Area from twelve offsets at 20 m interval

There are 12 offsets, so there are 11 equal strips of width $d = 20$ m, and the first offset is $O_1 = 0$ and the last is $O_{12} = 4.19$ m.

(i) Trapezoidal rule

$$A = d\left[\frac{O_1 + O_n}{2} + O_2 + O_3 + \dots + O_{n-1}\right]$$

Sum of the intermediate offsets ($O_2$ to $O_{11}$):

$$\sum O_{mid} = 50.62\ \text{m}$$ $$A = 20\left[\frac{0.00 + 4.19}{2} + 50.62\right] = 20 \times 52.715 = 1054.30\ \text{m}^2$$

(ii) Simpson's rule

Simpson's one-third rule needs an odd number of offsets (an even number of strips). Here there are 12 offsets, so apply Simpson's rule to the first 11 offsets (10 strips) and the trapezoidal rule to the last strip.

$$A = \frac{d}{3}\left[O_1 + O_{11} + 4(O_2 + O_4 + O_6 + O_8 + O_{10}) + 2(O_3 + O_5 + O_7 + O_9)\right] + d\,\frac{O_{11} + O_{12}}{2}$$

• Sum of even-numbered offsets: $O_2 + O_4 + O_6 + O_8 + O_{10} = 26.13$ m
• Sum of odd-numbered intermediate offsets: $O_3 + O_5 + O_7 + O_9 = 18.24$ m
• First and last offset of the Simpson part: $O_1 + O_{11} = 6.25$ m

$$\begin{aligned} A_{1-11} &= \frac{20}{3}\left[6.25 + 4(26.13) + 2(18.24)\right] = \frac{20}{3} \times 147.25 = 981.67\ \text{m}^2 \\ A_{last} &= 20 \times \frac{6.25 + 4.19}{2} = 104.40\ \text{m}^2 \\ A &= 981.67 + 104.40 = 1086.07\ \text{m}^2 \end{aligned}$$

Answer: Trapezoidal area = 1054.30 m²; Simpson's area = 1086.07 m².