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

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

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

x	2	2.5	3	3.5	4
y	5	11.625	22	36.875	57

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(5 +57)+32xx(11.625+36.875)+12xx(22)+14xx()]`

`int y dx=(2xx0.5)/45 [7xx(62) + 32xx(48.5) + 12xx(22) + 14xx(0)]`

`int y dx=(2xx0.5)/45 [(434) + (1552) + (264) + (0)]`

`int y dx=50`

Solution by Boole's Rule is `50`

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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*5=35`

`32y_1=32*11.625=372`

`12y_2=12*22=264`

`32y_3=32*36.875=1180`

`7y_4=7*57=399`

`int y dx=(2xx0.5)/45*(35+372+264+1180+399)`

`int y dx=(2xx0.5)/45*(2250)`

`int y dx=50`

Solution by Boole's Rule is `50`

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