1. Find Solution of an equation 1/x using Boole's rule
x1 = 1 and x2 = 2
Interval N = 8

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

`h = (b-a)/N`

`h = (2 - 1) / 8 = 0.125`

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

x	1	1.125	1.25	1.375	1.5	1.625	1.75	1.875	2
y	1	0.8889	0.8	0.7273	0.6667	0.6154	0.5714	0.5333	0.5

Method-1:
Using Boole's Rule
`int y dx=(2h)/45 [7(y_0 + y_8)+32(y_1+y_3+y_5+y_7)+12(y_2+y_6)+14(y_4)]`

`int y dx=(2xx0.125)/45 [7xx(1 +0.5)+32xx(0.8889+0.7273+0.6154+0.5333)+12xx(0.8+0.5714)+14xx(0.6667)]`

`int y dx=(2xx0.125)/45 [7xx(1.5) + 32xx(2.7649) + 12xx(1.3714) + 14xx(0.6667)]`

`int y dx=(2xx0.125)/45 [(10.5) + (88.4761) + (16.4571) + (9.3333)]`

`int y dx=0.6931`

Solution by Boole's Rule is `0.6931`

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Method-2:
Using Boole's Rule
`int y dx=(2h)/45 [7y_0+32y_1+12y_2+32y_3+14y_4+32y_5+12y_6+32y_7+7y_8]`

`7y_0=7*1=7`

`32y_1=32*0.8889=28.4444`

`12y_2=12*0.8=9.6`

`32y_3=32*0.7273=23.2727`

`14y_4=14*0.6667=9.3333`

`32y_5=32*0.6154=19.6923`

`12y_6=12*0.5714=6.8571`

`32y_7=32*0.5333=17.0667`

`7y_8=7*0.5=3.5`

`int y dx=(2xx0.125)/45*(7+28.4444+9.6+23.2727+9.3333+19.6923+6.8571+17.0667+3.5)`

`int y dx=(2xx0.125)/45*(124.7666)`

`int y dx=0.6931`

Solution by Boole's Rule is `0.6931`

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