Riemann sums and its tie to Gauss' trick for adding consecutive integers? I was thinking of how $\int x\ dx = \frac{x^2}{2} + C$ and thought about how I think it has to be tied to the idea of adding up consecutive integers. I then referred back to Gauss' trick, which says the sum of consecutive integers to $n$ is $\frac{n(n+1)}{2}$. Distribution of the $n$ term will be $\frac{n^2}{2} + \frac{n}{2}$. The first fraction seems very close to $\frac{x^2}{2}$ but I can't figure out its exact tie. I started thinking about Left Riemann Sum with rectangle width $1$, and that gives the same value as Gauss' trick. But what happens with $n > 10$ rectangles? How does the $\frac{n}{2}$ term vanish as we take the limit of the Riemann Sum as the # of rectangles goes to infinity? Or is this a coincidence?

Note that the Riemann sum is not the sum of the heights of rectangles.  Rather, it is the sum of all their areas.  We have
$$
\int_0^x t \,dt  \approx \sum_{i=0}^{n-1} f(x^*_i) \Delta x = 
\sum_{i=0}^{n-1} (ix/n) \frac{x}{n} = \frac{x^2}{n^2} \cdot \left(\frac{n(n-1)}{2} \right)
$$
Now, take this limit as $n \to \infty$.
