Area from eight offsets at 20 m interval

Given: 8 offsets at a common interval $d = 20$ m, so the length of the chain line = 7 × 20 = 140 m.

(i) Trapezoidal rule

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

Sum of the intermediate offsets = 26.27 m; (first + last)/2 = (0.00 + 6.55)/2 = 3.275 m.

$$\text{Area} = 20\times\left[3.275 + 26.27\right] = 590.900\ \text{m}^2$$

(ii) Simpson's one-third rule

There are 8 offsets, i.e. 7 (an odd number of) intervals, so Simpson's rule cannot be used for the whole length. It is applied to the first 6 intervals (offsets 0 to 6), and the last strip is found by the trapezoidal rule.

$$\text{Area}_1 = \frac{20}{3}\left[(0.00 + 4.30) + 4(12.28) + 2(9.69)\right] = 485.333\ \text{m}^2$$ $$\text{Area}_2 = 20\times\frac{4.30 + 6.55}{2} = 108.500\ \text{m}^2$$ $$\text{Total area} = 485.333 + 108.500 = 593.833\ \text{m}^2$$

Answer: trapezoidal rule = 590.90 m²; Simpson's rule = 593.83 m².