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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?

user166712
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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.

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davidlowryduda
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  • Hi, this is from calculus book: ""It's actually a theorem that if v is a continuous function of t, then the Riemann sum integral is independent of the partitions." So, for continuous functions only, the area converges independent of the partitions? – user166712 Jul 30 '14 at 21:58
  • No, not only for continuous functions. But it is true for continuous functions. It's also true for monotonic functions – davidlowryduda Jul 30 '14 at 22:03
  • Is this a particular theorem? – user166712 Jul 31 '14 at 21:25
  • @user: By a particular theorem, are you asking if it has a name? Not that I know of. Do you want me to provide a proof? – davidlowryduda Jul 31 '14 at 21:26
  • Yes please provide a proof of this. Also, another question: why are integrals defined in different ways? For example, my book talks about how the definite integral can be defined as the limit of the Riemann sums, but it also talks about how one can provide an upper and lower bound for the area, and one get bigger and the other smaller, until they both converge to the definite integral? @mixedmath – user166712 Jul 31 '14 at 21:45
  • Is one way of showing this is to use the upper and lower sums? If we can show that the limit of these as the mesh approaches infinity is the same, then by the squeeze principle, the area in between these sums (riemann sums) will converge to the area no matter the partition? – user166712 Jul 31 '14 at 22:01
  • Ah, I shouldn't be writing this at all. I'll direct you to more. – davidlowryduda Aug 01 '14 at 07:12