Area Under the Curve
Area bounded by curve and line
Grade 12
Question:
<p>The area bounded by the curve \(y = 2x - x^2\) and the straight line \(y = -x\) is given by</p>
<p>\(\dfrac{9}{2}\) sq. unit</p>
<p>\(\dfrac{43}{6}\) sq. unit</p>
<p>\(\dfrac{35}{6}\) sq. unit</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Find intersection points of the curve and line, then integrate the difference of upper and lower functions over the bounded region.
<p><strong>Step 1:</strong> Find intersection points by setting <strong>2x - x² = -x</strong></p><p>3x - x² = 0 → x(3 - x) = 0 → x = 0 or x = 3</p><p><strong>Step 2:</strong> Determine which function is above. At x = 1: curve gives y = 2(1) - 1² = 1; line gives y = -1. So the curve is above.</p><p><strong>Step 3:</strong> Set up the integral:</p><p>A = ∫₀³ [(2x - x²) - (-x)] dx = ∫₀³ (3x - x²) dx</p><p><strong>Step 4:</strong> Evaluate:</p><p>A = [3x²/2 - x³/3]₀³ = (27/2 - 9) - 0 = 27/2 - 18/2 = 9/2</p><p><strong>∴ Answer: A (9/2 square units)</strong></p>
Correct Answer: A