Find the approximated integral value of an equation 1/(x+1) using Simpson's 1/3 rule
a = 0 and b = 1
Interval n = 5

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

`a=0`

`b=1`

`Delta x =(b-a)/n=(1 - 0)/5=0.2`

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

`x`	`f(x)`
`x_0=0`	`f(x_(0))=f(0)=1`
`x_1=0.2`	`f(x_(1))=f(0.2)=0.8333`
`x_2=0.4`	`f(x_(2))=f(0.4)=0.7143`
`x_3=0.6`	`f(x_(3))=f(0.6)=0.625`
`x_4=0.8`	`f(x_(4))=f(0.8)=0.5556`
`x_5=1`	`f(x_(5))=f(1)=0.5`

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))+2f(x_(4))+f(x_(5))]`

`f(x_(0))=1`

`4f(x_(1))=4*0.8333=3.3333`

`2f(x_(2))=2*0.7143=1.4286`

`4f(x_(3))=4*0.625=2.5`

`2f(x_(4))=2*0.5556=1.1111`

`f(x_(5))=0.5`

`int f(x) dx=0.2/3*(1+3.3333+1.4286+2.5+1.1111+0.5)`

`=0.2/3*(9.873)`

`=0.6582`

Solution by Simpson's `1/3` Rule is `0.6582`

---

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



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

`=0.2/3 [(1 +0.5)+4xx(0.8333+0.625)+2xx(0.7143+0.5556)]`

`=0.2/3 [(1 +0.5)+4xx(1.4583)+2xx(1.2698)]`

`=0.2/3 [(1.5)+(5.8333)+(2.5397)]`

`=0.6582`

Solution by Simpson's `1/3` Rule is `0.6582`

---

---