3. Find Solution of an equation x^3-2x+1 using Trapezoidal rule
x1 = 2 and x2 = 4
Step value (h) = 0.5

Solution:
Equation is `f(x)=x^3-2x+1`.

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

x	2	2.5	3	3.5	4
y	5	11.625	22	36.875	57

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

`int y dx=0.5/2 [5 +57 + 2xx(11.625+22+36.875)]`

`int y dx=0.5/2 [5 +57 + 2xx(70.5)]`

`int y dx=0.5/2 [5 +57 + 141]`

`int y dx=50.75`

Solution by Trapezoidal Rule is `50.75`

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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=5`

`2y_1=2*11.625=23.25`

`2y_2=2*22=44`

`2y_3=2*36.875=73.75`

`y_4=57`

`int y dx=0.5/2*(5+23.25+44+73.75+57)`

`int y dx=0.5/2*(203)`

`int y dx=50.75`

Solution by Trapezoidal Rule is `50.75`

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