curves:</p><p>A = ∫₋₁¹ [(1 - 3x²) - (-2x²)] dx</p><p>A = ∫₋₁¹ (1 - x²) dx</p><p><strong>Step 4:</strong> Evaluate the integral:</p><p>A = [x - x³/3]₋₁¹</p><p>A = (1 - 1/3) - (-1 + 1/3)</p><p>A = (2/3) - (-2/3)</p><p>A = 4/3</p><p>∴ Answer: D (4/3 square units)</p>
Correct Answer: D