3. Find Solution using Trapezoidal 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 `y`

x	0	0.25	0.5	0.75	1
y	1	0.9896	0.9589	0.9089	0.8415

Method-1:
Using Trapezoidal Rule
`int y dx=h/2 [y_0+y_4+2(y_1+y_2+y_3)]`

`int y dx=0.25/2 [1 +0.8415 + 2xx(0.9896+0.9589+0.9089)]`

`int y dx=0.25/2 [1 +0.8415 + 2xx(2.8574)]`

`int y dx=0.25/2 [1 +0.8415 + 5.7148]`

`int y dx=0.9445`

Solution by Trapezoidal Rule is `0.9445`

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Method-2:
Using Trapezoidal Rule
`int y dx=h/2 [y_0+2y_1+2y_2+2y_3+y_4]`

`y_0=1`

`2y_1=2*0.9896=1.9792`

`2y_2=2*0.9589=1.9178`

`2y_3=2*0.9089=1.8178`

`y_4=0.8415`

`int y dx=0.25/2*(1+1.9792+1.9178+1.8178+0.8415)`

`int y dx=0.25/2*(7.5563)`

`int y dx=0.9445`

Solution by Trapezoidal Rule is `0.9445`

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