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. 

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 $L_{20}$.
