Formula
1. Boole's Rule
`int y dx =(2h)/45 [7(y_0 + y_n) + 32(y_1+y_3+y_5+...) + 12(y_2+y_6+y_10+...) + 14(y_4+y_8+y_12+...)]`
`int f(x) dx =(2Delta x )/45 [7(f(x_(0))+f(x_(n)))+32(f(x_(1))+f(x_(3))+f(x_(5))+...)+12(f(x_(2))+f(x_(6))+f(x_(10))+...)+14(f(x_(4))+f(x_(8))+f(x_(12))+...)]`
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Examples
1. Find the approximated integral value using Boole's rule
| x | f(x) |
| 1.4 | 4.0552 |
| 1.6 | 4.9530 |
| 1.8 | 6.0436 |
| 2.0 | 7.3891 |
| 2.2 | 9.0250 |
Solution:The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=1.4` | `f(x_(0))=4.0552` |
| `x_1=1.6` | `f(x_(1))=4.953` |
| `x_2=1.8` | `f(x_(2))=6.0436` |
| `x_3=2` | `f(x_(3))=7.3891` |
| `x_4=2.2` | `f(x_(4))=9.025` |
Method-1:Using Boole's Rule
`int f(x) dx =(2Delta x )/45 [7(f(x_(0))+f(x_(n)))+32(f(x_(1))+f(x_(3))+f(x_(5))+...)+12(f(x_(2))+f(x_(6))+f(x_(10))+...)+14(f(x_(4))+f(x_(8))+f(x_(12))+...)]`
`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*4.0552=28.3864`
`32f(x_(1))=32*4.953=158.496`
`12f(x_(2))=12*6.0436=72.5232`
`32f(x_(3))=32*7.3891=236.4512`
`7f(x_(4))=7*9.025=63.175`
`int f(x) dx=(2xx0.2)/45*(28.3864+158.496+72.5232+236.4512+63.175)`
`=(2xx0.2)/45*(559.0318)`
`=4.9692`
Solution by Boole's Rule is `4.9692`
Method-2:Using Boole's Rule
`int f(x) dx =(2Delta x )/45 [7(f(x_(0))+f(x_(n)))+32(f(x_(1))+f(x_(3))+f(x_(5))+...)+12(f(x_(2))+f(x_(6))+f(x_(10))+...)+14(f(x_(4))+f(x_(8))+f(x_(12))+...)]`
`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()]`
`=(2xx0.2)/45 [7xx(4.0552 +9.025)+32xx(4.953+7.3891)+12xx(6.0436)+14xx()]`
`=(2xx0.2)/45 [7xx(13.0802) + 32xx(12.3421) + 12xx(6.0436) + 14xx(0)]`
`=(2xx0.2)/45 [(91.5614) + (394.9472) + (72.5232) + (0)]`
`=4.9692`
Solution by Boole's Rule is `4.9692`
This material is intended as a summary. Use your textbook for detail explanation.
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