As you observed, the midpoint method is typically more accurate than the trapezoidal method. This is suggested by the composite error bounds, but they don't rule out the possibility that the trapezoidal method might be more accurate in some cases.
We can get a better understanding by examining the local errors for single-segment rules. Consider an interval $[a,b]$ and define interval length $h = b-a$ and midpoint $c = (a+b)/2.$ Note that $b-c = c-a = (b-a)/2 = h/2.$
The midpoint error is
$$E_M = f(c)h - \int_a^b f(x) \, dx = \int_a^b [f(c) - f(x)] \, dx.$$
Using a second-order Taylor approximation,
$$f(c) = f(x) + f'(x)(c-x) + \frac{1}{2} f''(\xi_x)(x-c)^2, $$
we see
$$E_M = -\int_a^b f'(x)(x-c) \, dx + \frac{1}{2}\int_a^b f''(\xi_x)(x-c)^2 \, dx.$$
Applying integration by parts to the first integral on the RHS we get
$$\int_a^b f'(x)(x-c) = \left.(x-c)f(x)\right|_a^b - \int_a^b f(x) \, dx = \frac{h}{2}[f(a) + f(b)] - \int_a^b f(x) \, dx .$$
Note that this result gives us the error $E_T$ for the trapezoidal method.
Hence,
$$E_M = -E_T + \frac{1}{2}\int_a^b f''(\xi_x)(x-c)^2 \, dx.$$
It is actually not that easy to find examples where $|E_T| < |E_M|$. Using the above result, we can surmise that this could happen when the midpoint method overestimates, $E_M > 0$, the trapezoidal method underestimates $E_T < 0,$ and we have high curvature in a small neighborhood of a point in the interval.
Here is a somewhat contrived example.
Consider the following function that meets those requirements. Note that the second derivative is piecewise continuous but bounded -- which does not degrade the composite $O(n^{-2})$ accuracy.
$$f(x) = \begin{cases} 0.25 + 0.75\exp(-200 x^2), &\mbox{if } -1 \leqslant x \leqslant 0 \\ 0.99 + 0.01 \cos(\pi x), &\mbox{if } \,\,\,\,\,\,\, 0 < x \leqslant 1 \end{cases}.$$
Then
$$\begin{align}\int_{-1}^1 f(x) \, dx &\approx 1.2870\\ M &\approx 2 \\ T &\approx 1.2300 \\ E_M &\approx 0.7130 \\ E_T &\approx 0.0570 \end{align}$$
In this context "Cauchy integral" has the meaning you know.
It is a fact that if a function is bounded and Cauchy integrable over $[a,b]$, then it is also Riemann integrable over that interval.
It seems that there is no elementary proof of this theorem.
The proof in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505, (theorem 1), could be considered elementary because plays only with Riemann sums but is an indigestible game.
Note that there exist unbounded functions Cauchy integrable.
Also the use of regular partitions is enough to define Riemann integral.
See Jingcheng Tong Partitions of the interval in the definition of Riemann integral, Int. Journal of Math. Educ. in Sc. and Tech. 32 (2001), 788-793 (theorem 2).
I repeat that the use of only left (or right) Riemann sums with only regular partitions doesn't work.
Best Answer
There is no "general" formula to get the result of a Riemann sum. However,