Find the approximated integral value of an equation 1/(x+1) using Boole's rule
a = 0 and b = 1
Interval n = 5Solution:Equation is `f(x)=(1)/(x+1)`
`a=0`
`b=1`
`Delta x =(b-a)/n=(1 - 0)/5=0.2`
The value of table for `x` and `f(x)`
| `x` | `f(x)` |
| `x_0=0` | `f(x_(0))=f(0)=1` |
| `x_1=0.2` | `f(x_(1))=f(0.2)=0.8333` |
| `x_2=0.4` | `f(x_(2))=f(0.4)=0.7143` |
| `x_3=0.6` | `f(x_(3))=f(0.6)=0.625` |
| `x_4=0.8` | `f(x_(4))=f(0.8)=0.5556` |
| `x_5=1` | `f(x_(5))=f(1)=0.5` |
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))+14f(x_(4))+7f(x_(5))]`
`7f(x_(0))=7*1=7`
`32f(x_(1))=32*0.8333=26.6667`
`12f(x_(2))=12*0.7143=8.5714`
`32f(x_(3))=32*0.625=20`
`14f(x_(4))=14*0.5556=7.7778`
`7f(x_(5))=7*0.5=3.5`
`int f(x) dx=(2xx0.2)/45*(7+26.6667+8.5714+20+7.7778+3.5)`
`=(2xx0.2)/45*(73.5159)`
`=0.6535`
Solution by Boole's Rule is `0.6535`
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_(5)))+32(f(x_(1))+f(x_(3)))+12(f(x_(2)))+14(f(x_(4)))]`
`=(2xx0.2)/45 [7xx(1 +0.5)+32xx(0.8333+0.625)+12xx(0.7143)+14xx(0.5556)]`
`=(2xx0.2)/45 [7xx(1.5) + 32xx(1.4583) + 12xx(0.7143) + 14xx(0.5556)]`
`=(2xx0.2)/45 [(10.5) + (46.6667) + (8.5714) + (7.7778)]`
`=0.6535`
Solution by Boole's Rule is `0.6535`
This material is intended as a summary. Use your textbook for detail explanation.
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