2. Find Solution using Boole's rule

x	f(x)
0.0	1.0000
0.1	0.9975
0.2	0.9900
0.3	0.9776
0.4	0.8604

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

x	0	0.1	0.2	0.3	0.4
y	1	0.9975	0.99	0.9776	0.8604

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=(2xx0.1)/45 [7xx(1 +0.8604)+32xx(0.9975+0.9776)+12xx(0.99)+14xx()]`

`int y dx=(2xx0.1)/45 [7xx(1.8604) + 32xx(1.9751) + 12xx(0.99) + 14xx(0)]`

`int y dx=(2xx0.1)/45 [(13.0228) + (63.2032) + (11.88) + (0)]`

`int y dx=0.3916`

Solution by Boole's Rule is `0.3916`

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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*0.9975=31.92`

`12y_2=12*0.99=11.88`

`32y_3=32*0.9776=31.2832`

`7y_4=7*0.8604=6.0228`

`int y dx=(2xx0.1)/45*(7+31.92+11.88+31.2832+6.0228)`

`int y dx=(2xx0.1)/45*(88.106)`

`int y dx=0.3916`

Solution by Boole's Rule is `0.3916`

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