Area Under the Curve
Area bounded by curves and lines
Grade 12
Question:
<p>The area of the region \(A = \{(x, y) : 0 \leq y \leq x|x| + 1 \text{ and } -1 \leq x \leq 1\}\) in sq. units, is:</p>
<p>\(\dfrac{2}{3}\)</p>
<p>2</p>
<p>\(\dfrac{4}{3}\)</p>
<p>\(\dfrac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Split the integral at x=0 to handle the absolute value in x|x|, where x|x| = x² for x≥0 and x|x| = -x² for x<0. The region is bounded above by the curve and below by y=0.
<p><strong>Step 1:</strong> Analyze the function y = x|x| + 1</p><p>For x ≥ 0: y = x·x + 1 = x² + 1</p><p>For x < 0: y = x·(-x) + 1 = -x² + 1</p><p><strong>Step 2:</strong> The region A is bounded by 0 ≤ y ≤ x|x| + 1 for -1 ≤ x ≤ 1, so we integrate the upper bound:</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 - (-2/3) = 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> Total area = 2/3 + 4/3 = 6/3 = 2 sq. units</p><p>∴ Answer: B (2 sq. units)</p>
Correct Answer: B