Find the approximated integral value using Weddle's rule

x	f(x)
0	1
1	1/2
2	1/3
3	1/4
4	1/5
5	1/6
6	1/7

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

`x`	`f(x)`
`x_0=0`	`f(x_(0))=1`
`x_1=1`	`f(x_(1))=0.5`
`x_2=2`	`f(x_(2))=0.3333`
`x_3=3`	`f(x_(3))=0.25`
`x_4=4`	`f(x_(4))=0.2`
`x_5=5`	`f(x_(5))=0.1667`
`x_6=6`	`f(x_(6))=0.1429`

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`

`5f(x_(1))=5*0.5=2.5`

`f(x_(2))=0.3333`

`6f(x_(3))=6*0.25=1.5`

`f(x_(4))=0.2`

`5f(x_(5))=5*0.1667=0.8333`

`f(x_(6))=0.1429`

`int f(x) dx=(3xx1)/10*[(1+2.5+0.3333+1.5+0.20.8333+0.1429)]`

`=(3xx1)/10*(6.5095)`

`=1.9529`

Solution by Weddle's Rule is `1.9529`

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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)))]`

`=(3xx1)/10 [(1 + 5xx0.5 + 0.3333 + 6xx0.25 + 0.2 + 5xx0.1667 + 0.1429)]`

`=(3xx1)/10 [6.5095]`

`=1.9529`

Solution by Weddle's Rule is `1.9529`

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