Question

Find the value of the integral `int_0^1 x^2/(1+x^3`𝒅𝒙 using Simpson’s (1/3)^𝒕𝒉 rule.

Answer 1

Let I = `int_0^1 x^2/(1+x^3)dx`

a=0 , b=1
Dividing limits into 4 parts i.e n = 4
`thereforeh=(b-a)/n=1/4=0.25`

𝒙𝟎=0	𝒙𝟏=0.25	𝒙𝟐=0.50	𝒙𝟑=0.75	𝒙𝟒=1.0
𝒚𝟎=0	𝒚𝟏=0.06153	𝒚𝟐=0.2222	𝒚𝟑=0.39560	𝒚𝟒=𝟎.𝟓

Simpson’s (𝟏/𝟑)^𝒓𝒅 rule :

`"I"=h/3[X+2E+40]`  --------------(2)

𝑿=𝒔𝒖𝒎 𝒐𝒇 𝒆𝒙𝒕𝒓𝒆𝒎𝒆 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔=𝒚_𝟎+𝒚_𝟒=𝟎+𝟎.𝟓=𝟎.𝟓
𝑬=𝒔𝒖𝒎 𝒐𝒇 𝒆𝒗𝒆𝒏 𝒃𝒂𝒔𝒆 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔= 𝒚_𝟐=𝟎.𝟐𝟐𝟐𝟐
𝑶=𝒔𝒖𝒎 𝒐𝒇 𝒐𝒅𝒅 𝒃𝒂𝒔𝒆 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔= 𝒚_𝟏+𝒚_𝟑=𝟎.𝟎𝟔𝟏𝟓𝟑+𝟎.𝟑𝟗𝟓𝟔𝟎=𝟎.𝟒𝟓𝟕𝟏𝟑

`"I"=(0.25)/3`(𝟎.𝟓+𝟐×𝟎.𝟐𝟐𝟐𝟐+𝟒×𝟎.𝟒𝟓𝟕𝟏𝟑) ……………(from 2)

∴ I = 0.23108