Midpoint rule vs trapezoidal rule accuracy? 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?
 A: 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.
A: You seem to be going by the accurate digits, but that is not what the cited claim is about.
Taking the first digits that are different between the results, there is 2835 for the midpoint method, 3147 for the exact value and 3771 for the trapezoidal method. The differences in that order are 312 and 624, the second exact the double of the first, as per the claim.

The accepted answer (@Parcly Taxel) is wrong in its dismissiveness of the error bounds and error estimates, at least in this case.
The single-segment errors for the method are
for the trapezoidal method
$$
\int_a^bf(x)\,dx - \frac{h}{2}(f(a)+f(b))=\int_a^bf[a,b,x](x-a)(x-b)\,dx
\\=-\frac{f''(\xi)}2\frac{(b-a)^3}{6}=-\frac{f''(\xi)(b-a)^3}{12}
$$
and for the midpoint method with $m=\frac{a+b}2$
$$
\int_a^bf(x)\,dx-hf(m)=\int_a^bf[m,x](x-m)\,dx\\
=\int_a^bf[m,x,2m-x](x-m)^2\,dx
\\
=\frac{f''(\xi)}2\frac{2m^3}{3}=\frac{f''(\xi)(b-a)^3}{24},
$$
establishing the factor 2 in the coefficients.
In the composite methods from both remainders one can extract sums $\sum_{k=1}^nf''(\xi_k)h$, $h=\frac{b-a}n$, usually with different midpoints $\xi_k$. These sums can be interpreted as Riemann sums giving the approximate value $f'(b)-f'(a)+O(h)$ or as an arithmetic mean resulting via intermediate-value theorem in $f''(\xi_{ivt})(b-a)$.
Deriving an upper bound from the second interpretation via the supremum norm of the second derivative will indeed somewhat hide the relation between the errors of the methods.
With the first interpretation however one can produce predictions for the errors of the methods. As
$$
(f'(b)-f'(a))=-\frac14+1=\frac34,
$$
the distance from midpoint method to exact value is approximately $\frac34\frac{h^2}{24}=\frac1{100\cdot 32}=0.0003125$ and from exact value to trapezoidal method twice of it, $0.000625$. This again is almost precisely what was obtained from the numerical method.

As another point-of-view, if $T(n)$ is the trapezoidal rule value for $n$ points and $M(n)$ the midpoint rule value, then $T(2n)$ is obtained by including the midpoints with equal weight, so $$T(2n)=\frac12(T(n)+M(n)).$$ The Simpson rule value $S(n)$ for $n$ segments is the Richardson extrapolation of the trapezoidal values,
$$
S(n)=\frac{4T(2n)-T(n)}{3}=\frac{T(n)+2M(n)}{3},
$$
so the better approximation is twice as close to the midpoint rule value than to the trapezoidal rule value.
