Setting $h=(b-a)/N$ where $b=1$, $a=0$, we can then rewrite the above integral as $$\begin{equation*} I=\int_0^1 f(x)dx\approx \frac{1}{N}\sum_{i=1}^Nf(x_{i-1/2}), \end{equation*}$$ where $x_{i-1/2}$ are the midpoint values of $x$. Introducing the concept of the average of the function $f$ for a given PDF $p(x)$ as $$\begin{equation*} \langle f \rangle = \sum_{i=1}^Nf(x_i)p(x_i), \end{equation*}$$ and identify $p(x)$ with the uniform distribution, viz. $p(x)=1$ when $x\in [0,1]$ and zero for all other values of $x$. The integral is is then the average of $f$ over the interval $x \in [0,1]$ $$\begin{equation*} I=\int_0^1 f(x)dx\approx \langle f \rangle. \end{equation*}$$