Find the approximated integral value of an equation x^3-2x+1 using Simpson's 1/3 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 Simpsons `1/3` Rule
`int f(x) dx=(Delta x )/3 (f(x_(0))+4(f(x_(1))+f(x_(3))+f(x_(5))+...+f(x_(n-1)))+2(f(x_(2))+f(x_(4))+f(x_(6))+...+f(x_(n-2)))+f(x_(n)))`
`int f(x) dx=(Delta x )/3 [f(x_(0))+4f(x_(1))+2f(x_(2))+4f(x_(3))+f(x_(4))]`
`f(x_(0))=5`
`4f(x_(1))=4*11.625=46.5`
`2f(x_(2))=2*22=44`
`4f(x_(3))=4*36.875=147.5`
`f(x_(4))=57`
`int f(x) dx=0.5/3*(5+46.5+44+147.5+57)`
`=0.5/3*(300)`
`=50`
Solution by Simpson's `1/3` Rule is `50`
Method-2:Using Simpsons `1/3` Rule
`int f(x) dx=(Delta x )/3 (f(x_(0))+4(f(x_(1))+f(x_(3))+f(x_(5))+...+f(x_(n-1)))+2(f(x_(2))+f(x_(4))+f(x_(6))+...+f(x_(n-2)))+f(x_(n)))`
`int f(x) dx=(Delta x )/3 [(f(x_(0))+f(x_(4)))+4(f(x_(1))+f(x_(3)))+2(f(x_(2)))]`
`=0.5/3 [(5 +57)+4xx(11.625+36.875)+2xx(22)]`
`=0.5/3 [(5 +57)+4xx(48.5)+2xx(22)]`
`=0.5/3 [(62)+(194)+(44)]`
`=50`
Solution by Simpson's `1/3` Rule is `50`
This material is intended as a summary. Use your textbook for detail explanation.
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