Areas Under Graphs and Accumulated Change
Higher
A rate graph tells you how fast a quantity accumulates. The area between the graph and the time axis combines the rate with the duration to find the total change.
Before you start: Real-Life and Travel Graphs, Rates of Change, Perimeter and Area and Rounding and Estimation.
1. Why area gives an accumulated quantity
At a constant speed of 5 m/s for 4 s, distance = 5 × 4 = 20 m. On a speed–time graph this is a rectangle of height 5 and width 4. When the rate changes, split the interval into pieces and add their contributions.
Multiply the units on the two axes: (m/s) × s = m; (L/min) × min = L; (£/h) × h = £. The result is the quantity added during the interval, not necessarily the final amount already present. Add any initial amount separately.
Make time units compatible first. For a constant flow of 3 L/min lasting 30 s, use 30 s = 0.5 min: volume = 3 × 0.5 = 1.5 L. Multiplying by 30 without converting would be wrong.
Not every area is distance. Beneath an ordinary distance–time graph, the units are metres × seconds (m·s), which are not metres. Its area does not give distance travelled. Always read both axes before choosing an interpretation.
2. Exact areas under straight sections
Speed rises uniformly from 0 to 6 m/s over 4 s and stays at 6 m/s until 10 s. Split where the rule changes, at 4 s.
Triangle: ½ × 4 × 6 = 12 m.
Rectangle: (10 − 4) × 6 = 6 × 6 = 36 m.
Total distance: 12 + 36 = 48 m.
These areas are exact for the stated straight-line model. The speed at 10 s is 6 m/s; that is a graph height, not the distance. The second interval is 6 s wide, not 10 s.
Choose an efficient partition
When speed increases uniformly from 2 to 8 m/s in 6 s, the region is one trapezium. Its parallel sides are the two vertical speeds; their perpendicular separation is the time interval.
Trapezium: ½ × (2 + 8) × 6 = 30 m. Alternatively, rectangle 2 × 6 = 12 m plus triangle ½ × 6 × (8 − 2) = 18 m gives 30 m. Use the fewest simple pieces that match the graph; do not force the earlier bent graph into a single trapezium.
For a trapezium, area = ½ × (first height + second height) × interval width. The mean of the endpoint rates works here because the graph between them is straight. It is only an approximation for a curve.
3. Velocity: displacement versus total distance
Velocity is +4 m/s for 0–2 s, then decreases uniformly to −4 m/s at 6 s. It crosses zero at 4 s. The object moves forwards first and backwards after 4 s.
Above the axis: rectangle 2 × 4 = 8 m, plus triangle ½ × 2 × 4 = 4 m, giving 12 m forwards.
Below the axis: triangle magnitude ½ × 2 × 4 = 4 m backwards.
Displacement: signed total = 12 − 4 = +8 m.
Total distance: add magnitudes = 12 + 4 = 16 m.
Displacement is the net position change, so +8 m means finishing 8 m in the chosen positive direction from the initial position. Distance counts all movement. Always split a region where velocity crosses zero before calculating total distance; opposite directions must not cancel.
4. Estimate a curved area with trapezia
| Time (min) | Flow rate (L/min) |
|---|---|
| 0 | 0 |
| 2 | 4 |
| 4 | 16 |
Estimate the volume entering a tank over 0–4 min. Three readings form two intervals, each 2 min wide. Replace each curved section with a straight top joining its endpoint readings.
First strip: ½ × (0 + 4) × 2 = 4 L.
Second strip: ½ × (4 + 16) × 2 = 20 L.
Estimated volume: 4 + 20 = 24 L.
The units are (L/min) × min = L. The graph height at 4 min, 16 L/min, is a rate and is not the volume. If the tank started with 10 L and none escaped, its final volume would be approximately 10 + 24 = 34 L.
The calculation is exact for the straight trapezia, but their tops do not follow the curve, so 24 L is an estimate of the curved area. On this supplied upward-bending curve, both straight tops lie above it: the shaded trapezia include extra area, so the estimate is too high. A table alone would not prove how the graph bends between readings.
Other sensible estimates
You can count complete grid squares and combine partial squares, remembering that each square represents horizontal interval × vertical rate. On this grid, one 1-minute by 4-L/min square represents 4 L. Smaller strips or squares can improve the fit, but the answer remains an estimate.
For an increasing non-negative rate, rectangles using left-end heights lie below the curve, while right-end rectangles lie above it. For trapezia, decide from how the curve bends relative to the straight tops, not just whether it increases. If the drawing does not justify an over/under conclusion, say that it cannot be determined from the information given.
Keep the arithmetic manageable: our curved estimates use at most four equal intervals, matching the Pearson guidance used for this course. Four intervals need five rate readings.
5. A monetary rate over time
A worker earns £12 per hour for the first 2 hours and £18 per hour for the next 3 hours. The graph shows pay rate, not total pay.
First rectangle = 12 × 2 = £24. Second rectangle = 18 × (5 − 2) = £54. Total gross pay = 24 + 54 = £78. Units: (£/h) × h = £.
Use the step at 2 h to split the area into two rectangles. A single trapezium would incorrectly assume that the rate rose smoothly from £12/h to £18/h. Financial arithmetic belongs in Number; the new idea here is interpreting area as accumulated money.
Think Like a Mathematician
Can two journeys have the same total distance but different final displacements?
Yes. Travelling 8 m forwards and 8 m backwards gives distance 16 m and displacement 0 m. Travelling 12 m forwards and 4 m backwards also gives distance 16 m, but displacement +8 m. The unsigned areas agree; the signed areas differ.
Common Mistakes
Using height instead of area: a rate at one instant is not an accumulated total.
Forgetting interval width: multiply each mean rate by its own duration, using compatible units.
Calling a curved estimate exact: the straight tops replace curved pieces, even when the endpoint readings are exact.
Subtracting backwards motion for total distance: subtract below-axis contributions for displacement; add their magnitudes for distance.
Assuming every area is distance: multiply the axis units. Distance–time area has distance × time units; flow-rate–time area has volume units.
Practice — areas and accumulated change
Use exact stated data unless asked to read or draw. Give the accumulated quantity with its units and say when the result is an estimate.
1. A speed graph is horizontal at 7 m/s for 5 s. Find the distance and explain why 7 is not the answer.
Rectangle = 7 × 5 = 35 m. The 7 m/s is a speed, not a distance; interval width matters.
2. Speed rises uniformly from 0 to 8 m/s in 3 s, stays at 8 m/s until 7 s, then falls uniformly to zero at 9 s. Find total distance.
Triangle ½ × 3 × 8 = 12 m; rectangle (7 − 3) × 8 = 32 m; triangle ½ × (9 − 7) × 8 = 8 m. Total = 52 m. These are exact areas for the stated straight-line model.
3. Speed increases uniformly from 3 to 9 m/s between t = 2 s and t = 6 s. Find distance in that interval using one trapezium.
Width = 6 − 2 = 4 s. Area = ½ × (3 + 9) × 4 = 24 m. The width is not 6 s.
4. An object moves at +2 m/s for 3 s, then at −1 m/s for 4 s. Find displacement and total distance.
Positive contribution = 2 × 3 = 6 m. Negative contribution = −1 × 4 = −4 m. Displacement = 6 − 4 = +2 m. Total distance = 6 + 4 = 10 m.
5. Use the signed graph in section 3. Why is taking the absolute value of its final displacement not enough to find total distance?
The net displacement is +8 m, so its absolute value is 8 m. But the object travels 12 m forwards and 4 m backwards: 16 m in total. Take the magnitude of each directional contribution before adding, not after cancellation.
6. A curved flow-rate graph has readings 2, 4, 10 and 20 L/min at times 0, 1, 2 and 3 min. Use three equal trapezia to estimate the volume. Can the table alone prove an overestimate?
Width is 1 min. Areas are ½(2 + 4) × 1 = 3 L, ½(4 + 10) × 1 = 7 L and ½(10 + 20) × 1 = 15 L. Total ≈ 25 L. Four readings form three intervals. It is an estimate because the graph is curved; the table alone does not show the bends between readings.
7. Use the supplied curved flow graph in section 4. Read the rates at 1 and 3 min and estimate its area with four width-1-min trapezia. Compare with 24 L.
Rates are approximately 1 and 9 L/min (allow readings within about 0.3 L/min). Using 0, 1, 4, 9, 16 gives ½(0+1) + ½(1+4) + ½(4+9) + ½(9+16) = 0.5 + 2.5 + 6.5 + 12.5 = 22 L. Allow roughly 21–23 L from sensible readings. The narrower trapezia fit this curve better; they still lie above it, so the estimate remains high.
8. A worker receives £15/h for 90 min, then £20/h for 2 h. Find total gross pay from the two rate–time rectangles.
90 min = 1.5 h. Pay = 15 × 1.5 + 20 × 2 = 22.50 + 40 = £62.50. (£/h) × h gives pounds.
9. A graph has vertical units m and horizontal units s. A student says its area gives distance. Explain the error. What units would a L/min against min graph give?
m × s = m·s, not metres: ordinary distance–time area is not distance. (L/min) × min = L, an accumulated volume.
10. An increasing positive rate is estimated with left-end rectangles. Is the estimate high or low? How many readings are needed for four trapezia?
Left-end rectangles lie below an increasing graph, so the estimate is low. Four trapezia need five endpoint readings. This rectangle conclusion follows from increase; the trapezium conclusion needs information about bending.
Frequently Asked Questions
Do I need calculus?
No. Use rectangles, triangles, trapezia and sensible graphical estimates. General curve shapes are covered in Quadratic Graphs and Cubic, Reciprocal and Exponential Graphs.
When is the answer exact?
When the stated graph is piecewise straight and its coordinates are exact, geometric areas give an exact result for that model. Replacing a curved boundary with straight strips gives an estimate. Reading coordinates from a drawing also introduces reading uncertainty.
Does an area tell me the final amount?
It gives the accumulated change over the chosen interval. For an inflow graph, add the starting volume to find the final volume, allowing for any outflow if one is specified.
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