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My child recently was asked a question in Calculus I on the conversion Riemann sums to their equivalent definite integrals. This got me wondering whether this conversion is in general unique.

  1. Given, $$\lim_{n\to\infty} \sum_{i=1}^{\infty}g(i,n),$$ and assuming the limit exists and is finite, can one uniquely decompose $g(i,n)$ into $f(x_i)$ and a $\Delta x$ in order to determine the corresponding definite integral $$\int_a^b f(x) dx?$$
  2. If the decomposition is in general not unique (a counter example would be appreciated), can relatively simple conditions be placed on $g(i,n)$ such that the decomposition is unique?
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  • $\begingroup$ You seem to be missing some parts of the definition. The limits $a$ and $b$ have nothing to do with $g$. You seem to be thinking of $g(i,n)$ as $f(a+\frac in(b-a))$ but do not say so. $\endgroup$ Commented Jan 14, 2022 at 3:39
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    $\begingroup$ Riemann sums (strictly speaking) are finite sums. Each $g(i,n)$ can be any $f(x)$ in the particular interval. $\endgroup$ Commented Jan 14, 2022 at 3:40
  • $\begingroup$ The way they typically pose these problems is that they don't give you $a$ or $b$, they ask you to figure it out from the form of $g(i,n)$. $\endgroup$ Commented Jan 14, 2022 at 3:45
  • $\begingroup$ And @herbsteinberg, Go Bulldogs. Yale 2003. $\endgroup$ Commented Jan 14, 2022 at 3:49
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    $\begingroup$ The easiest nonuniqueness to implement is $\int_a^b f(x) \,\mathrm{d}x = \int_{a+1}^{b+1} f(x-1) \,\mathrm{d}x$, obtained by using $f(a +\frac{i}{n}(b-a)) = f((a+1) + \frac{i}{n}((b+1)-(a+1))-1)$. $\endgroup$ Commented Jan 14, 2022 at 5:59

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I'm not convinced that such a decomposition exists, in the very general form stated in the OP. But there's an easy reason that such decompositions aren't unique: write down, for instance, the Riemann-sum definitions of $$ \int_0^1 \sin x\,dx \quad\text{and}\quad \int_3^4 \sin (x-3)\,dx. $$

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  • $\begingroup$ Thanks @GregMartin, you've proven my point that the conversion isn't unique. $\endgroup$ Commented Jan 15, 2022 at 17:44

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