4. Find Solution of an equation 2x^3-4x+1 using Boole's rule
x1 = 2 and x2 = 4
Step value (h) = 0.5

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

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

x	2	2.5	3	3.5	4
y	9	22.25	43	72.75	113

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

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

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

`int y dx=98`

Solution by Boole's Rule is `98`

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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*9=63`

`32y_1=32*22.25=712`

`12y_2=12*43=516`

`32y_3=32*72.75=2328`

`7y_4=7*113=791`

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

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

`int y dx=98`

Solution by Boole's Rule is `98`

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