Does the average of the lower estimate, $L_n$, and upper estimate, $R_n$, of an integral give the same or very similar approximation of an integral when using the Midpoint Rule or is there no relevant connection?
Where $f$ is increasing
$$L_n = \sum_{i=1}^n f(x_{i-1}) \Delta x$$
And
$$R_n = \sum_{i=1}^n f(x_{i}) \Delta x$$
Does
$$\sum_{i=1}^n f(\frac{x_{i-1} + x_{i}} {2}) \Delta x = \frac{L_n + R_n}{2}$$?
Best Answer
It's not true that $$f\left(\frac{x_{i-1} + x_i}{2}\right) = \frac{f(x_{i-1}) + f(x_i)}{2}$$ unless $f$ is linear. So your last equation is not true:
$$M_n := \sum_{i=1}^n f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x_i \neq \frac{L_n + R_n}{2}$$
In general, it's not even true that $M_n$ will be between $L_n$ and $R_n$. If the function is increasing, then you know that $L_n \leq M_n \leq R_n$.
Also, this is not the lower or upper integral. This the left end point and right end point Riemann sum. In general, one cannot say which of $L_n$, $R_n$, or $M_n$ gives a better approximation. However, if $f$ is integrable, then you know that
$$\lim_{n\to \infty} L_n = \lim_{n\to \infty} M_n = \lim_{n\to \infty} R_n = \int_a^b f$$