Area under a simple curve
Cut the region between y = f(x), the x-axis, x = a and x = b into thin vertical strips. One strip has height y and width dx, so its area is about y dx. Adding all strips gives
Area = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx (when f ≥ 0).
If the curve is given as x = g(y) and the region is between y = c and y = d with the y-axis, use horizontal strips: Area = ∫ from c to d of x dy.
When the curve goes below the x-axis
There ∫f dx is negative. Area = |∫ of the part below| + ∫ of the part above. Always find where the curve crosses the x-axis first.
Area under a straight line
For y = mx + c between x = a and x = b, the region is a trapezium or triangle, so you can check the integral with geometry. Example: ∫₀³ (x/2 + 1) dx = 21/4 = ½(1 + 2.5)(3).
Area inside a parabola
y² = 4ax opens to the right and is symmetric about the x-axis. The top half is y = 2√(ax). Area bounded by the parabola and x = k:
A = 2∫₀ᵏ 2√(ax) dx = (8/3)√a · k^(3/2).
For x² = 4ay (opens up), use y = x²/(4a) with vertical strips, or x = 2√(ay) with horizontal strips.
Area of a circle
x² + y² = a². Top half: y = √(a² − x²). First quadrant area:
∫₀ᵃ √(a² − x²) dx = [ (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) ]₀ᵃ = πa²/4.
Whole circle: 4 × πa²/4 = πa².
Area of an ellipse
x²/a² + y²/b² = 1. Top half: y = (b/a)√(a² − x²). First quadrant area = (b/a) × πa²/4 = πab/4.
Whole ellipse: πab. When a = b = r it becomes a circle, πr².
Key formulas and definitions
- Area = ∫ₐᵇ y dx (vertical strips), Area = ∫ from c to d of x dy (horizontal strips)
- ∫₀ᵃ √(a² − x²) dx = πa²/4
- Circle x² + y² = a²: area = πa²
- Ellipse x²/a² + y²/b² = 1: area = πab
- Parabola y² = 4ax cut by x = k: area = (8/3)√a · k^(3/2)
Worked examples
1. Find the area under y = 2x + 1 from x = 1 to x = 3.
∫₁³ (2x + 1) dx = [x² + x]₁³ = 12 − 2 = 10 square units.
2. Find the area bounded by y = x², the x-axis, x = 0 and x = 3.
∫₀³ x² dx = [x³/3]₀³ = 9 square units.
3. Find the area bounded by y² = 4x and the line x = 3.
Top half y = 2√x. A = 2∫₀³ 2√x dx = 4 × (2/3) × 3^(3/2) = 8√3 ≈ 13.86 square units.
4. Find the area bounded by x² = 4y, the y-axis and y = 1, y = 4 (first quadrant).
Horizontal strips: x = 2√y. A = ∫₁⁴ 2√y dy = (4/3)[y^(3/2)]₁⁴ = (4/3)(8 − 1) = 28/3 square units.
5. Find the area of the circle x² + y² = 16.
First quadrant: ∫₀⁴ √(16 − x²) dx = π(16)/4 = 4π. Whole: 16π square units.
6. Find the area of the ellipse x²/9 + y²/4 = 1.
a = 3, b = 2. Quarter = (2/3)∫₀³ √(9 − x²) dx = (2/3)(9π/4) = 3π/2. Whole = 4 × 3π/2 = 6π = πab.
7. Find the area between y = x − 2, the x-axis, x = 0 and x = 4.
Crosses at x = 2. ∫₀² (x − 2) dx = −2, ∫₂⁴ (x − 2) dx = 2. Area = 2 + 2 = 4 square units (not 0).
8. Find the area of the circle x² + y² = 4 that lies to the right of x = 1 (small cap).
A = 2∫₁² √(4 − x²) dx = 2[(x/2)√(4 − x²) + 2 sin⁻¹(x/2)]₁² = 2[π − (√3/2 + π/3)] = 4π/3 − √3 ≈ 2.46 square units.
Common mistakes
- Adding signed integrals when the curve crosses the x-axis, so + and − parts cancel. Split at the crossing and use absolute values.
- Forgetting to double (or ×4) after using symmetry.
- Using the wrong half of a curve: for y² = 4x the top half is y = 2√x, not y = 4x.
- Mixing up strips: with ∫ … dy the limits must be y-values, not x-values.