Find the approximated integral value using Midpoint Rule

x	f(x)
0.00	1.0000
0.25	0.9896
0.50	0.9589
0.75	0.9089
1.00	0.8415

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

`x`	`f(x)`
`x_0=0`	`f(x_(0))=1`
`x_1=0.25`	`f(x_(1))=0.9896`
`x_2=0.5`	`f(x_(2))=0.9589`
`x_3=0.75`	`f(x_(3))=0.9089`
`x_4=1`	`f(x_(4))=0.8415`

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))`

`=0.5xx(f((0.5+0)/2)+f((1+0.5)/2))`

`=0.5xx(f(0.25)+f(0.75))`

`=0.5xx(0.9896+0.9089)`

`=0.5xx(1.8985)`

`=0.4746`

Solution by Midpoint Rule of Riemann Sum is `0.4746`

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