3. Find Solution using Boole's 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 Boole's Rule
`int y dx=(2h)/45 [7(y_0 + y_4)+32(y_1+y_3)+12(y_2)+14()]`

`int y dx=(2xx0.25)/45 [7xx(1 +0.8415)+32xx(0.9896+0.9089)+12xx(0.9589)+14xx()]`

`int y dx=(2xx0.25)/45 [7xx(1.8415) + 32xx(1.8985) + 12xx(0.9589) + 14xx(0)]`

`int y dx=(2xx0.25)/45 [(12.8905) + (60.752) + (11.5068) + (0)]`

`int y dx=0.9461`

Solution by Boole's Rule is `0.9461`

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Method-2:
Using Boole's Rule
`int y dx=(2h)/45 [7y_0+32y_1+12y_2+32y_3+7y_4]`

`7y_0=7*1=7`

`32y_1=32*0.9896=31.6672`

`12y_2=12*0.9589=11.5068`

`32y_3=32*0.9089=29.0848`

`7y_4=7*0.8415=5.8905`

`int y dx=(2xx0.25)/45*(7+31.6672+11.5068+29.0848+5.8905)`

`int y dx=(2xx0.25)/45*(85.1493)`

`int y dx=0.9461`

Solution by Boole's Rule is `0.9461`

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