Question

Find the value of the integral `int_0^1 x^2/(1+x^3`𝒅𝒙 using Simpson’s (𝟑/𝟖)^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 (𝟑/𝟖)^𝒕𝒉 rule :

`"I"=(3h)/8[X+2T+3R]`  -------------(3)

𝑿=𝒔𝒖𝒎 𝒐𝒇 𝒆𝒙𝒕𝒓𝒆𝒎𝒆 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔=𝒚_𝟎+𝒚_𝟒=𝟎+𝟎.𝟓=𝟎.𝟓
𝑻=𝒔𝒖𝒎 𝒐𝒇 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒆 𝒐𝒇 𝒕𝒉𝒓𝒆𝒆 𝒃𝒂𝒔𝒆 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔= 𝒚_𝟑=𝟎.𝟑𝟗𝟓𝟔𝟎
𝑹= 𝒔𝒖𝒎 𝒐𝒇 𝒓𝒆𝒎𝒂𝒊𝒏𝒊𝒏𝒈 𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔 = 𝒚_𝟏+𝒚_𝟐=𝟎.𝟎𝟔𝟏𝟓𝟑+𝟎.𝟐𝟐𝟐𝟐=𝟎.𝟐𝟖𝟑𝟕𝟑

`"I"=(3xx0.25)/8(0.5+2xx0.39560+3xx0.28373)`

∴ I = 0.2008