Find the approximated integral value of an equation 1/(x+1) using Trapezoidal rule
a = 0 and b = 1
Interval n = 5

Solution:
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 Trapezoidal Rule


`int f(x) dx=(Delta x )/2 [f(x_(0))+2f(x_(1))+2f(x_(2))+2f(x_(3))+2f(x_(4))+f(x_(5))]`

`f(x_(0))=1`

`2f(x_(1))=2*0.8333=1.6667`

`2f(x_(2))=2*0.7143=1.4286`

`2f(x_(3))=2*0.625=1.25`

`2f(x_(4))=2*0.5556=1.1111`

`f(x_(5))=0.5`

`int f(x) dx=0.2/2*(1+1.6667+1.4286+1.25+1.1111+0.5)`

`=0.2/2*(6.9563)`

`=0.6956`

Solution by Trapezoidal Rule is `0.6956`

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Method-2:
Using Trapezoidal Rule


`int f(x) dx=(Delta x )/2 [f(x_(0))+f(x_(5))+2(f(x_(1))+f(x_(2))+f(x_(3))+f(x_(4)))]`

`=0.2/2 [1 +0.5 + 2xx(0.8333+0.7143+0.625+0.5556)]`

`=0.2/2 [1 +0.5 + 2xx(2.7282)]`

`=0.2/2 [1 +0.5 + 5.4563]`

`=0.6956`

Solution by Trapezoidal Rule is `0.6956`

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