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Please help me process this is any sane way.

If we look in Baby Rudin we find the integral as:

Common value of: $$\overline{\int_{a}^{b}}f \ dx = \inf U(P,f) \quad, \qquad \underline{\int_{a}^{b}} f \,dx = \sup L(P,f) $$ if these values exist and are equal. Here the $\inf$ and $\sup$ are taken over all partitions, where we have, in the above: $$ U(P,f) = \sum_{i=1}^nM_i\Delta x_i \quad, \qquad L(P,f) = \sum_{i=1}^nm_i\Delta x_i $$ where $m_i$ and $M_i$ are the $\sup$ and $\inf$ of $f$ on each respective subinterval.

When we calculate integrals, we dont see anything like this, and even when calculating Riemann sums, we put $\Delta x_i = \frac{b-a}{n}$ and let $n \to \infty$.

Question: How does this sum align with Rudins definition? This is supposed to be the $\inf$ over all partitions, how does fixing $\Delta x_i = \frac{b-a}{n}$ and letting $n \to \infty$ allow me to garantee that this infinite sum equals that $\inf / \sup$?. Also in calculations, people fix a point as the (often) rightmost point in that interval, but that is not the $\sup$ over that interval!! Whats going on here?

Willie Wong
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JuliusL33t
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    It is not clear what you are asking about: computing integrals, or the equivalence between the Riemann sum formulation and the Darboux sums formulation? – Alex M. Sep 09 '16 at 20:08
  • @AlexM.: He is asking this: Why we define integral via Darboux sum but calculate it via Riemann sum? – Paramanand Singh Sep 09 '16 at 20:12
  • @ParamanandSingh: If this is indeed what he is asking, then he is mistaken: we don't compute integrals via Riemann sums. In fact, I suggest the OP to compute $\int _0 ^\infty \Bbb e ^{-x^2} \Bbb d x$ via Riemann sums... It seems the OP does not correctly perceive the difference between computation and definition. – Alex M. Sep 09 '16 at 21:21
  • @AlexM.: We don't usually compute integrals as a limit of Riemann sum, but we can calculate simple integrals like $\int_{a}^{b}x^{n},dx$ or $\int_{a}^{b}\sin x,dx$ using limit of Riemann sum approach. OP is struggling to identify the link between limit of Riemann sums and $\sup/\inf$ of Darboux sums thanks to Baby Rudin. – Paramanand Singh Sep 09 '16 at 21:27

2 Answers2

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I would advise you to look at this answer and carefully read both definitions $1$ and $2$ of Riemann integral and also the link of the integral with Darboux sums. The definition of Riemann integral which we use to calculate integral of functions like $x^{2}$ is based on Definition 2. And we take a partition whose sub-intervals are of equal length and form a Riemann sum and take its limit as the number of sub-intervals tends to $\infty$.

Baby Rudin does not handle the topic of Riemann integrals nicely and I would also advise you to consider studying Apostol's Mathematical Analysis on which my linked answer is based. In case you don't have a copy of Apostol's book, then you may also have a look at my blog posts on Riemann integral (which are also based on Apostol's book). You have to understand the link between a Darboux sum and a Riemann sum and the linked answer tries to provide some details in that direction.

Community
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Paramanand Singh
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Indeed, using that definition of the integral would drive most mathematicians insane. Fortunately, there is that wonderful theorem called "the Leibniz-Newton formula" ($\int _a ^b f = F(b) - F(a)$ whenever $F' = f$), which tells us that computing a Riemann integral is the same as computing an antiderivative and evaluating it in the endpoints. Now you are in the position to understand why it is cherished as one of the most important theorems in mathematics.

If the Lebniz-Newton formula does not easily lead to a result, we usually combine it with various properties of the Riemann integral, such as the one describing how it changes under diffeomorphisms (we call this a "change of variables").

By now, you should have understood what we do: we define the concept of "definite integral" based upon Riemann or Darboux sums, prove some practical properties using it, and then use these practical properties in concrete problems, not the original concept itself.

Alex M.
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