Find the approximated integral value using Weddle's rule

x	f(x)
0	1
1	0.5
2	0.2
3	0.1
4	0.0588
5	0.0385
6	0.027

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))=0.2`
`x_3=3`	`f(x_(3))=0.1`
`x_4=4`	`f(x_(4))=0.0588`
`x_5=5`	`f(x_(5))=0.0385`
`x_6=6`	`f(x_(6))=0.027`

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))=0.2`

`6f(x_(3))=6*0.1=0.6`

`f(x_(4))=0.0588`

`5f(x_(5))=5*0.0385=0.1925`

`f(x_(6))=0.027`

`int f(x) dx=(3xx1)/10*[(1+2.5+0.2+0.6+0.05880.1925+0.027)]`

`=(3xx1)/10*(4.5783)`

`=1.3735`

Solution by Weddle's Rule is `1.3735`

---

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 + 0.2 + 6xx0.1 + 0.0588 + 5xx0.0385 + 0.027)]`

`=(3xx1)/10 [4.5783]`

`=1.3735`

Solution by Weddle's Rule is `1.3735`

---

---