Why do the infinitely many infinitesimal errors from each term of an infinite Riemann sum still add up to only an infinitesimal error? Ok, so after extensive research on the topic of how we deal with the idea of an infinitesimal amount of error, I learned about the standard part function as a way to deal with discarding this infinitesimal difference $\Delta x$ by rounding off to the nearest real number, which is zero. I've never taken nonstandard analysis before, but here's my question.
When you take a Riemann sum, you are approximating an area by rectangles, and each of those rectangles has an error in approximating the actual area under the curve for the corresponding part of the graph.  As $\Delta x$ becomes infinitesimal, the width of these rectangles becomes infinitesimal, so each error becomes infinitesimal.  But since there are infinitely many rectangles in that case, why is it that the total error from all of them still infinitesimal? In other words, shouldn't an infinite amount of infinitesimals add up to a significant amount?
 A: If I've understood the question correctly, here is a heuristic explanation.  Note that this is not rigorous, since to make your question rigorous you have to give some precise definition of what you mean by "the exact area", which is not at all easy to define in general.
Let us assume we are integrating a continuous function $f(x)$ from $0$ to $1$ by using a Riemann sum with infinitesimal increment $\Delta x$.  Let us also assume for simplicity that $f$ is increasing (the general case works out essentially the same way but is a little more complicated to talk about).  So we are approximating "the area under $f$" by replacing the region under the graph of $f$ from $x=c$ to $x=c+\Delta x$ by a rectangle of height $f(c)$, for $1/\Delta x$ different values of $c$.  Now since $f$ is increasing, the difference between our rectangle of height $f(c)$ and the actual area under the graph of $f$ from $c$ to $c+\Delta x$ is at most $\Delta x(f(c+\Delta x)-f(c))$.  But since $f$ is (uniformly) continuous, $f(c+\Delta x)-f(c)$ is infinitesimal.  So our error is an infinitesimal quantity times $\Delta x$.
So although we are adding up $1/\Delta x$ (an infinite number) different errors to get the total error, each individual error is not just infinitesimal but infinitesimally smaller than $\Delta x$.  So it is reasonable to expect that the sum of all of the errors is still infinitesimal.
A: Let's compute!
Suppose we integrate $\int_0^1 x \, \mathrm{d}x$ by pick a Riemann sum with $H$ equally sized intervals.
The smallest and largest value of $x$ in the $i$-th interval is $\frac{i-1}{H}$ and $\frac{i}{H}$ respectively, and that interval would contribute $\frac{i-1}{H} \cdot \frac{1}{H}$ and $\frac{i}{H} \cdot \frac{1}{H}$ respectively to the total sum, so the maximum possible error from ths interval is $\frac{1}{H^2}$.
However, there are only $H$ intervals — the error therefore cannot be any larger than $\frac{1}{H^2} \cdot H = \frac{1}{H}$, an infinitesimal.
And for fun,we compute the integral using the upper bounds:
$$ \begin{align*}
\int_0^1 x \, \mathrm{d}x
&= \text{std} \sum_{i=1}^H \left( \frac{i}{H} \right) \cdot \frac{1}{H}
\\&= \text{std} \left( \frac{1}{H^2} \sum_{i=1}^H i \right)
\\&= \text{std} \left( \frac{1}{H^2} \frac{H(H+1)}{2} \right)
\\&= \text{std} \left( \frac{1}{2} + \frac{1}{2H} \right)
\\&= \frac{1}{2}
\end{align*}
$$
You can see how the errors of 
