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## Main Question or Discussion Point

Hi There Everyone

I am studying undergraduate calculus in first year. My question regards the rules for identifying a limit sum as a Riemann sum and therefore a definite integral. The book we are using says that when choosing [tex] \inline \large c_{i} [/tex] for some [tex] f(x) [/tex], if [tex] \inline \large x_{i - 1} < c_{i} < x_{i} [/tex], then the sum is indeed a Riemann sum for [tex] f(x) [/tex] over an interval.

Allow me to include an example afterwhich I will pose my question in clarity.

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Express the limit [tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3} [/tex] as a definite integral

Solution:

We want to interpret the sum as a Riemann sum for [tex] f(x)=(1+x)^{1/3} [/tex]. The factor 2/n suggests that the interval of integration has length 2 and is partitioned into n equal subintervals, each of length 2/n. Let [tex] \inline \large c_{i} = (2i - 1)/n [/tex] for i = 1, 2, 3, ..., n. As [tex] n \to \infty, c_{1} = 1/n \to 0 [/tex] and [tex] c_{n} = (2n -1)n \to 2 [/tex]. Thus, the interval is [0, 2], and the points of the partition are [tex] x_{i} = 2i/n [/tex]. Observe that [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] for each i, so that the sum is indeed a Riemann sum for [tex] f(x) [/tex] over [0, 2]. Since f is continuous on that interval, it is integrable there, and

[tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3} = \int_{0}^{2} (1+x)^{1/3} dx[/tex]

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My question is, is it necessary for the condition [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] be met for the sum to be converted to a definite integral. The reason I ask is that as [tex] n \to \infty [/tex] adding or subtracting any constant from the index does not change the sum at infinity in any case because if we choose [tex] c_{i} = (2i- 20000000)/n [/tex] the big number vanishes when [tex] n \to \infty [/tex] so that the sum is the same what ever the big number is. But then [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] is no longer true and by the definition of a Riemann sum, [tex] c_{i} [/tex] must lie with in the subinterval [tex] [x_{i-1}, x_{i} ][/tex]

I am studying undergraduate calculus in first year. My question regards the rules for identifying a limit sum as a Riemann sum and therefore a definite integral. The book we are using says that when choosing [tex] \inline \large c_{i} [/tex] for some [tex] f(x) [/tex], if [tex] \inline \large x_{i - 1} < c_{i} < x_{i} [/tex], then the sum is indeed a Riemann sum for [tex] f(x) [/tex] over an interval.

Allow me to include an example afterwhich I will pose my question in clarity.

========================

Express the limit [tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3} [/tex] as a definite integral

Solution:

We want to interpret the sum as a Riemann sum for [tex] f(x)=(1+x)^{1/3} [/tex]. The factor 2/n suggests that the interval of integration has length 2 and is partitioned into n equal subintervals, each of length 2/n. Let [tex] \inline \large c_{i} = (2i - 1)/n [/tex] for i = 1, 2, 3, ..., n. As [tex] n \to \infty, c_{1} = 1/n \to 0 [/tex] and [tex] c_{n} = (2n -1)n \to 2 [/tex]. Thus, the interval is [0, 2], and the points of the partition are [tex] x_{i} = 2i/n [/tex]. Observe that [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] for each i, so that the sum is indeed a Riemann sum for [tex] f(x) [/tex] over [0, 2]. Since f is continuous on that interval, it is integrable there, and

[tex] \lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n} \left( 1 + \frac{2i - 1}{n}\right)^{1/3} = \int_{0}^{2} (1+x)^{1/3} dx[/tex]

==========================

My question is, is it necessary for the condition [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] be met for the sum to be converted to a definite integral. The reason I ask is that as [tex] n \to \infty [/tex] adding or subtracting any constant from the index does not change the sum at infinity in any case because if we choose [tex] c_{i} = (2i- 20000000)/n [/tex] the big number vanishes when [tex] n \to \infty [/tex] so that the sum is the same what ever the big number is. But then [tex] x_{i-1} = (2i-2)/n < c_{i} < 2i/n = x_{i}[/tex] is no longer true and by the definition of a Riemann sum, [tex] c_{i} [/tex] must lie with in the subinterval [tex] [x_{i-1}, x_{i} ][/tex]