Find the approximated integral value of an equation 1/x using Boole's rule
a = 1 and b = 2
Step value (h) = 0.25

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

`a=1`

`b=2`

The value of table for `x` and `f(x)`

`x`	`f(x)`
`x_0=1`	`f(x_(0))=f(1)=1`
`x_1=1.25`	`f(x_(1))=f(1.25)=0.8`
`x_2=1.5`	`f(x_(2))=f(1.5)=0.6667`
`x_3=1.75`	`f(x_(3))=f(1.75)=0.5714`
`x_4=2`	`f(x_(4))=f(2)=0.5`

Method-1:
Using Boole's Rule


`int f(x) dx=(2Delta x )/45 [7f(x_(0))+32f(x_(1))+12f(x_(2))+32f(x_(3))+7f(x_(4))]`

`7f(x_(0))=7*1=7`

`32f(x_(1))=32*0.8=25.6`

`12f(x_(2))=12*0.6667=8`

`32f(x_(3))=32*0.5714=18.2857`

`7f(x_(4))=7*0.5=3.5`

`int f(x) dx=(2xx0.25)/45*(7+25.6+8+18.2857+3.5)`

`=(2xx0.25)/45*(62.3857)`

`=0.6932`

Solution by Boole's Rule is `0.6932`

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Method-2:
Using Boole's Rule


`int f(x) dx=(2Delta x )/45 [7(f(x_(0))+f(x_(4)))+32(f(x_(1))+f(x_(3)))+12(f(x_(2)))+14()]`

`=(2xx0.25)/45 [7xx(1 +0.5)+32xx(0.8+0.5714)+12xx(0.6667)+14xx()]`

`=(2xx0.25)/45 [7xx(1.5) + 32xx(1.3714) + 12xx(0.6667) + 14xx(0)]`

`=(2xx0.25)/45 [(10.5) + (43.8857) + (8) + (0)]`

`=0.6932`

Solution by Boole's Rule is `0.6932`

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