Find the approximated integral value of an equation x^3-2x+1 using Trapezoidal rule
a = 2 and b = 4
Step value (h) = 0.5

Solution:
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 Trapezoidal Rule


`int f(x) dx=(Delta x )/2 [f(x_(0))+2f(x_(1))+2f(x_(2))+2f(x_(3))+f(x_(4))]`

`f(x_(0))=5`

`2f(x_(1))=2*11.625=23.25`

`2f(x_(2))=2*22=44`

`2f(x_(3))=2*36.875=73.75`

`f(x_(4))=57`

`int f(x) dx=0.5/2*(5+23.25+44+73.75+57)`

`=0.5/2*(203)`

`=50.75`

Solution by Trapezoidal Rule is `50.75`

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Method-2:
Using Trapezoidal Rule


`int f(x) dx=(Delta x )/2 [f(x_(0))+f(x_(4))+2(f(x_(1))+f(x_(2))+f(x_(3)))]`

`=0.5/2 [5 +57 + 2xx(11.625+22+36.875)]`

`=0.5/2 [5 +57 + 2xx(70.5)]`

`=0.5/2 [5 +57 + 141]`

`=50.75`

Solution by Trapezoidal Rule is `50.75`

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