If given $ \int\limits_a^b f \left( x \right) \mathrm{d}x $
and the function contains both intervals of concavity up and concavity down within the limits of integration $[a,b]$, and given that:
on an interval of concavity down $[r,s]$
$ Trapezoid(n) < \int\limits_r^s f \left( x \right) \mathrm{d}x < Midpoint(n) $,
on an interval of concavity up $[p,q]$
$ Midpoint(n) < \int\limits_p^q f \left( x \right) \mathrm{d}x < Trapezoid(n) $,
could we generalize that:
$ Midpoint(n) <\int\limits_a^b f \left( x \right) \mathrm{d}x < Trapezoid(n) $ or
$ Trapezoid(n) < \int\limits_a^b f \left( x \right) \mathrm{d}x< Midpoint(n) $
?
It is certainly false that the integral must lie between the midpoint and trapezoid approximations for every $n$. For example, take the function $\cos x$ on $[0,4\pi]$ with $n=1$: the two approximations are both $4\pi$, but the integral equals $0$. (If you don't like $n=1$, take $\cos mx$ for any positive integer $m$ and $n=m$ - same result.)