Intuition concerning Riemann Sums

512 Views Asked by At

I have just started learning integrals, and I want to know the following:

In the definition of a riemann integral, it states that the interval that the integral is to be evaluated, is partitioned into smaller intervals. My question is that, how do we know, regardless of the partitions we choose, that the limit will converge to the same value? In other words, how do we know the partitions are independent of the area limit?

1

There are 1 best solutions below

7
On BEST ANSWER

In short, we don't know that. We can give weird functions that don't have that property. We call those functions "nonintegrable", because they can't be integrated in this way. (Perhaps one might call them non-Riemann integrable, as there are other approaches to integration).

On the other hand, if the limit as the partitions become finer exists, then we say the function is "integrable" (Riemann integrable). A good question to have now is to try to decide which functions are integrable and which aren't. One of the first things you should prove (if your course cares about proofs) or learn (otherwise) is that continuous functions are integrable.

A good parallel might be to think about derivatives. How do we know a priori that, as we let $h \to 0$ (in the $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ definition) that the derivative is well-defined? How do we know it's independent of sequences of decreasing $h$ values? We don't! But if the limit exists, we call the function differentiable there.


For more information, check out these MSE questions.

How to prove that continuous functions are Riemann-integrable?

Proof that monotone functions are integrable with the classical definition of the Riemann Integral

And you ask why Riemann integration is sometimes defined as lower and upper bounds of partitions, and sometimes on limits of partitions. You should try to prove that these are the same definition, just parsing the definition of limits. In particular, show that they imply each other.