Find the approximated integral value of an equation 1/x using Weddle's rule
a = 1 and b = 2
Interval n = 6

Solution:
Equation is `f(x)=(1)/(x)`

`a=1`

`b=2`

`Delta x =(b-a)/n=(2 - 1)/6=0.1667`

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

`x`	`f(x)`
`x_0=1`	`f(x_(0))=f(1)=1`
`x_1=1.1667`	`f(x_(1))=f(1.1667)=0.8571`
`x_2=1.3333`	`f(x_(2))=f(1.3333)=0.75`
`x_3=1.5`	`f(x_(3))=f(1.5)=0.6667`
`x_4=1.6667`	`f(x_(4))=f(1.6667)=0.6`
`x_5=1.8333`	`f(x_(5))=f(1.8333)=0.5455`
`x_6=2`	`f(x_(6))=f(2)=0.5`

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*0.8571=4.2857`

`f(x_(2))=0.75`

`6f(x_(3))=6*0.6667=4`

`f(x_(4))=0.6`

`5f(x_(5))=5*0.5455=2.7273`

`f(x_(6))=0.5`

`int f(x) dx=(3xx0.1667)/10*[(1+4.2857+0.75+4+0.62.7273+0.5)]`

`=(3xx0.1667)/10*(13.863)`

`=0.6931`

Solution by Weddle's Rule is `0.6931`

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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)))]`

`=(3xx0.1667)/10 [(1 + 5xx0.8571 + 0.75 + 6xx0.6667 + 0.6 + 5xx0.5455 + 0.5)]`

`=(3xx0.1667)/10 [13.863]`

`=0.6931`

Solution by Weddle's Rule is `0.6931`

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