Formula

1. Midpoint Rule

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Examples
1. Find the approximated integral value using Midpoint Rule

x	f(x)
1.4	4.0552
1.6	4.9530
1.8	6.0436
2.0	7.3891
2.2	9.0250

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

`x`	`f(x)`
`x_0=1.4`	`f(x_(0))=4.0552`
`x_1=1.6`	`f(x_(1))=4.953`
`x_2=1.8`	`f(x_(2))=6.0436`
`x_3=2`	`f(x_(3))=7.3891`
`x_4=2.2`	`f(x_(4))=9.025`

Method-1:
Using Midpoint Rule of Riemann Sum
`int f(x) dx=Delta x xx(f((x_2+x_0)/2)+f((x_4+x_2)/2))`

`=0.4xx(f((1.8+1.4)/2)+f((2.2+1.8)/2))`

`=0.4xx(f(1.6)+f(2))`

`=0.4xx(4.953+7.3891)`

`=0.4xx(12.3421)`

`=2.4684`

Solution by Midpoint Rule of Riemann Sum is `2.4684`

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