Find the approximated integral value of an equation 1/(x+1) using Simpson's 3/8 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 Simpson's `3/8` Rule



`int f(x) dx=(3Delta x )/8 [f(x_(0))+3f(x_(1))+3f(x_(2))+2f(x_(3))+3f(x_(4))+f(x_(5))]`

`f(x_(0))=1`

`3f(x_(1))=3*0.8333=2.5`

`3f(x_(2))=3*0.7143=2.1429`

`2f(x_(3))=2*0.625=1.25`

`3f(x_(4))=3*0.5556=1.6667`

`f(x_(5))=0.5`

`int f(x) dx=(3xx0.2)/8 *(1+2.5+2.1429+1.25+1.6667+0.5)`

`=(3xx0.2)/8 *(9.0595)`

`=0.6795`

Solution by Simpson's `3/8` Rule is `0.6795`

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Method-2:
Using Simpson's `3/8` Rule



`int f(x) dx=(3Delta x )/8 [(f(x_(0))+f(x_(5)))+2(f(x_(3)))+3(f(x_(1))+f(x_(2))+f(x_(4)))]`

`=(3xx0.2)/8 [(1 +0.5)+2xx(0.625)+3xx(0.8333+0.7143+0.5556)]`

`=(3xx0.2)/8 [(1 +0.5)+2xx(0.625)+3xx(2.1032)]`

`=(3xx0.2)/8 [(1.5)+(1.25)+(6.3095)]`

`=0.6795`

Solution by Simpson's `3/8` Rule is `0.6795`

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