Nine offsets at 5 m spacing: how many intervals and why does Simpson's rule apply to the whole length?

8 intervals, even, so Simpson applies to the whole length

9 intervals, odd, so Simpson cannot be used

8 intervals, but Simpson needs an odd count

9 intervals, and Simpson works for any count

The first and last offsets are both 0. What does the trapezoidal formula reduce to?

Area = d x (sum of intermediate offsets)

Area = d x (sum of all offsets)/2

Area = d/3 x (sum of intermediate offsets)

Area = 2d x (sum of intermediate offsets)

Odd-numbered offsets (1st, 3rd, 5th, 7th intermediate)

Only the first and last offsets

Trapezoidal 206.50 m^2, Simpson 207.67 m^2

Trapezoidal 207.67 m^2, Simpson 206.50 m^2

Trapezoidal 413.0 m^2, Simpson 207.67 m^2

Short answer

With a common interval of 5 m and 8 strips, Simpson’s rule gives 207.67 m², the trapezoidal rule 206.50 m².

The following offsets were taken from a chain line to a hedge.

Distance (m)	0	5	10	15	20	25	30	35	40
Offset (m)	0	2.5	5	7.5	8.8	7.5	6.5	3.5	0

Calculate the area enclosed between the chain line and hedge by (i) Simpson's rule (ii) the trapezoidal rule.

Given: 9 offsets at a common interval $d = 5$ m, so the length of the chain line = 8 × 5 = 40 m.

The number of intervals (8) is even, so the rule applies to the whole length:

\text{Area} = \frac{d}{3}\left[(o_0 + o_n) + 4(\text{odd offsets}) + 2(\text{even offsets})\right]

Sum of the odd-numbered offsets = 21.00 m; sum of the even-numbered (intermediate) offsets = 20.30 m.

\text{Area} = \frac{5}{3}\left[(0.00 + 0.00) + 4(21.00) + 2(20.30)\right] = 207.667\ \text{m}^2

\text{Area} = d\left[\frac{o_0 + o_n}{2} + o_1 + o_2 + \dots + o_{n-1}\right]

Sum of the intermediate offsets = 41.30 m; (first + last)/2 = (0.00 + 0.00)/2 = 0.000 m.

\text{Area} = 5\times\left[0.000 + 41.30\right] = 206.500\ \text{m}^2

Answer: Simpson's rule = 207.67 m²; trapezoidal rule = 206.50 m².

✓ Answer checked by Claude Opus 5.5 (high effort)