Find the approximated integral value of an equation 2x^3-4x+1 using Simpson's 1/3 rule
a = 2 and b = 4
Step value (h) = 0.5

Solution:
Equation is `f(x)=2x^3-4x+1`

`a=2`

`b=4`

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

`x`	`f(x)`
`x_0=2`	`f(x_(0))=f(2)=9`
`x_1=2.5`	`f(x_(1))=f(2.5)=22.25`
`x_2=3`	`f(x_(2))=f(3)=43`
`x_3=3.5`	`f(x_(3))=f(3.5)=72.75`
`x_4=4`	`f(x_(4))=f(4)=113`

Method-1:
Using Simpsons `1/3` Rule



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

`f(x_(0))=9`

`4f(x_(1))=4*22.25=89`

`2f(x_(2))=2*43=86`

`4f(x_(3))=4*72.75=291`

`f(x_(4))=113`

`int f(x) dx=0.5/3*(9+89+86+291+113)`

`=0.5/3*(588)`

`=98`

Solution by Simpson's `1/3` Rule is `98`

---

Method-2:
Using Simpsons `1/3` Rule



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

`=0.5/3 [(9 +113)+4xx(22.25+72.75)+2xx(43)]`

`=0.5/3 [(9 +113)+4xx(95)+2xx(43)]`

`=0.5/3 [(122)+(380)+(86)]`

`=98`

Solution by Simpson's `1/3` Rule is `98`

---

---