Find the area between the chain lines the 1st and last ordinate and boundary by

1. Trapezoidal rule

2. Simpson’s rule

3. Co-ordinate rule (average ordinate rule)

$$ $$ Area by the Trapezoidal rule:

Area of the interval $ {d_1}$ of 5m $ {A_1} = {d_1}[{\frac{y_1 + y_2}{2} + y_2 + y_3 + y_4 + y_5}] $

$ = 5[{\frac{2.5 + 3.5}{2} + 3.8 + 8.4 + 7.5 + 10.5}] $

= 180.5 $ m^2 $

Area of the interval $ {d_2} $ of 10m

$ {A_2} = {d_2}[{\frac{y_7 + y_11}{2} + y_8 + y_9 + y_10}] $

$ = 10[{\frac{9.3 + 8.4}{2} + 5.8 + 7.8 + 6.9}] $

= 293.5 $ m^2 $

Total area A = $ A_1 + A_2 = 180.5+293.5 = 475 m^2 $

Area by Simpson's rule:

$ A_1 = {\frac{d_1}{3}}[{(y_1 + y_5) + 4(y_2 + y_4) + 2(y_3)}] + {\frac{d_1}{2}(y_5 + y_6)} $

$ = {\frac{5}{3}}[{(2.5 + 10.5) + 4(3.8 + 7.5) + 2(8.4)}] + {\frac{5}{2}(10.5 + 9.3)} $

= 174.5 $ m^2 $

$ A_2 = {\frac{d_2}{3}}[{(y_1 + y_5) + 4(y_2 + y_4) + 2(y_3)}] $

$ = {\frac{10}{3}}[{(9.3 + 8.4) + 4(5.8 + 6.9) + 2(7.8)}] $

= 280.33 $ m^2 $

Total area A = $ A_1 + A_2 = 174.5 + 280.33 = 454.83 m^2 $

Area by coordinate rule

For $ 1^{st} $ section

1. Number of interval $ (N_1 )=5 $

2. Length of interval $ (d_1 )=5m $

3. length = $ N_1*d_1=5*5=25m $

$ A_1 = [{\frac{2.5+3.8+8.4+7.5+10.5+9.3}{N_1 + 1}}] = 25[{\frac{42}{5+1}}] = 175 {m^2} $

For $ 2^{nd} $ section

1. Number of interval $ (N_2 )=4 $

2. Length of interval $ (d_2 )=10m $

3. length = $ N_2*d_2=4*10=40m $

$ A_2 = [{\frac{9.3+5.8+7.8+6.9+8.4}{N_2 + 1}}] = 40[{\frac{38.2}{4+1}}] = 305.6 {m^2} $

Total area A = $ A_1 + A_2 = 175 + 305.6 = 480.6 m^2 $