written 6.2 years ago by
teamques10
★ 67k
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modified 6.2 years ago
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Area by the Trapezoidal rule:
Ch.in m |
Interval (d) |
ordinate |
Offset in m |
0 |
|
Y1 |
2.5 |
5 |
|
Y2 |
3.8 |
10 |
d1 = 5 |
Y3 |
8.4 |
15 |
|
Y4 |
7.5 |
20 |
|
Y5 |
10.5 |
25 |
|
Y6 |
9.3 |
25 |
|
Y7 |
9.3 |
35 |
|
Y8 |
5.8 |
45 |
d2 = 10 |
Y9 |
7.8 |
55 |
|
Y10 |
6.9 |
65 |
|
Y11 |
8.4 |
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:
Ch.in m |
Interval (d) |
ordinate |
Offset in m |
0 |
|
Y1 |
2.5 |
5 |
|
Y2 |
3.8 |
10 |
d1 = 5 |
Y3 |
8.4 |
15 |
|
Y4 |
7.5 |
20 |
|
Y5 |
10.5 |
25 |
|
Y6 |
9.3 |
25 |
|
Y1 |
9.3 |
35 |
|
Y2 |
5.8 |
45 |
d2 = 10 |
Y3 |
7.8 |
55 |
|
Y4 |
6.9 |
65 |
|
Y5 |
8.4 |
$ 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
Number of interval $ (N_1 )=5 $
Length of interval $ (d_1 )=5m $
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
Number of interval $ (N_2 )=4 $
Length of interval $ (d_2 )=10m $
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 $