Area under a curve
A farmer measuring a field with a curved boundary cuts it into very thin strips and adds their areas; the integral does exactly this.
The area bounded by y = f(x), the x-axis and the lines x = a, x = b (with f ≥ 0) is A = ∫ₐᵇ f(x) dx. If f < 0 take the modulus. For regions bounded by the y-axis use A = ∫ x dy.
Rule 1: Above the axis — A = ∫ₐᵇ y dx.
Rule 2: Below the axis — A = |∫ₐᵇ y dx|; when the curve crosses the axis split at the crossing points.
In the figure : The shaded area under y = x² from 0 to 3 is ∫₀³ x² dx = 9.
How is the formula obtained? : A = ∫₀³ x² dx = [x³/3]₀³ = 9 − 0 = 9 square units.
- Easy : Area under y = x² from 0 to 3 is 9.
- Medium : Area under y = sin x from 0 to π is [−cos x]₀^π = 2.
- Exam-type : Area between y = x² − 4 and the x-axis: the curve is below the axis from −2 to 2, area = ∫₋₂² (4 − x²) dx = 32/3.
Area is never negative.
Giving a negative area for a region below the axis.
Sketch first; split at x-intercepts. Memory clue : "Sketch, split, sum."
Do not integrate across a sign change in one go; areas cancel.
CET-style example : Area under y = x² from x = 0 to 3 is (A) 9 (B) 27 (C) 3 (D) 18
Answer : (A). Time-saving method : x³/3 at 3.
Spot the difference
| Region | Area | ||
|---|---|---|---|
| Curve above x-axis | ∫ y dx | ||
| Curve below x-axis | ∫ y dx |
Practice questions
- 1Area under y = x from 0 to 4.
- 2Area under y = x² from 0 to 3.
- 3Area under y = sin x from 0 to π.
- 4Area under y = eˣ from 0 to 1.
- 5Area under y = 1/x from 1 to e.
- 6Area between y = x² − 4 and the x-axis.
- 7Area of the circle x² + y² = a² by integration.
- 8Area of the ellipse x²/a² + y²/b² = 1.
Show answers (try them yourself first)
- 18.
- 29.
- 32.
- 4e − 1 ≈ 1.718.
- 51.
- 632/3 ≈ 10.67.
- 74∫₀ᵃ √(a² − x²) dx = πa².
- 8πab.
Area between two curves
The path between two rivers has two banks; its area is the area under the upper bank minus the area under the lower bank.
If f(x) ≥ g(x) on [a, b], the area between the curves is ∫ₐᵇ [f(x) − g(x)] dx. Find a and b where the curves meet.
Rule 1: Intersection — Solve f(x) = g(x) for the limits.
Rule 2: Subtract — Upper curve minus lower curve.
How is the formula obtained? : y = x and y = x²: x = x² ⇒ x = 0, 1. Area = ∫₀¹ (x − x²) dx = ½ − ⅓ = 1/6.
- Easy : Between y = x and y = x² is 1/6.
- Medium : Between y = 2x and y = x²: meet at 0 and 2; ∫₀² (2x − x²) dx = 4 − 8/3 = 4/3.
- Exam-type : Area enclosed by y = x² and y = 4: ∫₋₂² (4 − x²) dx = 32/3.
Check which curve is upper on the interval.
Subtracting the upper from the lower and getting a negative area.
Test one point inside the interval to see which curve is on top. Memory clue : "Top minus bottom."
Curves may cross more than twice; split the interval at each crossing.
CET-style example : Area between y = x and y = x² is (A) 1/6 (B) 1/3 (C) 1/2 (D) 1/12
Answer : (A). Time-saving method : ∫₀¹ (x − x²) dx.
Spot the difference
| Case | Formula |
|---|---|
| f above g | ∫(f − g) dx |
| g above f | ∫(g − f) dx |
Practice questions
- 1Points where y = x and y = x² meet.
- 2Area between y = x and y = x².
- 3Points where y = 2x and y = x² meet.
- 4Area between y = 2x and y = x².
- 5Area between y = x² and y = 4.
- 6Area between y = x² and y = x + 2.
- 7Area between y² = 4x and y = x.
- 8Area between y = x and y = x³ from 0 to 1.
Show answers (try them yourself first)
- 1x = 0 and 1.
- 21/6.
- 3x = 0 and 2.
- 44/3.
- 532/3.
- 6x² = x + 2 ⇒ x = −1, 2; ∫(x + 2 − x²) = 9/2.
- 7x = 0, 4; ∫₀⁴ (2√x − x) dx = 32/3 − 8 = 8/3.
- 8½ − ¼ = 1/4.