Find the approximated integral value of an equation 1/(x+1) using Midpoint 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`

`x`

`x_0=0`

`x_1=0.2`

`x_2=0.4`

`x_3=0.6`

`x_4=0.8`

`x_5=1`

Method-1:
Using Midpoint Rule of Riemann Sum


`int f(x) dx=Delta x xx(f((x_0+x_1)/2)+f((x_1+x_2)/2)+f((x_2+x_3)/2)+f((x_3+x_4)/2)+f((x_4+x_5)/2))`

`=Delta x xx(f((0+0.2)/2)+f((0.2+0.4)/2)+f((0.4+0.6)/2)+f((0.6+0.8)/2)+f((0.8+1)/2))`

`=Delta x xx(f(0.1)+f(0.3)+f(0.5)+f(0.7)+f(0.9))`

`=0.2xx(0.9091+0.7692+0.6667+0.5882+0.5263)`

`=0.2xx(3.4595)`

`=0.6919`

Solution by Midpoint Rule of Riemann Sum is `0.6919`

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