Area under lines
Area under y = mx + c is ∫ (mx + c) dx between the limits. If the line crosses the x-axis, split there and add the parts.
Why lines are the easy case
A straight line y = mx + c gives a simple integral: ∫ (mx + c) dx = mx²/2 + cx. The region under a line is a triangle or a trapezium, so you can check every answer with a geometry formula. The board still wants the integral in your answer; the geometry check is only for you.
Example 1: a trapezium
Find the area of the region bounded by the line y = 2x + 1, the x-axis and the lines x = 1 and x = 3.
- On [1, 3], y = 2x + 1 is positive, so the region is above the x-axis.
- A = ∫ (2x + 1) dx from 1 to 3
- = [x² + x] from 1 to 3
- = (9 + 3) − (1 + 1) = 12 − 2 = 10
- Check: a trapezium with parallel sides y(1) = 3 and y(3) = 7 and width 2 has area (3 + 7)/2 × 2 = 10.
Area = 10 sq units.
Example 2: a line that crosses the x-axis
Find the area of the region bounded by the line y = 2x − 4, the x-axis and the lines x = 0 and x = 4.
- The line meets the x-axis where 2x − 4 = 0, at x = 2. It is below the axis on [0, 2] and above it on [2, 4].
- A₁ = ∫ (2x − 4) dx from 0 to 2 = [x² − 4x] from 0 to 2 = 4 − 8 = −4, so this part has area 4.
- A₂ = ∫ (2x − 4) dx from 2 to 4 = [x² − 4x] from 2 to 4 = 0 − (−4) = 4
- Area = |A₁| + A₂ = 4 + 4 = 8
- Warning: ∫ (2x − 4) dx from 0 to 4 = 0, which is not the area.
Area = 8 sq units.
Example 3: a triangle with one side on the x-axis
Using integration, find the area of the triangle with vertices A(−1, 0), B(1, 4) and C(3, 0).
- Side AB: slope = (4 − 0)/(1 + 1) = 2, so y = 2(x + 1) = 2x + 2.
- Side BC: slope = (0 − 4)/(3 − 1) = −2, so y = −2(x − 3) = 6 − 2x.
- AC lies on the x-axis. So the area is the area under AB (x = −1 to 1) plus the area under BC (x = 1 to 3): Area = ∫ (2x + 2) dx from −1 to 1 + ∫ (6 − 2x) dx from 1 to 3
- First part = [x² + 2x] from −1 to 1 = 3 − (−1) = 4
- Second part = [6x − x²] from 1 to 3 = 9 − 5 = 4
- Area = 4 + 4 = 8. Check: (1/2) × base 4 × height 4 = 8.
Area of the triangle = 8 sq units.
Area under a line, step by step
- 1Sketch the line, the x-axis and the boundary lines x = a and x = b. Shade the region.
- 2Find where the line meets the x-axis: put y = 0, so x = −c/m.
- 3★If that point lies between a and b, split the interval there.
- 4Integrate y = mx + c over each part separately.
- 5Add the absolute values of the parts. Check with the triangle or trapezium formula.
★ key step
How to draw the figure
- Draw both axes with a clear scale, such as 1 cm for 1 unit.
- Plot the vertices, or the end points of the line segment, and label them with their coordinates.
- Join them with a ruler and write the equation of each line beside it.
- Draw boundary lines such as x = a and x = b as dotted lines.
- Shade the region and show one vertical strip in it.
A farmer measuring a triangular corner plot beside a road, or an engineer finding the side area of a sloping ramp at a railway station, uses these same areas under lines.
2 marks: “Using integration, find the area bounded by the line y = 3x, the x-axis and the line x = 2.” A = ∫ 3x dx from 0 to 2 = [3x²/2] from 0 to 2 = 6 sq units. Check: triangle (1/2) × 2 × 6 = 6.
3 marks: “Find the area of the region bounded by the line y = x − 3, the x-axis and the lines x = 1 and x = 5.” The line crosses the x-axis at x = 3. |∫ (x − 3) dx from 1 to 3| = |−2| = 2 and ∫ (x − 3) dx from 3 to 5 = 2. Area = 2 + 2 = 4 sq units.
3 marks: “Sketch the graph of y = |x + 1| and evaluate ∫ |x + 1| dx from −3 to 2. What area does this integral give?” |x + 1| = −(x + 1) for x < −1 and x + 1 for x ≥ −1. ∫ −(x + 1) dx from −3 to −1 = 2 and ∫ (x + 1) dx from −1 to 2 = 9/2. Total 13/2: the area, in sq units, bounded by y = |x + 1|, the x-axis and the lines x = −3 and x = 2.
Line crosses? Then split. Never integrate straight across a point where the line cuts the x-axis.