The area enclosed by a curve and the x-axis is found using the absolute value of the definite integral (pay attention to the sign).
Area Formulas
When f(x) ≥ 0 on the interval [a, b]: S = ∫ₐᵇ f(x)dx
If the sign of f(x) changes, divide the interval at the points where the sign changes and add the absolute values of each part.
S = ∫ₐᵇ |f(x)| dx
Example ①
Find the area enclosed by y = x² − 1 and the x-axis.
Answer:x²−1 = 0 → x = ±1。On the interval [−1, 1], f(x) ≤ 0。
S = −∫₋₁¹ (x²−1)dx = −[x³/3−x]₋₁¹ = −((1/3−1)−(−1/3+1)) = −(−4/3) = 4/3
Find the area enclosed by y = x² − 1 and the x-axis.
Answer:x²−1 = 0 → x = ±1。On the interval [−1, 1], f(x) ≤ 0。
S = −∫₋₁¹ (x²−1)dx = −[x³/3−x]₋₁¹ = −((1/3−1)−(−1/3+1)) = −(−4/3) = 4/3
Example ②
Find the area enclosed by y = x³ − x and the x-axis.
Answer:x³−x = x(x+1)(x−1) = 0 → x = −1, 0, 1。
f ≥ 0 on [−1, 0], and f ≤ 0 on [0, 1]。
S = ∫₋₁⁰ (x³−x)dx + |∫₀¹ (x³−x)dx|
= [x⁴/4−x²/2]₋₁⁰ + |[x⁴/4−x²/2]₀¹| = (0−1/4) + |1/4−1/2| = ...
S = |−1/4| + |−1/4| = 1/2
Find the area enclosed by y = x³ − x and the x-axis.
Answer:x³−x = x(x+1)(x−1) = 0 → x = −1, 0, 1。
f ≥ 0 on [−1, 0], and f ≤ 0 on [0, 1]。
S = ∫₋₁⁰ (x³−x)dx + |∫₀¹ (x³−x)dx|
= [x⁴/4−x²/2]₋₁⁰ + |[x⁴/4−x²/2]₀¹| = (0−1/4) + |1/4−1/2| = ...
S = |−1/4| + |−1/4| = 1/2
Key Points
- Area = ∫|f(x)|dx: to remove the absolute value, divide at points where the sign changes
- On parts where f(x) ≤ 0, use −∫f(x)dx to make the area positive
- Check the sign intervals by sketching the graph
Practice Problems
- Find the area enclosed by y = x² − 4 and the x-axis.
- Find the area enclosed by y = x(x−2) and the x-axis.
- Find the total area enclosed by y = x³ − 3x and the x-axis.
Answers and Explanations
- Answer:32/3
Explanation: Intersections x=±2。[−2,2]でf≤0。S=−∫₋₂²(x²−4)dx=−[x³/3−4x]₋₂²=−((8/3−8)−(−8/3+8))=−(−32/3)=32/3。 - Answer:4/3
Explanation: x(x−2)=0→x=0,2。[0,2]でf≤0。S=−∫₀²(x²−2x)dx=−[x³/3−x²]₀²=−(8/3−4)=4/3。 - Answer:9/2 (using symmetry)
Explanation: x³−3x=x(x²−3)=0→x=0,±√3。Since it is an odd function, the areas on [−√3,0] and [0,√3] are symmetric. S=2∫₀^√3|x³−3x|dx=2∫₀^√3(3x−x³)dx=2[3x²/2−x⁴/4]₀^√3=2(9/2−9/4)=2×9/4=9/2。