1. Chapter at a Glance
Earlier you found areas of triangles, rectangles and circles using formulas. But what is the area inside a parabola, or inside an ellipse? Definite integrals answer this: the area under a curve is the sum of infinitely many thin strips. This chapter uses integration to find areas bounded by simple curves — straight lines, circles, parabolas and ellipses in standard form — and the coordinate axes or vertical/horizontal lines. These are classic 5-mark board questions.
- Area under y = f(x) between x = a and x = b using vertical strips.
- Area using horizontal strips: ∫ x dy.
- Handling curves below the x-axis (take absolute value).
- Areas of a circle, an ellipse and regions of a parabola.
- Using symmetry to reduce work.
2. Area by Strips
Divide the region into thin vertical strips of width dx and height y = f(x). Each strip has area ≈ y dx; adding all strips gives
- Vertical strips: A = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx (region between the curve, the x-axis and x = a, x = b).
- Horizontal strips: A = ∫꜀ᵈ x dy = ∫꜀ᵈ g(y) dy (region between the curve, the y-axis and y = c, y = d).
- If the curve lies below the axis, the integral is negative: take |∫ f(x) dx|.
- If the curve crosses the axis, split at the crossing points and add absolute values.
3. Area of a Circle
Find the area enclosed by the circle x² + y² = a².
Solution: By symmetry, area = 4 × (area in the first quadrant) = 4∫₀ᵃ √(a² − x²) dx = 4[(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a)]₀ᵃ = 4 × (a²/2)(π/2) = πa².
4. Area of an Ellipse
Find the area enclosed by x²/a² + y²/b² = 1.
Solution: In the first quadrant y = (b/a)√(a² − x²). Area = 4(b/a)∫₀ᵃ √(a² − x²) dx = 4(b/a)(πa²/4) = πab.
Check: when a = b the ellipse is a circle and πab = πa² ✓.
5. Regions of a Parabola
Find the area of the region bounded by y² = 4x and the line x = 3.
Solution: Symmetric about the x-axis: area = 2∫₀³ 2√x dx = 4[(2/3)x^(3/2)]₀³ = (8/3)(3√3) = 8√3 sq units.
Find the area bounded by x² = 4y, the y-axis and y = 2, y = 4 (first quadrant).
Solution: x = 2√y. Area = ∫₂⁴ 2√y dy = (4/3)[y^(3/2)]₂⁴ = (4/3)(8 − 2√2) = (32 − 8√2)/3 sq units.
6. Lines and Curves Below the Axis
Find the area bounded by y = x³, the x-axis and x = −2, x = 1.
Solution: x³ < 0 on (−2, 0). Area = |∫₋₂⁰ x³ dx| + ∫₀¹ x³ dx = |−4| + ¼ = 17/4. (Directly ∫₋₂¹ x³ dx = −15/4 would be wrong as an area.)
Find the area under the line y = 3x + 2 between x = −1 and x = 1 (above and below the axis).
Solution: The line meets the axis at x = −2/3. Area = |∫₋₁^(−2/3) (3x + 2) dx| + ∫₋₂/₃¹ (3x + 2) dx = |−1/6| + 25/6 = 13/3.
7. Solved Problems (Board Pattern)
Find the area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Solution: ∫₂⁴ 3√x dx = 2[x^(3/2)]₂⁴ = 2(8 − 2√2) = 16 − 4√2.
Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² = 32.
Solution: They meet at (4, 4). Area = ∫₀⁴ x dx + ∫₄^(4√2) √(32 − x²) dx = 8 + [ (x/2)√(32 − x²) + 16 sin⁻¹(x/(4√2)) ]₄^(4√2) = 8 + (16·π/2 − (8 + 16·π/4)) = 8 + 8π − 8 − 4π = 4π.
Find the area of the ellipse x²/16 + y²/9 = 1.
Solution: πab = π × 4 × 3 = 12π.
8. Master Formula Table
| Region | Area |
|---|---|
| Under y = f(x), x = a to b | ∫ₐᵇ |f(x)| dx |
| Left of x = g(y), y = c to d | ∫꜀ᵈ |g(y)| dy |
| Circle x² + y² = a² | πa² |
| Ellipse x²/a² + y²/b² = 1 | πab |
| y² = 4ax and x = c | 2∫₀ᶜ 2√(ax) dx = (8/3)√a c^(3/2) |
| Key integral | ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2)sin⁻¹(x/a) |
9. Master Summary Table
| Step | Action |
|---|---|
| 1 | Sketch the curve(s) and shade the region |
| 2 | Find intersection points / limits |
| 3 | Choose vertical (dx) or horizontal (dy) strips |
| 4 | Use symmetry (multiply by 2 or 4) |
| 5 | Integrate; take absolute values for parts below the axis |
Chart — areas for a = 4, b = 3:
10. Mind Map
- y dx
- x dy
- Circle πa²
- Ellipse πab
- Symmetry
- x^(3/2) terms
- Below axis: |·|
- Sketch first
11. Tips and Tricks
- Always sketch — it shows symmetry and whether parts lie below the axis.
- For parabolas opening sideways (y² = 4ax), vertical strips with y = 2√(ax) and a factor 2 for symmetry work well.
- Remember πab for ellipses to check your integration answer.
- Adding signed integrals across the axis (areas cancel wrongly).
- Forgetting the symmetry factor.
- Wrong limits — find intersection points first.
12. Important One-liners
- Area is always positive.
- Area of a circle by integration = πa².
- Area of an ellipse = πab.
- Vertical strips: ∫y dx; horizontal strips: ∫x dy.
- Use symmetry about axes to reduce calculation.
13. Exam-pattern Questions with Detailed Answers
The area of the circle x² + y² = 4 is:
- 2π
- 4π
- 8π
- 16π
Show answer
πr² with r = 2: 4π.
Answer: (B)
Find the area under y = x² from x = 0 to x = 3.
Show answer
[x³/3]₀³ = 9.
Answer: 9 sq units
Find the area of the ellipse x²/25 + y²/4 = 1.
Show answer
π × 5 × 2 = 10π.
Answer: 10π sq units
Find the area bounded by y = sin x and the x-axis from x = 0 to x = 2π.
Show answer
∫₀^π sin x dx + |∫_π^(2π) sin x dx| = 2 + 2 = 4.
Answer: 4 sq units
Find the area bounded by y² = x, x = 1, x = 4 and the x-axis in the first quadrant.
Show answer
∫₁⁴ √x dx = (2/3)(8 − 1) = 14/3.
Answer: 14/3
Find the area bounded by x = y², the y-axis and the lines y = 1 and y = 2.
Show answer
∫₁² y² dy = (8 − 1)/3 = 7/3.
Answer: 7/3
Find the area of the region bounded by the line y = 3x + 2, the x-axis and x = 1, x = 3.
Show answer
∫₁³ (3x + 2)dx = [3x²/2 + 2x]₁³ = (27/2 + 6) − (3/2 + 2) = 16.
Answer: 16 sq units
Find the area of the quarter circle x² + y² = 16 in the first quadrant using integration.
Show answer
∫₀⁴ √(16 − x²) dx = [(x/2)√(16 − x²) + 8 sin⁻¹(x/4)]₀⁴ = 8(π/2) = 4π.
Answer: 4π
Find the area of the region bounded by y² = 4ax and its latus rectum.
Show answer
Latus rectum x = a. Area = 2∫₀ᵃ 2√(ax) dx = 4√a (2/3) a^(3/2) = 8a²/3.
Answer: 8a²/3
Find the area of the smaller region cut off from the ellipse x²/9 + y²/4 = 1 by the line x/3 + y/2 = 1.
Show answer
In the first quadrant, area under ellipse = (1/4)π(3)(2) = 3π/2; under the line = ½ × 3 × 2 = 3. Smaller region = 3π/2 − 3 = (3/2)(π − 2).
Answer: (3/2)(π − 2) sq units
Find the area bounded by y = |x + 3|, the x-axis and x = −6, x = 0.
Show answer
Two triangles each with base 3 and height 3: 2 × 4.5 = 9. (By integration: ∫₋₆⁻³ −(x + 3)dx + ∫₋₃⁰ (x + 3)dx = 9/2 + 9/2.)
Answer: 9 sq units
Find the area of the region {(x, y) : y² ≤ 4x, 0 ≤ x ≤ 1}, and compare it with the area of the enclosing rectangle of width 1 and height 4.
Show answer
Area = 2∫₀¹ 2√x dx = 8/3. Rectangle area = 1 × 4 = 4. The parabolic region is 2/3 of the enclosing rectangle — a result known to Archimedes.
Answer: 8/3 (two-thirds of 4)
14. Quick Revision Summary
- Area = ∫|y| dx (vertical strips) or ∫|x| dy (horizontal strips).
- Split at axis crossings; areas are positive.
- Circle πa², ellipse πab (by integration with symmetry).
- Parabola regions: use y = 2√(ax) and symmetry factor 2.
- Always sketch and find limits first.