Applications of Integrals
'If differentiation describes the shape, integration measures the space it occupies.'
1. Chapter Overview
This chapter applies DEFINITE INTEGRALS to compute the AREA of regions bounded by curves. The key idea: the area between two curves f(x) and g(x) from x = a to x = b is the integral of the absolute difference of the functions. Special cases include the area of a circle and ellipse using integration. These techniques form the foundation for calculating volumes, arc lengths, and surface areas in higher mathematics.
2. Area Under a Curve
- The area bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b is: A = ∫ₐᵇ |f(x)| dx
- If f(x) ≥ 0 on [a, b], A = ∫ₐᵇ f(x) dx.
- If f(x) changes sign, split the integral at the points where f(x) = 0.
'Area above the x-axis is positive. Area below is negative — but we take the ABSOLUTE value for total area.'
3. Area Between Two Curves
- Area between y = f(x) and y = g(x) where f(x) ≥ g(x) on [a, b]: A = ∫ₐᵇ [f(x) − g(x)] dx
- In general: A = ∫ₐᵇ |f(x) − g(x)| dx
Worked Example 1
Problem: Find the area of the region bounded by y = x² and y = x. Solution: Step 1: Find intersection points: x² = x ⇒ x² − x = 0 ⇒ x(x − 1) = 0 ⇒ x = 0, x = 1. Step 2: On [0, 1], x ≥ x² (check at x = 0.5: 0.5 ≥ 0.25). Step 3: A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = (1/2 − 1/3) − 0 = 1/6 square units.
4. Area of a Circle Using Integration
- Circle: x² + y² = r². By symmetry, area of quadrant = ∫₀ʳ √(r² − x²) dx.
- Total area = 4 × ∫₀ʳ √(r² − x²) dx = 4 × [πr²/4] = πr².
'Integration confirms what we already know — the area of a circle is πr² — but shows HOW the formula is derived from first principles.'
Worked Example 2
Problem: Using integration, find the area of the circle x² + y² = 16. Solution: r = 4. Area = π(4)² = 16π square units. Derivation: A = 4∫₀⁴ √(16 − x²) dx = 4 × (π·16/4) = 16π.
5. Area of an Ellipse Using Integration
- Ellipse: x²/a² + y²/b² = 1. y = ±(b/a)√(a² − x²).
- Area = 4 × ∫₀ᵃ (b/a)√(a² − x²) dx = (4b/a) × (πa²/4) = πab.
'The area of an ellipse is πab — a beautiful generalisation of the circle (where a = b = r gives πr²).'
6. Area in Terms of y
- Sometimes it is easier to integrate with respect to y.
- Area between x = f(y) and x = g(y) from y = c to y = d: A = ∫𝄤ᵈ |f(y) − g(y)| dy
Worked Example 3
Problem: Find the area bounded by the parabola y² = 4ax and its latus rectum x = a. Solution: y² = 4ax ⇒ y = ±2√(ax). Latus rectum: x = a. By symmetry, consider upper half and double. A = 2∫₀ᵃ 2√(ax) dx = 4√a ∫₀ᵃ √x dx = 4√a × [2/3 x³/²]₀ᵃ = (8√a/3)(a³/²) = 8a²/3 square units.
7. Comparison Table: Area Formulas
| Region | Formula |
|---|---|
| Between y = f(x) and x-axis, x = a to x = b | ∫ₐᵇ |
| Between y = f(x) and y = g(x) | ∫ₐᵇ |
| Between x = f(y) and y-axis, y = c to y = d | ∫𝄤ᵈ |
| Between x = f(y) and x = g(y) | ∫𝄤ᵈ |
| Circle: x² + y² = r² | πr² |
| Ellipse: x²/a² + y²/b² = 1 | πab |
8. Common Mistakes
- Forgetting absolute value: When a curve goes below the x-axis, the integral gives NEGATIVE area. Always use |f(x)|.
- Wrong intersection points: ALWAYS verify that the intersection points you found satisfy BOTH equations.
- Not checking which curve is on top: Between two curves, integrate (upper − lower), not the other way around.
- Limits of integration: The limits are x-values (for dx integrals) or y-values (for dy integrals). Don't mix them up.
9. CBSE Exam Focus
- Area under a curve — simple curves like y = x², y = √x
- Area between two curves — finding intersection points, determining upper/lower
- Area of circle/ellipse using integration
- Symmetry — using symmetry to simplify calculations
- Word problems — regions bounded by curves, lines, and axes
10. Self-Test
Q1: Find the area bounded by y = x² + 2, the x-axis, x = 0, and x = 3. A1: A = ∫₀³ (x² + 2) dx = [x³/3 + 2x]₀³ = (9 + 6) − 0 = 15 square units.
Q2: Find the area of the region bounded by y² = x and y = x. A2: Intersection: x = x² ⇒ x = 0, x = 1. On [0, 1], √x ≥ x (check x = 0.25: 0.5 > 0.25). A = ∫₀¹ (√x − x) dx = [2x³/²/3 − x²/2]₀¹ = (2/3 − 1/2) = 1/6 square units.
Q3: Using integration, find the area of the ellipse x²/9 + y²/4 = 1. A3: a = 3, b = 2. Area = πab = π(3)(2) = 6π square units.
Q4: Find the area bounded by the curve y = sin x, the x-axis, and x = 0 to x = 2π. A4: sin x ≥ 0 on [0, π] and sin x ≤ 0 on [π, 2π]. A = ∫₀^π sin x dx + |∫_π^{2π} sin x dx| = [−cos x]₀^π + |[−cos x]_π^{2π}| = (1 + 1) + |(−1) − 1| = 2 + 2 = 4 square units.
Q5: Find the area between the curves y = 4 − x² and y = x² − 4. A5: Intersection: 4 − x² = x² − 4 ⇒ 2x² = 8 ⇒ x = ±2. On [−2, 2], (4 − x²) ≥ (x² − 4). A = ∫₋₂² [(4 − x²) − (x² − 4)] dx = ∫₋₂² (8 − 2x²) dx = [8x − 2x³/3]₋₂² = (16 − 16/3) − (−16 + 16/3) = (32/3) − (−32/3) = 64/3 square units.
11. Conclusion
Applications of integrals show that INTEGRATION MEASURES:
- AREA: 'The space enclosed by curves — the definite integral adds up infinitely many thin rectangles.'
- SYMMETRY: 'Use it to simplify calculations — compute one quadrant and multiply.'
- CONFIRMATION: 'Integration confirms known formulas (πr², πab) and derives new ones.'
'Integration is the tool that lets us measure any shape — from simple rectangles to the most complicated curves.'
