This is a comprehensive set of exercises on definite integrals, area, and volume.
Example ① (Application of Area)
Find the area enclosed by the curve y = x³ − x and its tangent line (the tangent line at x = 1).
Solution:f(x) = x³−x。f'(x) = 3x²−1。f'(1) = 2、f(1) = 0。
Tangent line: y = 2(x−1) = 2x−2。
Intersection points: x³−x = 2x−2 → x³−3x+2 = 0 → (x−1)²(x+2) = 0。x = 1, −2。
S = ∫₋₂¹ |(x³−x)−(2x−2)| dx = ∫₋₂¹ |x³−3x+2| dx = ∫₋₂¹ (x−1)²(x+2) dx
On [−2, 1], (x−1)²(x+2) ≥ 0（(x−1)² ≥ 0、x+2 ≥ 0）
= ∫₋₂¹ (x³−3x+2) dx = [x⁴/4−3x²/2+2x]₋₂¹ = (1/4−3/2+2)−(4−6−4) = 3/4+6 = 27/4
Find the area enclosed by the curve y = x³ − x and its tangent line (the tangent line at x = 1).
Solution:f(x) = x³−x。f'(x) = 3x²−1。f'(1) = 2、f(1) = 0。
Tangent line: y = 2(x−1) = 2x−2。
Intersection points: x³−x = 2x−2 → x³−3x+2 = 0 → (x−1)²(x+2) = 0。x = 1, −2。
S = ∫₋₂¹ |(x³−x)−(2x−2)| dx = ∫₋₂¹ |x³−3x+2| dx = ∫₋₂¹ (x−1)²(x+2) dx
On [−2, 1], (x−1)²(x+2) ≥ 0（(x−1)² ≥ 0、x+2 ≥ 0）
= ∫₋₂¹ (x³−3x+2) dx = [x⁴/4−3x²/2+2x]₋₂¹ = (1/4−3/2+2)−(4−6−4) = 3/4+6 = 27/4
Review
- Definite integral: F(b)−F(a) (no constant of integration is needed)
- Area: ∫|f(x)|dx or ∫(upper−lower)dx
- 1/6 formula: quadratic curve and the x-axis → |a|(β−α)³/6
- Volume of a solid of revolution: π∫{f(x)}²dx
Practice Problems
- Find the area between y = x³ and y = x (for the part where x ≥ 0).
- Find the volume obtained by rotating y = x² − 2x about the x-axis (over the interval [0, 2]).
- Find the total area enclosed by y = x and y = x³.
Solutions and Explanations
- Answer:1/4
Explanation: x³=x→x=0,1 (for x≥0). On [0,1], x≥x³。∫₀¹(x−x³)dx=[x²/2−x⁴/4]₀¹=1/2−1/4=1/4。 - Answer:16π/15
Explanation: V=π∫₀²(x²−2x)²dx=π∫₀²(x⁴−4x³+4x²)dx=π[x⁵/5−x⁴+4x³/3]₀²=π(32/5−16+32/3)=π(96/15−240/15+160/15)=16π/15。 - Answer:1/2
Explanation: x=x³→x(x²−1)=0→x=0,±1。By symmetry, 2×∫₀¹(x−x³)dx=2×1/4=1/2。