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

Solution:
Equation is `f(x)=2x^3-4x+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)=9`
`x_1=2.5`	`f(x_(1))=f(2.5)=22.25`
`x_2=3`	`f(x_(2))=f(3)=43`
`x_3=3.5`	`f(x_(3))=f(3.5)=72.75`
`x_4=4`	`f(x_(4))=f(4)=113`

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))=9`

`2f(x_(1))=2*22.25=44.5`

`2f(x_(2))=2*43=86`

`2f(x_(3))=2*72.75=145.5`

`f(x_(4))=113`

`int f(x) dx=0.5/2*(9+44.5+86+145.5+113)`

`=0.5/2*(398)`

`=99.5`

Solution by Trapezoidal Rule is `99.5`

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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 [9 +113 + 2xx(22.25+43+72.75)]`

`=0.5/2 [9 +113 + 2xx(138)]`

`=0.5/2 [9 +113 + 276]`

`=99.5`

Solution by Trapezoidal Rule is `99.5`

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