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. 

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, $L_{1200} \leq L_{20}$.
