Formula

1. Weddle's Rule `int y dx = (3h)/10 [(y_0+5y_1+y_2+6y_3+y_4+5y_5+y_6)+(y_6+5y_7+y_8+6y_9+y_10+5y_11+y_12)+...]` `int f(x) dx=(3Delta x )/10 [(f(x_(0))+5f(x_(1))+f(x_(2))+6f(x_(3))+f(x_(4))+5f(x_(5))+f(x_(6)))+(f(x_(6))+5f(x_(7))+f(x_(8))+6f(x_(9))+f(x_(10))+5f(x_(11))+f(x_(12)))+...]`

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Examples
1. Find the approximated integral value using Weddle's rule

x	f(x)
7.47	1.93
7.48	1.95
7.49	1.98
7.50	2.01
7.51	2.03
7.52	2.06
7.53	2.09

Solution:
The value of table for `x` and `f(x)`

`x`	`f(x)`
`x_0=7.47`	`f(x_(0))=1.93`
`x_1=7.48`	`f(x_(1))=1.95`
`x_2=7.49`	`f(x_(2))=1.98`
`x_3=7.5`	`f(x_(3))=2.01`
`x_4=7.51`	`f(x_(4))=2.03`
`x_5=7.52`	`f(x_(5))=2.06`
`x_6=7.53`	`f(x_(6))=2.09`

Method-1:
Using Weddle's Rule


`int f(x) dx=(3Delta x )/10 [(f(x_(0))+5f(x_(1))+f(x_(2))+6f(x_(3))+f(x_(4))+5f(x_(5))+f(x_(6)))]`

`f(x_(0))=1.93`

`5f(x_(1))=5*1.95=9.75`

`f(x_(2))=1.98`

`6f(x_(3))=6*2.01=12.06`

`f(x_(4))=2.03`

`5f(x_(5))=5*2.06=10.3`

`f(x_(6))=2.09`

`int f(x) dx=(3xx0.01)/10*[(1.93+9.75+1.98+12.06+2.0310.3+2.09)]`

`=(3xx0.01)/10*(40.14)`

`=0.1204`

Solution by Weddle's Rule is `0.1204`

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Method-2:
Using Weddle's Rule


`int f(x) dx=(3Delta x )/10 [(f(x_(0))+5f(x_(1))+f(x_(2))+6f(x_(3))+f(x_(4))+5f(x_(5))+f(x_(6)))]`

`=(3xx0.01)/10 [(1.93 + 5xx1.95 + 1.98 + 6xx2.01 + 2.03 + 5xx2.06 + 2.09)]`

`=(3xx0.01)/10 [40.14]`

`=0.1204`

Solution by Weddle's Rule is `0.1204`

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