2. Find Solution using Trapezoidal rule

x	f(x)
0.0	1.0000
0.1	0.9975
0.2	0.9900
0.3	0.9776
0.4	0.8604

Solution:
The value of table for `x` and `y`

x	0	0.1	0.2	0.3	0.4
y	1	0.9975	0.99	0.9776	0.8604

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

`int y dx=0.1/2 [1 +0.8604 + 2xx(0.9975+0.99+0.9776)]`

`int y dx=0.1/2 [1 +0.8604 + 2xx(2.9651)]`

`int y dx=0.1/2 [1 +0.8604 + 5.9302]`

`int y dx=0.3895`

Solution by Trapezoidal Rule is `0.3895`

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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*0.9975=1.995`

`2y_2=2*0.99=1.98`

`2y_3=2*0.9776=1.9552`

`y_4=0.8604`

`int y dx=0.1/2*(1+1.995+1.98+1.9552+0.8604)`

`int y dx=0.1/2*(7.7906)`

`int y dx=0.3895`

Solution by Trapezoidal Rule is `0.3895`

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