Identifying left- and right-Riemann sums of $\int_9^{14}e^{-x^4}\ dx$ 
My attempt: 

Relooking at it, I think $L_{20}$ would be the highest, so like
$R_{1200} < L_{1200} < L_{20}$, but I have no way to justify it, any help is appreciated. 
 A: No matter what $n$ and $m$ are, $R_n<L_m$ based on your knowledge that $R_n<A$ and $A<L_n$. So $R_{1200}$ should be the smallest of the three: $0.33575$
Now both $L_{20}$ and $L_{1200}$ overestimate the value of $A$. Informally, $L_{1200}$ is closer to $A$, because $A=\lim_{n\to\infty}L_{n}$. The function is not particularly weird enough for $L_{1200}$ to break the downward trend of $L_n$ towards $A$ as $n\to\infty$. So this much understanding suggests $L_{1200}$ is the smaller of the remaining two numbers.
A little more formally, $L_{20}$ is the area of a certain $20$ rectangles. And $L_{1200}$ is the area of a certain $1200$ rectangles. Since $20$ divides into $1200$, we can in fact place sets of $60$ of the rectangles from $L_{1200}$ inside each rectangle from $L_20$. Since the function is deacreasing, the $60$ rectangles will fit inside the one rectangle with room to spare. So again, $L_{1200}$ should be less than $L_{20}$.
A: The main ideas here are that, for a decreasing function,


*

*Estimating using left endpoints (i.e., $L_{20}, L_{1200}$) will always overestimate the true value of the integral. 

*Estimating using right endpoints will always underestimate the value of the integral.
These two tell you that $R_m < I < L_n$ for any integers $m, n$.


*

*Using more subintervals (i.e., larger $n$) will always produces a better estimate -- they'll produce estimates closer to the actual value of the integral.


So, if $m < n$, we have
$$R_m < \underbrace{R_n < I < L_n}_{\text{more subintervals}} < L_m$$
which agrees exactly with what you have! (in the body of the question, not the image)
A: You're last thought is correct. One rigorous justification goes as follows:
$\displaystyle L_{20} = \sum_{i=0}^{19}\frac{5}{20}f\left(9+\frac{5i}{20}\right)$, and
$\displaystyle L_{1200} = \sum_{i=0}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.
Now, we are going to split up the $L_{1200}$ sum into groups of 60. In particular,
$\displaystyle L_{1200} = \sum_{i=0}^{59}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right) + \sum_{i=60}^{119}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right) + \ldots + \sum_{i=1140}^{1199}\frac{5}{1200}f\left(9+\frac{5i}{1200}\right)$.
Now, by monotonicity of the function $f$ we have that this sum is less than
$\displaystyle\sum_{i=0}^{59}\frac{5}{1200}f\left(9\right) + \sum_{i=60}^{119}\frac{5}{1200}f\left(9+\frac{5\cdot 60}{1200}\right) + \ldots + \sum_{i=1140}^{1199}\frac{5}{1200}f\left(9+\frac{5\cdot 1140}{1200}\right) \\
=60\cdot\frac{5}{1200}f\left(9\right) + 60\cdot\frac{5}{1200}f\left(9 + \frac{5}{20}\right) + \ldots + 60\cdot\frac{5}{1200}f\left(9 + \frac{5\cdot 19}{20}\right) =: L_{20}$.
Therefore, $L_{1200} \leq L_{20}$.
