Geometric representation of Euler-Maclaurin Summation Formula

When reading Tom Apostol's expository article (or the free link), I was expecting more diagrams to come that follow the figure below (or this from the Wolfram project). It was a disappointment not seeing further decomposition of the "curved triangular pieces". The same goes for the textbooks and other materials I've read. My Question:

What is the correct analysis to demonstrate the 'geometry' of the successive orders of the Euler-Maclaurin formula?

Is there any existing source with such diagrams (visualization) similar to the above but of higher order?

I'm actually not sure whether my proposal is a valid idea (that such a demonstration is possible). Below is a tentative description of what I mean by the decomposition of pieces as the terms of Euler-Maclaurin formula.

Basically it is approximating the definite integral $F(b)-F(a)$ where $b>a$, unfolding in the direction "opposite" to Euler-Maclaurin formula itself. $$\int_{a}^b f(x) \,\mathrm{d}x \approx F_0 + F_1 + F_2 + \cdots$$ The approximation is done by columns of unit width (summation of $f(k)$, discretely sampled points) plus corrections to get close to the curve (via derivatives of end points). $$\int_{a}^b f(x) \,\mathrm{d}x \approx \overbrace{\sum_{k = a}^b f(k) - \frac{f(a)+f(b)}2 }^{ \text{unit columns centered} } - \underbrace{ \frac{f'(b) - f'(a)}{12} - 0 }_{ \textstyle\genfrac{}{}{0pt}{}{\text{net result of} }{ \text{triangular & parabolic top} } } - \overbrace{ \frac{f'''(b) - f'''(a)}{720} - 0}^{ \text{cubic & guartic top} } - \cdots$$

The correct analysis would have to account for both the emergence of Bernoulli numbers and the individual orders (or correction)... and the remainder if possible.

For simplicity let $a,b \in \mathbb{N}$ and the summation is in unit steps as usual, and pretend that things will scale nicely.

1. The zero-th order approx for the integral is the columns of unit width $$F_0 = \sum_{k=a}^b f(k)$$ Note that each column of height $f(k)$ is left-aligned. For example, the first column for $k = a$ occupies $x \in [a, a+1]$. This means the last column at $k = b$ is entirely outside of the range of integration.
2. The 2nd term (still zero-th order) shifts all the columns by $\frac12$ to make them centered. That is, each columm of height $f(k)$ now sits at $x \in [a-\frac12, a+\frac12]$. The last column is now half-outside, and so is the first column. Thus we remove half of each of them. $$F_1 = -\frac{f(a) + f(b)}2$$
3. The 3rd term (1st order, linear) correction is supposed to be shaving off a triangular top from each column (note that the outermost are half-columns at $k = b$ and $k = a$). The width of each triangle is $f'(k)$ and the width is unity. $\color{brown}{\textit{Somehow}}$ there's a telescoping of terms along with cancellation with the 3rd order correction and we end up with $$F_2 = - \frac16 \frac{f'(b) - f'(a)}2$$
4. The 4th term (2nd order, parabolic) correction is supposed to be something involving both $f''(k)$ and $f'(k)$. It should be a parabolic piece on top of the triangular (linear) correction of the previous order, similar to what one sees in the quadrature of parabola. Note that there's a fixed ratio of $\frac13$ (or $\frac23$) between the parabolic area and the rectangle it spans. $\color{brown}{\textit{Somehow}}$ there's telescoping and cancellation with the 2nd as well as the 4th order so that $$F_3 = 0$$
5. The 5th term (3rd order, cubic) is supposed to some kind of moon crest on top of the parabolic correction, and $\color{brown}{\textit{somehow}}$ after cancellation it will end up being $$F_4 = -\frac{f'''(b) - f'''(a)}{720}$$
6. So on and so forth ....

Anyway, in the end, one keep the summation $F_0$ on one side of the equation by itself, removing the terms $F_1$, $F_2$, $F_3$, etc to the side of the integral, as seen earlier. $$F_0 \approx \int_{a}^b f(x) \,\mathrm{d}x - F_1 - F_2 - \cdots$$