Area Under the Curve
Area bounded by exponential and linear curves
Grade 12
Question:
<p>The area (in sq. units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is:</p>
<p>\(\log_e 2 + \dfrac{3}{2}\)</p>
<p>\(\dfrac{3}{2}\)</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{3}{2} - \dfrac{1}{\log_e 2}\)</p>
Step-by-Step Solution
Key Concept: The region in the first quadrant is bounded by y = 2x (a line through origin) and y = |x+1| (a V-shaped curve). Since x ≥ 0 in the first quadrant, |x+1| = x+1, so we need to find where 2x intersects x+1 and integrate the difference.
<p><strong>Step 1:</strong> Identify the curves in the first quadrant. For x ≥ 0: y = 2x and y = x + 1 (since |x+1| = x+1)</p><p><strong>Step 2:</strong> Find intersection point: 2x = x + 1 → x = 1, y = 2</p><p><strong>Step 3:</strong> Determine which curve is on top. At x = 0: y = 2x gives 0, y = x+1 gives 1. At x = 1: both equal 2. So x+1 is above 2x for 0 ≤ x ≤ 1</p><p><strong>Step 4:</strong> Calculate area: A = ∫₀¹ [(x+1) - 2x] dx = ∫₀¹ (1-x) dx = [x - x²/2]₀¹ = 1 - 1/2 = <strong>1/2</strong></p><p>∴ Answer: D (assuming D = 1/2 sq. units)</p>
Correct Answer: D