The area between two curves y = f(x) and y = g(x) is found using ∫|f(x)−g(x)|dx.
Area Between Two Curves
When f(x) ≥ g(x): S = ∫ₐᵇ (f(x) − g(x))dx
Example ①
Find the area enclosed by y = x² and y = x.
Answer:Intersections: x² = x → x = 0, 1。On [0, 1], x ≥ x²。
S = ∫₀¹ (x−x²)dx = [x²/2−x³/3]₀¹ = 1/2−1/3 = 1/6
Find the area enclosed by y = x² and y = x.
Answer:Intersections: x² = x → x = 0, 1。On [0, 1], x ≥ x²。
S = ∫₀¹ (x−x²)dx = [x²/2−x³/3]₀¹ = 1/2−1/3 = 1/6
Example ②
Find the area enclosed by y = x² − 2x and y = x.
Answer:Intersections: x²−2x = x → x²−3x = 0 → x = 0, 3。On [0, 3], x ≥ x²−2x。
S = ∫₀³ (x−(x²−2x))dx = ∫₀³ (3x−x²)dx = [3x²/2−x³/3]₀³ = 27/2−9 = 9/2
Find the area enclosed by y = x² − 2x and y = x.
Answer:Intersections: x²−2x = x → x²−3x = 0 → x = 0, 3。On [0, 3], x ≥ x²−2x。
S = ∫₀³ (x−(x²−2x))dx = ∫₀³ (3x−x²)dx = [3x²/2−x³/3]₀³ = 27/2−9 = 9/2
Example ③ (1/6 Formula)
When y = ax² + bx + c intersects the x-axis at two points (α, 0) and (β, 0) (α < β):
S = |a|/6 × (β−α)³
Example: y = x²−x−2 (α=−1, β=2). S = 1/6 × 3³ = 9/2
When y = ax² + bx + c intersects the x-axis at two points (α, 0) and (β, 0) (α < β):
S = |a|/6 × (β−α)³
Example: y = x²−x−2 (α=−1, β=2). S = 1/6 × 3³ = 9/2
Key Points
- Find the intersections first to determine the interval of integration
- Check which curve is higher (upper − lower)
- For a quadratic curve and the x-axis, the 1/6 formula is fast
Practice Problems
- Find the area between y = x² and y = 2x − x².
- Find the area between y = x² − 3x and y = −x² + x.
- Find the area enclosed by y = x² − 4 and the x-axis using the 1/6 formula.
Answers and Explanations
- Answer:1/3
Explanation: x²=2x−x²→2x²−2x=0→x=0,1。∫₀¹(2x−2x²)dx=[x²−2x³/3]₀¹=1−2/3=1/3。 - Answer:8/3
Explanation: x²−3x=−x²+x→2x²−4x=0→x=0,2。∫₀²(−2x²+4x)dx=[−2x³/3+2x²]₀²=−16/3+8=8/3。 - Answer:32/3
Explanation: x²−4=0→x=±2。Using the 1/6 formula: 1×(2−(−2))³/6=64/6=32/3。