2. Find Solution of an equation 1/(x+1) using Trapezoidal rule
x1 = 0 and x2 = 1
Interval N = 5

Solution:
Equation is `f(x)=(1)/(x+1)`.

`h = (b-a)/N`

`h = (1 - 0) / 5 = 0.2`

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

x	0	0.2	0.4	0.6	0.8	1
y	1	0.8333	0.7143	0.625	0.5556	0.5

Method-1:
Using Trapezoidal Rule
`int y dx=h/2 [y_0+y_5+2(y_1+y_2+y_3+y_4)]`

`int y dx=0.2/2 [1 +0.5 + 2xx(0.8333+0.7143+0.625+0.5556)]`

`int y dx=0.2/2 [1 +0.5 + 2xx(2.7282)]`

`int y dx=0.2/2 [1 +0.5 + 5.4563]`

`int y dx=0.6956`

Solution by Trapezoidal Rule is `0.6956`

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Method-2:
Using Trapezoidal Rule
`int y dx=h/2 [y_0+2y_1+2y_2+2y_3+2y_4+y_5]`

`y_0=1`

`2y_1=2*0.8333=1.6667`

`2y_2=2*0.7143=1.4286`

`2y_3=2*0.625=1.25`

`2y_4=2*0.5556=1.1111`

`y_5=0.5`

`int y dx=0.2/2*(1+1.6667+1.4286+1.25+1.1111+0.5)`

`int y dx=0.2/2*(6.9563)`

`int y dx=0.6956`

Solution by Trapezoidal Rule is `0.6956`

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