Find the approximated integral value of an equation 1/x using Trapezoidal rule
a = 1 and b = 2
Step value (h) = 0.25

Solution:
Equation is `f(x)=(1)/(x)`

`a=1`

`b=2`

The value of table for `x` and `f(x)`

`x`	`f(x)`
`x_0=1`	`f(x_(0))=f(1)=1`
`x_1=1.25`	`f(x_(1))=f(1.25)=0.8`
`x_2=1.5`	`f(x_(2))=f(1.5)=0.6667`
`x_3=1.75`	`f(x_(3))=f(1.75)=0.5714`
`x_4=2`	`f(x_(4))=f(2)=0.5`

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))=1`

`2f(x_(1))=2*0.8=1.6`

`2f(x_(2))=2*0.6667=1.3333`

`2f(x_(3))=2*0.5714=1.1429`

`f(x_(4))=0.5`

`int f(x) dx=0.25/2*(1+1.6+1.3333+1.1429+0.5)`

`=0.25/2*(5.5762)`

`=0.697`

Solution by Trapezoidal Rule is `0.697`

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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)))]`

`=0.25/2 [1 +0.5 + 2xx(0.8+0.6667+0.5714)]`

`=0.25/2 [1 +0.5 + 2xx(2.0381)]`

`=0.25/2 [1 +0.5 + 4.0762]`

`=0.697`

Solution by Trapezoidal Rule is `0.697`

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