Area from offsets, 10 m interval

Given: 4 offsets at a common interval $d = 10$ m, so the length of the chain line = 3 × 10 = 30 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 = 16.80 m; (first + last)/2 = (4.60 + 6.40)/2 = 5.500 m.

$$\text{Area} = 10\times\left[5.500 + 16.80\right] = 223.000\ \text{m}^2$$

(ii) Simpson's one-third rule

There are 4 offsets, i.e. 3 (an odd number of) intervals, so Simpson's rule cannot be used for the whole length. It is applied to the first 2 intervals (offsets 0 to 2), and the last strip is found by the trapezoidal rule.

$$\text{Area}_1 = \frac{10}{3}\left[(4.60 + 9.60) + 4(7.20)\right] = 143.333\ \text{m}^2$$ $$\text{Area}_2 = 10\times\frac{9.60 + 6.40}{2} = 80.000\ \text{m}^2$$ $$\text{Total area} = 143.333 + 80.000 = 223.333\ \text{m}^2$$

Answer: trapezoidal rule = 223.00 m²; Simpson's rule = 223.33 m².

Alternative, Simpson's three-eighths rule (valid for exactly 3 intervals): Area $= \frac{3d}{8}[o_0 + 3o_1 + 3o_2 + o_3] = \frac{3 \times 10}{8}[4.6 + 3(7.2) + 3(9.6) + 6.4] = 230.250\ \text{m}^2$.