Question

What are the disadvantages of the Simpson’s 3/8 rule compared with the
Simpson’s 1/3 rule?

Answer 1

So, I was studying Numerical Integration from Hildebrand where I came across Simpson's Rule and 3/8 rule. Now, Simpson's rule has an error of order 5 and degree of precision 3, which is same as 3/8 rule. The difference lies in the Lagrange polynomials we use to interpolate.

In 3/8 rule we need an extra computation point, hence I feel it's computationally insufficient as compared to Simpson's rule. Or if we restate Simpson's rule is better. My question lies, does 3/8 rule have any advantage?

Simpson's Rule: ∫baf(x)dx≈b−a6[f(a)+4f(a+b2)+f(b)] with error as 190(b−a2)5|f4(z)|

3/8 Rule: ∫baf(x)dx≈(b−a)8[f(a)+3f(2a+b3)+3f(a+2b3)+f(b)] with error as ∣∣(b−a)56480f4(z)∣∣

PLEASE MARK ME AS THE BRAINIEST...