<p>Given region \(A = \{(x, y): 0 \leq y \leq x|x| + 1 \text{ and } -1 \leq x \leq 1\}\). The area of region \(A\) is:</p>
Step-by-Step Solution
Key Concept: The function y = x|x| + 1 is even in its absolute value component, so split the integral at x = 0 and use the symmetry: for x ≥ 0, x|x| = x², and for x < 0, x|x| = -x². The total area is the sum of integrals over [-1,0] and [0,1].
<p><strong>Step 1:</strong> Rewrite x|x| piecewise:</p><p>For x ∈ [0,1]: x|x| = x·x = x²</p><p>For x ∈ [-1,0]: x|x| = x·(-x) = -x²</p><p><strong>Step 2:</strong> Set up the area integral:</p><p>Area = ∫₋₁⁰ (-x² + 1)dx + ∫₀¹ (x² + 1)dx</p><p><strong>Step 3:</strong> Evaluate the first integral:</p><p>∫₋₁⁰ (-x² + 1)dx = [-x³/3 + x]₋₁⁰ = 0 - (-(-1/3) - 1) = 0 - (1/3 - 1) = 2/3</p><p><strong>Step 4:</strong> Evaluate the second integral:</p><p>∫₀¹ (x² + 1)dx = [x³/3 + x]₀¹ = (1/3 + 1) - 0 = 4/3</p><p><strong>Step 5:</strong> Add the areas:</p><p>Total Area = 2/3 + 4/3 = 6/3 = 2</p><p>∴ Answer: B</p>
Correct Answer: B