According to a wikipedia and mathworld, the trapezoidal rule is:
$$ \int_a^b f(x)\,dx \approx h\left[\frac{f(a) + f(b)}{2} \right], $$
where $h = (b-a)$.
If you apply this rule to a function $f(\mathbf{q})$ of a linear interpolation over an $n$-dimensional vector from $\mathbf{q}_1$ to $\mathbf{q}_2$ (in $\mathbb{R}^n$), i.e., $$ \int_0^1 f(\mathbf{q}_1+x(\mathbf{q}_2-\mathbf{q}_1)) \, dx, $$ does it still hold that $h=(b-a)=1$? Or is it instead in the dimensionality of the interpolation, thus $h=\|\mathbf{q}_2-\mathbf{q}_1\|$? Or is it something else altogether?
$$I=\int_0^1 f({\vec q_1}+x({\vec q_2-\vec q_1}))dx=\frac{f_0({\vec q_1}+x({\vec q_2-\vec q_1}))}{x}|_0^1$$ where $\int f(x)=f_0(x)$ So, $$I=f_0({\vec q_2})-f_0({\vec q_1})=\int_{\vec q_1}^{\vec q_2}f(x)dx\approx g({\vec q_2-\vec q_1})\left[\frac{f({\vec q_1})+f({\vec q_2})}{2}\right]$$ If $f({\vec v})=f(|{\vec v}|)$ or simply if f is a scalar then $g({\vec v})=|v|$ If not then $g({\vec v})=v$