The area under a graph can represent a real physical quantity — distance, displacement, or total change — depending on what the axes show. At GCSE you estimate it using the trapezium rule: split the region into vertical strips, treat each strip as a trapezium, and sum the areas of all the strips. More strips always gives a closer estimate.
What does the area under a graph represent?
The physical meaning depends on the axes:
- Velocity–time graph: area under the line = distance (or displacement) travelled in that time interval.
- Acceleration–time graph: area under the curve = change in velocity.
- Flow rate graph (litres per second against seconds): area = total volume that has flowed.
When the graph involves real quantities, always include units in your area answer. If the y-axis is in m/s and the x-axis is in seconds, the area is in m.
What is the trapezium rule?
The trapezium rule divides the area under a curve into vertical strips of equal width h. Each strip is approximately a trapezium — the two parallel sides are the y-values at the left and right edges, and the height of each trapezium is h.
Area of one trapezium = ½ × h × (y_left + y_right)
Summing all strips gives the trapezium rule formula for n strips:
Area ≈ ½h[(y₀ + yₙ) + 2(y₁ + y₂ + … + yₙ₋₁)]
where h is the strip width, y₀ is the y-value at the left boundary, yₙ is the y-value at the right boundary, and y₁ … yₙ₋₁ are the y-values at the internal boundaries between strips.
How do you apply the trapezium rule step by step?
Worked example: Estimate the area under y = x² between x = 0 and x = 4 using strips of width 1.
- List the x-values: 0, 1, 2, 3, 4 (five values, giving four strips).
- Calculate the y-values:
| x | y = x² |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
| 4 | 16 |
- Identify the boundary values: y₀ = 0, y₄ = 16 (outer); y₁ = 1, y₂ = 4, y₃ = 9 (inner).
- Apply the formula with h = 1:
Area ≈ ½ × 1 × [(0 + 16) + 2(1 + 4 + 9)] = ½ × [16 + 2 × 14] = ½ × [16 + 28] = ½ × 44 = 22
The exact area (found using calculus: [x³/3] from 0 to 4) is 64/3 ≈ 21.33. Our estimate of 22 is close, and slightly high.
How do you tell whether your estimate is an overestimate or underestimate?
The shape of the curve determines whether the trapezium rule overestimates or underestimates:
| Curve shape in the strip | Trapezium vs actual strip | Estimate type |
|---|---|---|
| Concave (bending upward, like y = x²) | Top of trapezium lies above the curve | Overestimate |
| Convex (bending downward, like y = √x) | Top of trapezium lies below the curve | Underestimate |
For y = x² the curve bends upward (concave) throughout, so all four trapezia are above the actual curve — the estimate of 22 overestimates the true area of 21.33. ✓
How do you increase the accuracy of the estimate?
Use more strips (smaller h). As the strips get narrower, each trapezium more closely matches the actual curved strip beneath it, and the total area estimate converges to the true value.
- 4 strips (h = 1): estimate ≈ 22
- 8 strips (h = 0.5): estimate ≈ 21.5
- 16 strips (h = 0.25): estimate closer still to 21.33
GCSE questions always specify the number of strips or the strip width; you will not need to choose.
What are the most common errors?
- Miscounting the number of y-values. n strips require n + 1 y-values (including both endpoints). Four strips need y₀ through y₄ — that is five values, not four.
- Doubling the outer values. The formula doubles the inner y-values only. y₀ and yₙ are added once each, not multiplied by 2.
- Using x-values instead of y-values. You read the y-values from the graph for each x-value. If the question gives a table of values, read from the column labelled y (or f(x)).
- Omitting the ½. The area of a trapezium is ½(a + b)h. Forgetting to halve at the end doubles the answer.
Frequently asked questions
Does the trapezium rule give an exact answer?
No — it gives an approximation. The only exception is when the curve between two adjacent x-values is actually a straight line (i.e. the graph is already piecewise linear). In that case the trapezium fits the region exactly and the rule gives the true area.
What if the graph dips below the x-axis?
If part of the graph is below the x-axis, the y-values in that region are negative. The trapezium formula still works — negative y-values produce negative areas, which reduce the total. This is consistent with signed area: area below the x-axis is subtracted from area above. Some GCSE questions ask for the "total distance" on a velocity–time graph, in which case you take the absolute value of any negative section before adding.
Can the strips have different widths?
The standard trapezium rule uses equal-width strips, which is what GCSE questions always present. Unequal strips are possible (you calculate each trapezium's area individually) but you will not be asked to use them at GCSE.
How does this link to integration at A-level?
At GCSE you estimate the area numerically using the trapezium rule. At A-level, integration finds the exact area under a curve algebraically. The trapezium rule is a numerical approximation to the definite integral — more strips make it converge to the same answer that integration gives exactly.
For rate-of-change and graph work at GCSE with Professor Pi, visit aitutors.me.