Let $a, b \in \mathbb{R}, a<b$. Suppose that $f$ is continuous and strictly decreasing over $[a, b]$. Let $L_n$ be the left Riemann sum, $R_n$ the right Riemann sum, and $M_n$ the midpoint (Riemann) sum for $f$ on $[a, b]$ with $n ∈ \mathbb{Z}^{+}$ subintervals, respectively.
(b) Prove that $R_n < M_n < L_n$
I know how we can prove this by using a graph, but I don't know how to prove this numerically. do we need to factor out 1/n ??
I think this is immediate, since $f$ is decreasing so for any $\xi_i\in (x_{i-1}, x_i)$ we then have $f(x_i)<f(\xi_i)<f(x_{i-1})$ then $$ R_n=\sum_{i=1}^nf(x_i)(x_i-x_{i-1})<M_n=\sum_{i=1}^nf(\xi_i)(x_i-x_{i-1})<L_n=\sum_{i=1}^nf(x_{i-1})(x_i-x_{i-1})$$ it is clear now? Since $f$ is Riemann integrable, all of this sums converges to $\int_a^b fdx$.