Find the approximated integral value of an equation 2x^3-4x+1 using Boole's rule
a = 2 and b = 4
Step value (h) = 0.5

Solution:
Equation is `f(x)=2x^3-4x+1`

`a=2`

`b=4`

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

`x`	`f(x)`
`x_0=2`	`f(x_(0))=f(2)=9`
`x_1=2.5`	`f(x_(1))=f(2.5)=22.25`
`x_2=3`	`f(x_(2))=f(3)=43`
`x_3=3.5`	`f(x_(3))=f(3.5)=72.75`
`x_4=4`	`f(x_(4))=f(4)=113`

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*9=63`

`32f(x_(1))=32*22.25=712`

`12f(x_(2))=12*43=516`

`32f(x_(3))=32*72.75=2328`

`7f(x_(4))=7*113=791`

`int f(x) dx=(2xx0.5)/45*(63+712+516+2328+791)`

`=(2xx0.5)/45*(4410)`

`=98`

Solution by Boole's Rule is `98`

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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()]`

`=(2xx0.5)/45 [7xx(9 +113)+32xx(22.25+72.75)+12xx(43)+14xx()]`

`=(2xx0.5)/45 [7xx(122) + 32xx(95) + 12xx(43) + 14xx(0)]`

`=(2xx0.5)/45 [(854) + (3040) + (516) + (0)]`

`=98`

Solution by Boole's Rule is `98`

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