Find the approximated integral value using Boole's rule
| x | f(x) |
| 0.0 | 1.0000 |
| 0.1 | 0.9975 |
| 0.2 | 0.9900 |
| 0.3 | 0.9776 |
| 0.4 | 0.8604 |
Solution:The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=0` | `f(x_(0))=1` |
| `x_1=0.1` | `f(x_(1))=0.9975` |
| `x_2=0.2` | `f(x_(2))=0.99` |
| `x_3=0.3` | `f(x_(3))=0.9776` |
| `x_4=0.4` | `f(x_(4))=0.8604` |
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*1=7`
`32f(x_(1))=32*0.9975=31.92`
`12f(x_(2))=12*0.99=11.88`
`32f(x_(3))=32*0.9776=31.2832`
`7f(x_(4))=7*0.8604=6.0228`
`int f(x) dx=(2xx0.1)/45*(7+31.92+11.88+31.2832+6.0228)`
`=(2xx0.1)/45*(88.106)`
`=0.3916`
Solution by Boole's Rule is `0.3916`
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()]`
`=(2xx0.1)/45 [7xx(1 +0.8604)+32xx(0.9975+0.9776)+12xx(0.99)+14xx()]`
`=(2xx0.1)/45 [7xx(1.8604) + 32xx(1.9751) + 12xx(0.99) + 14xx(0)]`
`=(2xx0.1)/45 [(13.0228) + (63.2032) + (11.88) + (0)]`
`=0.3916`
Solution by Boole's Rule is `0.3916`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
