Find the approximated integral value of an equation x^3-2x+1 using Boole's rule
a = 2 and b = 4
Step value (h) = 0.5Solution:Equation is `f(x)=x^3-2x+1`
`a=2`
`b=4`
The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=2` | `f(x_(0))=f(2)=5` |
| `x_1=2.5` | `f(x_(1))=f(2.5)=11.625` |
| `x_2=3` | `f(x_(2))=f(3)=22` |
| `x_3=3.5` | `f(x_(3))=f(3.5)=36.875` |
| `x_4=4` | `f(x_(4))=f(4)=57` |
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*5=35`
`32f(x_(1))=32*11.625=372`
`12f(x_(2))=12*22=264`
`32f(x_(3))=32*36.875=1180`
`7f(x_(4))=7*57=399`
`int f(x) dx=(2xx0.5)/45*(35+372+264+1180+399)`
`=(2xx0.5)/45*(2250)`
`=50`
Solution by Boole's Rule is `50`
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()]`
`=(2xx0.5)/45 [7xx(5 +57)+32xx(11.625+36.875)+12xx(22)+14xx()]`
`=(2xx0.5)/45 [7xx(62) + 32xx(48.5) + 12xx(22) + 14xx(0)]`
`=(2xx0.5)/45 [(434) + (1552) + (264) + (0)]`
`=50`
Solution by Boole's Rule is `50`
This material is intended as a summary. Use your textbook for detail explanation.
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