Verification of a basic Riemann summation problem: $f(x) = 1+x$ where $x\in [-1, 2]$ This is the very first time I solve a problem involving Riemann's summation so I would like to verify whether I get it correctly. Problem statement (hopefully, I have translated the problem statement in a comprehensible way.):

Find an integral sum $\sigma_n$ for a function $f(x) = 1+x$ on the interval $x\in [-1, 2]$. Choose points so that they are in the middle of each partition.

I've started by considering the interval. Suppose that we split the interval into $n$ parts. That means that each interval is of length:
$$
\Delta x = {3\over n}
$$
The interval begins at $x = -1$, so we might find the points $\zeta_k$ lying in the middle of each interval. So that means:
$$
\zeta_k = -1 + \frac{3k + 3(k-1)}{2n}
$$
Here $-1$ is the beginning of the range of $x$, and $\frac{3k + 3(k-1)}{2n}$ is chosen because we need a center point in each split. Here is a sandbox for the above.
Now by we want to find a sum:
$$
\begin{align}
\sigma_n &= \sum_{k=1}^nf(\zeta_k)\Delta x \\
&= \sum_{k=1}^n \left(1 + \frac{3k + 3(k-1)}{2n} -1\right)\Delta x \\
&= \sum_{k=1}^n \frac{3k + 3(k-1)}{2n}\Delta x \\
&= \sum_{k=1}^n \frac{3k + 3(k-1)}{2n}\cdot {3\over n} \\
&= {9\over 2n^2}\sum_{k=1}^n (2k-1) \\
&= {9\over 2n^2}\left({2n(n+1)\over 2}-n\right)\\
&= {9\over 2n^2}n^2 = {9\over 2}
\end{align}
$$
Here the sum yields a value. But as far as I understood if $\sigma_n$ is a function of $n$ then we have to take the limit of $\sigma_n$ as $n\to\infty$.
Does my writing make sense at all?

Your work looks correct.  In your particular example, $\sigma_n$ is independent of $n$, but if you had taken points on the left (or on the right) side of each partition interval, instead of in the middle, or if you had a non-linear function such as $f(x)=1+x^2$, then only in the limit as $n\to\infty$ would the Riemann sum give the exact value for the integral.  It is worth trying those as other exercises to see what I am talking about.
