Area under a curve above the x-axis
A farmer measures a field with a curved boundary by cutting it into thin vertical strips and adding their areas. The definite integral does this addition exactly.
The area bounded by the curve y = f(x), the x-axis and the lines x = a and x = b is the definite integral of y over x from a to b, provided the curve lies above the x-axis.
Formula: Area = ∫ from a to b of y dx = ∫ from a to b of f(x) dx
In the figure: The shaded area between y = x², the x-axis and the line x = 3 equals ∫ from 0 to 3 of x² dx = 9 square units.
Rule 1: Set up the integral — Find the limits from the lines or the intersection points.
- Easy: Area under y = x² from x = 0 to 3 is [x³/3] from 0 to 3 = 9.
- Medium: Area under y = sin x from 0 to π is [−cos x] from 0 to π = 2.
- Exam-type: Area under y = √x from 0 to 4 is [(2/3) x^(3/2)] from 0 to 4 = 16/3.
Rule 2: Closed curves — Use symmetry to take a quarter and multiply.
- Easy: Area under y = 4 − x² between x = −2 and 2 is 2 × (8 − 8/3) = 32/3.
- Medium: Area under y = 2x + 1 from x = 0 to 3 is [x² + x] from 0 to 3 = 12.
- Exam-type: Area under y = 1/x from x = 1 to e is 1.
Always draw a rough sketch and decide whether the curve is above or below the axis.
Using wrong limits. They are the x-values of the boundary lines, not the y-values.
Sketch, set limits, integrate.
Practice questions
- 1Find the area under y = 2x + 1 from x = 0 to x = 3.
- 2Find the area under y = x³ from x = 0 to x = 2.
- 3Find the area under y = x² from x = 1 to x = 3.
- 4Find the area under y = cos x from x = 0 to x = π/2.
- 5Find the area under y = eˣ from x = 0 to x = 1.
- 6Find the area under y = 1/x from x = 1 to x = e.
- 7Find the area under y = √x from x = 0 to x = 4.
- 8Find the area between y = 4 − x² and the x-axis from x = −2 to x = 2.
Show answers (try them yourself first)
- 1[x² + x] from 0 to 3 = 12
- 2[x⁴/4] from 0 to 2 = 4
- 3[x³/3] from 1 to 3 = 9 − 1/3 = 26/3
- 4sin(π/2) − 0 = 1
- 5e − 1
- 6ln e − ln 1 = 1
- 716/3
- 82 ∫ from 0 to 2 of (4 − x²) dx = 2 (8 − 8/3) = 32/3
Curve below the axis and area along the y-axis
Money owed and money earned are both amounts, but they have opposite signs in a ledger. Area is always positive, so a part of a curve below the axis must be counted with its sign turned.
Where the curve is below the x-axis the integral is negative, but the area is its modulus. If the curve crosses the axis, split the interval at the crossing point and add the positive areas. The area between a curve x = g(y), the y-axis and the lines y = c and y = d is the integral of x with respect to y.
Formula: Area = |∫ from a to c of f(x) dx| + |∫ from c to b of f(x) dx|, and Area along y-axis = ∫ from c to d of x dy
Rule 1: Split at the crossing — Find where y = 0.
- Easy: For y = x² − 4 from x = −2 to 2 the curve is below the axis, so the area is |−32/3| = 32/3.
- Medium: For y = sin x from 0 to 2π the integral is 0, but the area is 2 + 2 = 4.
- Exam-type: For y = 2x − 4 from x = 0 to 4 the line crosses at x = 2. The area is 4 + 4 = 8, although the integral is 0.
Rule 2: Integrate along the y-axis — Use x as a function of y.
- Easy: Area bounded by x = y², the y-axis and y = 0 to y = 3 is ∫ from 0 to 3 of y² dy = 9.
- Medium: Area bounded by x = 4 − y² and the y-axis is ∫ from −2 to 2 of (4 − y²) dy = 32/3.
- Exam-type: Area under y = |x| from x = −2 to 2 is 2 × 2 = 4.
Area is never negative. If the answer comes out negative, take the modulus after checking the sketch.
Using the integral over a crossing without splitting, which cancels the positive and negative parts.
Below the axis flips the sign, crossing needs a split.
Practice questions
- 1Find the area between y = x² − 4 and the x-axis from x = −2 to x = 2.
- 2Find the area between y = sin x and the x-axis from x = 0 to x = 2π.
- 3Find the area between y = 2x − 4 and the x-axis from x = 0 to x = 4.
- 4Find the area between y = x³ and the x-axis from x = −1 to x = 1.
- 5Find the area bounded by x = y², the y-axis and the lines y = 0 and y = 3.
- 6Find the area bounded by x = 4 − y² and the y-axis.
- 7Find the area bounded by x = y², the y-axis and the lines y = 1 and y = 2.
- 8Find the area between y = |x| and the x-axis from x = −2 to x = 2.
Show answers (try them yourself first)
- 1|−32/3| = 32/3
- 22 + 2 = 4
- 34 + 4 = 8
- 42 × (1/4) = 1/2
- 5[y³/3] from 0 to 3 = 9
- 6[4y − y³/3] from −2 to 2 = 32/3
- 7[y³/3] from 1 to 2 = 7/3
- 82 × (½ × 2 × 2) = 4
Circle, ellipse and parabola regions
A round table, an oval track and a satellite dish all have curved edges. Their areas come from the same method: find the area of one quarter and multiply by four.
A circle, an ellipse and a parabola are symmetric, so the area of the whole figure is a multiple of one part. The area of the ellipse reduces to that of the circle when a = b.
Formula: Circle x² + y² = a²: area = πa². Ellipse x²/a² + y²/b² = 1: area = πab. Parabola y² = 4ax cut by x = a: area = (8/3) a²
Rule 1: Use symmetry — Integrate the first quadrant part and multiply.
- Easy: The circle x² + y² = 25 has area 25π.
- Medium: The ellipse x²/9 + y²/4 = 1 has area π × 3 × 2 = 6π.
- Exam-type: The parabola y² = 4x and the line x = 1 enclose an area (8/3) × 1² = 8/3.
Rule 2: Different cut — For y² = 4ax cut by x = h, area = (8/3) √a · h^(3/2).
- Easy: For y² = 16x, a = 4, and the cut x = 4 gives (8/3) × 2 × 8 = 128/3.
- Medium: First quadrant area under y = 3√x from 0 to 4 is 16.
- Exam-type: First quadrant of the ellipse x²/25 + y²/4 = 1 has area π × 5 × 2 / 4 = 5π/2.
In πab the numbers a and b are the semi-axes, the square roots of the denominators.
Using the denominators 9 and 4 directly as a and b in the formula πab.
Circle is pi r squared, ellipse is pi a b.
Practice questions
- 1Find the area of the circle x² + y² = 16.
- 2Find the area of the ellipse x²/16 + y²/9 = 1.
- 3Find the area of the ellipse 4x² + 9y² = 36.
- 4Find the area of the part of the circle x² + y² = 9 in the first quadrant.
- 5Find the area of the part of the ellipse x²/25 + y²/4 = 1 in the first quadrant.
- 6Find the area enclosed by y² = 4x and the line x = 1.
- 7Find the area enclosed by y² = 16x and the line x = 4.
- 8Find the area under y = 3√x from x = 0 to x = 4.
Show answers (try them yourself first)
- 116π
- 2π × 4 × 3 = 12π
- 3x²/9 + y²/4 = 1 gives π × 3 × 2 = 6π
- 49π/4
- 5π × 5 × 2 / 4 = 5π/2
- 68/3
- 7128/3
- 816