Formula

1. Trapezoidal Rule `int y dx = h/2 (y_0 + 2 (y_1 + y_2 + y_3 + ... + y_(n-1)) + y_n)` `int f(x) dx=(Delta x )/2 (f(x_(0))+2(f(x_(1))+f(x_(2))+f(x_(3))+...+f(x_(n-1)))+f(x_(n)))`

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Examples
1. Find the approximated integral value using Trapezoidal 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 Trapezoidal Rule


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

`f(x_(0))=4.0552`

`2f(x_(1))=2*4.953=9.906`

`2f(x_(2))=2*6.0436=12.0872`

`2f(x_(3))=2*7.3891=14.7782`

`f(x_(4))=9.025`

`int f(x) dx=0.2/2*(4.0552+9.906+12.0872+14.7782+9.025)`

`=0.2/2*(49.8516)`

`=4.9852`

Solution by Trapezoidal Rule is `4.9852`

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Method-2:
Using Trapezoidal Rule


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

`=0.2/2 [4.0552 +9.025 + 2xx(4.953+6.0436+7.3891)]`

`=0.2/2 [4.0552 +9.025 + 2xx(18.3857)]`

`=0.2/2 [4.0552 +9.025 + 36.7714]`

`=4.9852`

Solution by Trapezoidal Rule is `4.9852`

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