A curved boundary can be difficult to measure directly, but a table of heights gives you a practical starting point. Join neighbouring heights with straight lines and the area becomes a row of trapezia. You can calculate each small area, add them and then ask how trustworthy the approximation is.
The key is to count the gaps as well as the heights. Five ordinates give four strips. This lesson uses non-negative heights and explicitly stated spacing. Exact areas, where needed, are supplied as comparison values; you do not need integration to complete the examples.
Why the first and last heights count once
An ordinate is a vertical height from the baseline to the boundary. Between two adjacent ordinates y₀ and y₁, separated horizontally by h, the trapezium area is h(y₀ + y₁) ÷ 2. Here the two parallel sides are the vertical ordinates. The horizontal strip width is their perpendicular separation.
For several equally spaced strips, every interior ordinate belongs to two neighbouring trapezia. The first and last ordinates each belong to just one. That produces the compact rule: estimated area = (h ÷ 2)[first height + last height + 2 × sum of interior heights]. If there are n strips between x = a and x = b, the width is h = (b − a) ÷ n and there are n + 1 ordinates.
The formula is an organised sum of trapezia, not a reason to skip checking the table. First confirm that all x intervals are equal and that the listed heights describe the intended boundary. When both axes measure lengths in metres, the resulting area is in square metres. More generally, the answer carries the product of the horizontal and vertical units.
Worked example 1: four strips under a curved graph
Estimate the area between y = x² + 1, the x-axis and x = 0 to x = 4, using four equal strips. Both coordinates use the same arbitrary length unit. Use the table below.
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| y = x² + 1 | 1 | 2 | 5 | 10 | 17 |
The total width is 4 and there are four strips, so h = 1. The endpoint heights are 1 and 17. The interior heights are 2, 5 and 10, whose sum is 17. Estimated area = ½[1 + 17 + 2(2 + 5 + 10)] = ½ × 52 = 26 square units.
Check by calculating each trapezium separately. The four areas are (1 + 2)/2 = 1.5, (2 + 5)/2 = 3.5, (5 + 10)/2 = 7.5 and (10 + 17)/2 = 13.5. Their sum is 26. This second route catches a lost endpoint or an interior height accidentally counted only once.
The exact area is supplied as 76/3 square units, approximately 25.3333. The estimate is higher by 26 − 76/3 = 2/3 square unit. Percentage error relative to the exact value is [(2/3) ÷ (76/3)] × 100 = 2.63%, to two decimal places. Using 26 as the denominator would measure a different comparison.
Worked example 2: compare a coarser estimate
Use the same curve and interval, but only two strips. The ordinates are at x = 0, 2 and 4, with heights 1, 5 and 17. Now h = (4 − 0) ÷ 2 = 2.
Estimated area = (2 ÷ 2)[1 + 17 + 2(5)] = 28 square units. Direct addition gives 2(1 + 5)/2 + 2(5 + 17)/2 = 6 + 22 = 28. The absolute error is 28 − 76/3 = 8/3 square units, and the percentage error is [(8/3) ÷ (76/3)] × 100 = 10.53%.
For this smooth upward-curving graph, the shorter straight joins follow the curve more closely, and the four-strip estimate is better. Both estimates are too high because each joining chord lies above the curve between its endpoints. That explanation concerns the curve’s shape, not just whether its values increase.
Do not turn this comparison into a promise that doubling strips always provides a fixed accuracy for any dataset. A boundary may change rapidly between measured points, and extra measurements may carry error. The table alone cannot reveal every bend between its entries. A claim such as “accurate to 1%” needs evidence beyond the number of strips.
Worked example 3: catch unequal spacing
A second dataset gives horizontal positions 0, 1 and 3 m and boundary heights 2, 4 and 8 m. Someone uses a common width of 1 m for the whole table. Explain why this fails and calculate the trapezium estimate correctly.
The intervals are 1 m and 2 m, so there is no single common width. Use each actual width: first area = 1(2 + 4)/2 = 3 m²; second area = 2(4 + 8)/2 = 12 m². Total = 15 m². A common-width calculation with h = 1 would give 9 m² and would silently shorten the second strip.
For this example the three points lie on the straight line y = 2x + 2. If the boundary is stated to be that straight line between the points, the trapezia describe it exactly and 15 m² is the exact area. With only isolated measurements and no boundary-shape information, call 15 m² an estimate. The same arithmetic can carry a different certainty depending on the evidence.
Read a problem before choosing the bracket
If the instruction says “use six ordinates,” draw six short marks and count the five gaps. If it says “use six strips,” draw seven boundary ordinates. Label the x-values under those marks; this makes an incorrect strip width visible before you start the weighted sum.
On a printed graph, read the axis scales rather than assuming one square equals one unit. Do not treat the visual height of a trapezium on the page as its numerical ordinate. If the last height is zero, include it as an endpoint anyway. Zero does not remove the last strip. These simple checks often matter more than calculator speed.
Three practice tasks with solved answers
Practice 1: add the internal heights
At x = 0, 2, 4 and 6 m, boundary heights are 3, 5, 4 and 2 m. Estimate the area with three trapezia.
Answer: h = 2 m. Area = (2/2)[3 + 2 + 2(5 + 4)] = 23 m². Individual areas 8, 9 and 6 m² give the same total. Four heights correctly produce three strips.
Practice 2: a fractional strip width
For x = 1.0, 1.5, 2.0 and 2.5, heights are 1.0, 1.5, 2.5 and 4.0. Both axes use the same length unit. Estimate the area.
Answer: h = 0.5. Area = (0.5/2)[1 + 4 + 2(1.5 + 2.5)] = 3.25 square units. Individual areas 0.625, 1.0 and 1.625 sum to 3.25.
Practice 3: assess the error
Four equal strips cover a horizontal length of 8 cm. Heights in centimetres are 2, 3, 5, 4 and 2. Estimate the area and compare with a supplied exact area of 27 cm².
Answer: h = 8/4 = 2 cm. Area = (2/2)[2 + 2 + 2(3 + 5 + 4)] = 28 cm². The estimate is high by 1 cm², giving percentage error = (1/27) × 100 = 3.70%. Dividing 8 by five ordinates would incorrectly set the strip width.
Frequently asked questions
Are strips and ordinates the same thing?
No. Ordinates are boundary heights; strips are the intervals between successive ordinates. n strips require n + 1 ordinates.
Why double the internal heights?
Each interior height is a side of two adjacent trapezia. Endpoints belong to only one trapezium, so they appear once.
Can I use the compact formula with unequal spacing?
Not with one common h. Calculate each trapezium using its actual width and add the separate areas.
Is the trapezium estimate always too large?
No. It depends on the boundary shape. The upward-curving example here gives an overestimate, but other curves can give an underestimate or mixed errors.
Which denominator belongs in percentage error?
Use the supplied exact or accepted reference value: absolute difference divided by that value, multiplied by 100. State whether the estimate is high or low separately.
What if the final ordinate is zero?
Keep it as the endpoint. Its contribution to the sum is zero, but the neighbouring strip still has a width and usually a nonzero area.
Can I report an exact area from tabulated points?
Only if additional information makes the straight-line model exact. Otherwise the trapezium sum approximates the unknown boundary between the points.
Background references
The original examples and diagrams above were independently calculated. These public educational references support the underlying methods; they do not establish a current Kenyan grade allocation.
Continue your revision
For the wider collection, visit the Mathematics topical revision collection. Resource access may require login or membership; check the listed contents and access conditions. This lesson is self-contained and does not promise a particular downloadable file.
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