1. Find Solution of an equation 1/x using Boole's rule
x1 = 1 and x2 = 2
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 |
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y | 1 | 0.8889 | 0.8 | 0.7273 | 0.6667 | 0.6154 | 0.5714 | 0.5333 | 0.5 |
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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 = 0.6931`
Solution by Boole's Rule is `0.6931`
2. 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 |
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y | 9 | 22.25 | 43 | 72.75 | 113 |
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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 = 98`
Solution by Boole's Rule is `98`
This material is intended as a summary. Use your textbook for detail explanation.
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