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I found that Riemann mid sum for arbitrary partition is $S(P, f)=\dfrac{1}{2}\sum_{i=1}^n(x_i^2-x_{i-1}^2)=\dfrac{b^2}{2}$, and now I have to show that Riemann sum approaches this value, i.e. chosen points and partition doesn't affect as long norm $|P|$ is small enough.

Law Neutral
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  • Not every Riemann sum of $f(x)=x$ on $[0,b]$ is equal to $\frac12b^2$. – Did May 09 '16 at 18:50
  • @Did I meant for middle sum, or is it wrong too? – Law Neutral May 09 '16 at 18:55
  • It's not true for ever riemann sum. Maybe the best that you can do is define some sequence of subintervals $(I_n)$ so that for the lengths$\mathcal{L}(I_n)$, $\lim_{n \to \infty} \mathcal{L}(I_n) \to 0$ . In this case, all but finitely many of the corresponding Riemann sums will be within $\epsilon$ of $\frac{1}{2}b^2$ for any $\epsilon>0$. – Andres Mejia May 09 '16 at 19:00
  • That might on the other hand, be a totally useless endeavor. Just pick a partition so that when the mesh is sufficiently small, the riemann sum over that partition will be within $\epsilon$ of $b^2/2$. – Andres Mejia May 09 '16 at 19:03
  • Every sum using the middle points to evaluate the function is equal to $\frac12b^2$, yes. If this is what you have in mind, this should be clearly stated in the question. – Did May 09 '16 at 19:15
  • @LawNeutral The second half of the question doesn't seem to agree with the first paragraph. Perhaps it would help to supply your definition of "integral" and "Riemann Sum." It is certainly not true that every Riemann sum evaluates to $\frac{1}{2}b^2$. – Andres Mejia May 09 '16 at 19:20
  • $Andres Mejia, I edited the question, but English is not my native language so I may have some trouble. – Law Neutral May 09 '16 at 19:28
  • I think the real problem is the term "for every" – Andres Mejia May 09 '16 at 19:41
  • Damn. I forgot to edit the title too. – Law Neutral May 09 '16 at 20:15

1 Answers1

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Well, IF you found that for every partition, the corresponding riemann sum equals $\frac {b^2}2$, then for any $\delta>0$ we already have that

$$ |P|<\delta \implies 0=|S(P,f)-I|<\varepsilon $$

Since the consequent is always true.

Also, just a remark: If it wasn't the case that for every riemann sum, the 'limit' (i.e, your $I$ here), the integral $\int_0^b f$ would not exist, as, per definition, the symbol $\int_0^b f $ is defined to be $I$ when this is equal for each and every Riemann sum (else, we say $f$ is not integrable over $[0,b]$, and it would make no sense to talk about $f$'s integral).

YoTengoUnLCD
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  • What? Riemann integrability is not defined as the fact that every Riemann sum is equal to the same value (if it was, nearly only the constant functions would be integrable). Please avoid comforting the OP in this false belief. – Did May 09 '16 at 18:50
  • @Did Whoops, that may have been terrible wording there, I meant the $I$'s should be the same for each Riemann sum. (Actually, re-reading my answer, I believe I was clear...) – YoTengoUnLCD May 09 '16 at 18:51
  • I actually also think this is fairly clear, it doesn't seem to be a "false comfort." – Andres Mejia May 09 '16 at 19:02
  • @AndresMejia Please be constructive: the exercise is false as stated, so, either we help the OP correct it or we mislead them. – Did May 09 '16 at 19:16
  • Ah, I see your point. I thought you were suggesting that the current answer had the same problem as the way the question was asked. – Andres Mejia May 09 '16 at 19:18
  • I still don't see your point, Did, I believe you were referring to my "Remark" paragraph, is there anything wrong with that? – YoTengoUnLCD May 09 '16 at 20:00
  • @Yo Really? Quotes: "I meant for middle sum". "Every sum using the middle points to evaluate the function is equal to $\frac12b^2$, yes. If this is what you have in mind, this should be clearly stated in the question." The point seems pretty clearly stated to me. – Did May 09 '16 at 20:19