$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\la{\lambda}$If $f$ is an arbitrary Lebesgue-integrable function, then, as is done in a definition of the Lebesgue integral, it makes sense to group together close values of the function $f$, rather than close values of its argument.
So, say, for any Lebesgue-integrable function $f\colon[0,1]\to\R$, any real $a$, and any real $h>0$,
$$\Big|\int_0^1 f(x)\,dx-\sum_{k\in\Z}(a+k+1/2)h\,\la(E_{f,a,h,k})\Big|\le\frac h2,$$
where $E_{f,a,h,k}:=f^{-1}((a+kh,a+(k+1)h])$ and
$\la$ is the Lebesgue measure.
Of course, if $f$ is monotonic, then the sets $E_{f,a,h,k}$ will be intervals. In this case, the bound $\frac h2$ will be just as effective as the Koksma bound $V(f)D_n^*$ in Gerry Myerson's answer (provided the set $\{x_1,\dots,x_n\}$ is of the minimal discrepancy, $\frac1{2n}$ -- see Theorem 1.4 on p. 91).
If $f$ is not monotonic (think e.g. of $f(x)=\sin mx$ for large $m$), then the bound $\frac h2$ can be much better than the Koksma bound $V(f)D_n^*$.
As for the calculation of the variation $V(f)$ (when it is finite, even when $f$ is not monotonic), it seems to be of about the same complexity as the calculation of the $\la(E_{f,a,h,k})$'s.
Also, it seems that the bound $\frac h2$ will be more effective, or significantly more effective, than the bound $M(D_n^*)$ in Gerry Myerson's answer, even when the set $\{x_1,\dots,x_n\}$ is of the minimal discrepancy. For instance, when $f(x)=x^p$ for a real $p>0$ and all $x\in[0,1]$, $a=0$, and $h=1/n$, then the bound $\frac h2$ is $\frac1{2n}$, whereas the bound $M(D_n^*)$ is $\sim \frac p{2n}$ if $p\ge1$ and $=\frac1{(2n)^p}$ if $p\in(0,1]$ (again, provided the set $\{x_1,\dots,x_n\}$ is of the minimal discrepancy, $\frac1{2n}$).