calculusdefinite integralsintegrationnumerical methods
If we compute the exact value of $\int_1^2\frac1x\,dx$ we get $\ln2=0.693147\dots$ If we use the trapezoidal rule with $10$ intervals we get $0.693771$, and the midpoint rule with $10$ intervals gives $0.692835$.
Here it seems as if the trapezoidal rule is more accurate than the midpoint rule, even though we are told that the absolute value of the midpoint rule's error is half that of the trapezoidal rule, so the midpoint rule should be more accurate. Why?
Simpson's Rule is exact for polynomials of degree 3 or less, so the error term can't depend on any derivative less than the 4th. Put another way: the 3rd derivative of a 3rd degree polynomial is nonzero, so it can't be involved in the error term for Simpson, since the error for 3rd degree polynomials is zero.
You should be careful with this expression:
$$ {\rm err} = -\frac{b-a}{12}h^2f''(\mu) \tag{1} $$
The meaning is: there is a point $\mu \in (a,b)$ such that the error is given by this expression. To show this is true I calculate $S(h)$ for various values of $h$ and the absolute error $\epsilon$. I then find the value of $\mu$ guaranteed by Eq. (1), that is, the value of $\mu$ such that ${\rm err} = \epsilon$
Best Answer
The error bounds for classical numerical integration rules constrain nothing about the actual errors incurred when applying them to any function. It may be that the trapezoidal rule produces an exact result for a certain number of intervals while the midpoint rule does not, it may be the other way around, etc. – anything is possible when comparing the actual errors.
Instead the error bounds only give the asymptotic behaviour of the error as the number of intervals increases.