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For convenience let the interval be $[0,1]$, and let's look at the case $f(t) = t^p$. We have $E_n(t^p) = 0$ for $p \le 1$, while by Faulhaber's formula we have $$\eqalign{E_n(t^p) &= \dfrac{1}{p+1} + \dfrac{1}{2n} - \dfrac{1}{n^{p+1}} \sum_{i=1}^{n} i^p\cr &= - \sum_{j=0}^{p-2} {p \choose j} \dfrac{B_{p-j}}{j+1} n^{j-p} }$$ where $B_{p-j}$ is a Bernoulli number. For any polynomial $f$, $E_n(f)$ is a linear combination of these, and for its asymptotic behaviour as $n \to \infty$ we take the least negative power of $n$ whose coefficient does not cancel out: $E_n(f) \sim c n^{-k}$ for some nonzero constant $c$ and some integer $k \ge 2$. In particular its absolute value is monotonically decreasing for sufficiently large $n$.