I am trying to prove the following lemma for the Simpson's rule.
Suppose $g\in C^4([-1,1],\mathbb{R})$. Then $$\bigg|\int_{-1}^1g(x)dx-\frac{g(-1)+4g(0)+g(1)}{3}\bigg|\leq\frac{1}{90}\|g^{(4)}\|_\infty.$$
Here $\|\!\cdot\!\|_\infty$ is the supremum norm defined by $\|f\|_\infty:=\sup_{x\in[-1,1]}|f(x)|$.
My attempt: Applying the Euler–Maclaurin sum formula, we get $$\int_0^1g(x)dx-\frac{2g(0)+g(1)}{3}=\frac{g(1)-g(0)}{6}-\frac{g'(1)-g'(0)}{12}-\frac{1}{6}\int_0^1B_3(x)g^{(3)}(x)dx,$$ where $B_3(x):=x^3-3x^2/2+x/2$ is the third Bernoulli polynomial.
Now integrate by parts to get $$\int_0^1B_3(x)g^{(3)}(x)dx=\frac{x^2(x-1)^2}{4}g^{(3)}(x)\bigg|_0^1-\int_0^1\frac{x^2(x-1)^2}{4}g^{(4)}(x)dx.$$ The first term equals $0$, while for the second we have, by the mean value theorem for integrals, $$\int_0^1\frac{x^2(x-1)^2}{4}g^{(4)}(x)dx=\bigg(\int_0^1\frac{x^2(x-1)^2}{4}dx\bigg)g^{(4)}(\xi)=\frac{1}{120}g^{(4)}(\xi)$$ for some $\xi\in[0,1]$.
The problem is how to estimate the other two terms. If this approach were to work, then we must show that $$\bigg|\frac{g(1)-g(0)}{6}-\frac{g'(1)-g'(0)}{12}\bigg|\leq\frac{1}{240}\|g^{(4)}\|_\infty.$$ But I don't know how to relate these to the fourth derivative. I tried Lagrange's mean value theorem but got stuck. How should I proceed?
This appears as an exercise VI.6.7. from Analysis II by Amann and Escher, where the whole section is devoted to applications of the Euler–Maclaurin sum formula. So if you have other proofs, please make use of this formula. Thanks in advance!