3. Find Solution using Simpson's 1/3 rule

x	f(x)
0.00	1.0000
0.25	0.9896
0.50	0.9589
0.75	0.9089
1.00	0.8415

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

x	0	0.25	0.5	0.75	1
y	1	0.9896	0.9589	0.9089	0.8415

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

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

`int y dx=0.25/3 [(1 +0.8415)+4xx(0.9896+0.9089)+2xx(0.9589)]`

`int y dx=0.25/3 [(1 +0.8415)+4xx(1.8985)+2xx(0.9589)]`

`int y dx=0.25/3 [(1.8415)+(7.594)+(1.9178)]`

`int y dx=0.9461`

Solution by Simpson's `1/3` Rule is `0.9461`

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

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

`y_0=1`

`4y_1=4*0.9896=3.9584`

`2y_2=2*0.9589=1.9178`

`4y_3=4*0.9089=3.6356`

`y_4=0.8415`

`int y dx=0.25/3*(1+3.9584+1.9178+3.6356+0.8415)`

`int y dx=0.25/3*(11.3533)`

`int y dx=0.9461`

Solution by Simpson's `1/3` Rule is `0.9461`

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