Find the approximated integral value using Boole's rule
| x | f(x) |
| 0.00 | 1.0000 |
| 0.25 | 0.9896 |
| 0.50 | 0.9589 |
| 0.75 | 0.9089 |
| 1.00 | 0.8415 |
Solution:The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=0` | `f(x_(0))=1` |
| `x_1=0.25` | `f(x_(1))=0.9896` |
| `x_2=0.5` | `f(x_(2))=0.9589` |
| `x_3=0.75` | `f(x_(3))=0.9089` |
| `x_4=1` | `f(x_(4))=0.8415` |
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*0.9896=31.6672`
`12f(x_(2))=12*0.9589=11.5068`
`32f(x_(3))=32*0.9089=29.0848`
`7f(x_(4))=7*0.8415=5.8905`
`int f(x) dx=(2xx0.25)/45*(7+31.6672+11.5068+29.0848+5.8905)`
`=(2xx0.25)/45*(85.1493)`
`=0.9461`
Solution by Boole's Rule is `0.9461`
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()]`
`=(2xx0.25)/45 [7xx(1 +0.8415)+32xx(0.9896+0.9089)+12xx(0.9589)+14xx()]`
`=(2xx0.25)/45 [7xx(1.8415) + 32xx(1.8985) + 12xx(0.9589) + 14xx(0)]`
`=(2xx0.25)/45 [(12.8905) + (60.752) + (11.5068) + (0)]`
`=0.9461`
Solution by Boole's Rule is `0.9461`
This material is intended as a summary. Use your textbook for detail explanation.
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