2. Find Solution of an equation 1/(x+1) using Simpson's 1/3 rule
x1 = 0 and x2 = 1
Interval N = 5

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

`h = (b-a)/N`

`h = (1 - 0) / 5 = 0.2`

The value of table for `x` and `y`

x	0	0.2	0.4	0.6	0.8	1
y	1	0.8333	0.7143	0.625	0.5556	0.5

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

`int y dx=h/3 [(y_0+y_5)+4(y_1+y_3)+2(y_2+y_4)]`

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

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

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

`int y dx=0.6582`

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

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

`int y dx=h/3 [y_0+4y_1+2y_2+4y_3+2y_4+y_5]`

`y_0=1`

`4y_1=4*0.8333=3.3333`

`2y_2=2*0.7143=1.4286`

`4y_3=4*0.625=2.5`

`2y_4=2*0.5556=1.1111`

`y_5=0.5`

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

`int y dx=0.2/3*(9.873)`

`int y dx=0.6582`

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

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