Find the approximated integral value of an equation 1/(x+1) using Right endpoint approximation
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 Right endpoint approximation (Right Riemann Sum)


`int f(x) dx=Delta x xx(f(x_(1))+f(x_(2))+f(x_(3))+f(x_(4))+f(x_(5)))`

`=0.2xx(0.8333+0.7143+0.625+0.5556+0.5)`

`=0.2xx(3.2282)`

`=0.6456`

Solution by Right endpoint approximation (Right Riemann Sum) is `0.6456`

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