- Problem Statement
- Express the limit ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac2n\ (1+\frac {2-1}{n})^\frac13## as a definite integral. <Mentor note: the part in parentheses should be ##1 + \frac{2i - 1} n##.>

- Relevant Equations
- As ##n\rightarrow\infty, c_1=\frac1n=\rightarrow\ \ 0##

Express the limit ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac2n\ (1+\frac {2i-1}{n})^\frac13##

This is worked example but I would like to ask about the points I don't understand in the book.

"We want to intepret the sum as a Riemann sum for ##f(x)=(1+x)^3## The factor ##\frac2n## suggests that the interval of the integration has a length 2 and is partitioned into n equal subintervals, each of length 2/n. Let ##c_i=(2i-1)/n~~for~~ i=1,2, 3 \ldots, n##. As ##n_{\rightarrow\infty}\ c_1=1/n \ \rightarrow 0\ \ and \ \ c_n (2n-1)/n \ \rightarrow\ 2##. Thus the interval is [0,2], and the points of the partition are ##x_i=2i/n## observe that ## x_{i-1}=(2i-2)/n<c_i <2i/n=x_i## for each i, so that the sum is indeed a riemann sum for f(x) over [0,2]. Since f(x) is continuous on that interval it is integrable there, and

## lim_{n\rightarrow} \sum_{i=1}^n \frac 2n(1+ {\frac {2i-1}n)}^\frac13=\int_0^2(1+x)^\frac13 dx##

Why it is said in the solution ##c_i=(2i-1)/n## but not ##1+c_i=(2i-1)/n##

Source: Calculus, A Complete Course by Robert A. Adams.

Thank you.

This is worked example but I would like to ask about the points I don't understand in the book.

"We want to intepret the sum as a Riemann sum for ##f(x)=(1+x)^3## The factor ##\frac2n## suggests that the interval of the integration has a length 2 and is partitioned into n equal subintervals, each of length 2/n. Let ##c_i=(2i-1)/n~~for~~ i=1,2, 3 \ldots, n##. As ##n_{\rightarrow\infty}\ c_1=1/n \ \rightarrow 0\ \ and \ \ c_n (2n-1)/n \ \rightarrow\ 2##. Thus the interval is [0,2], and the points of the partition are ##x_i=2i/n## observe that ## x_{i-1}=(2i-2)/n<c_i <2i/n=x_i## for each i, so that the sum is indeed a riemann sum for f(x) over [0,2]. Since f(x) is continuous on that interval it is integrable there, and

## lim_{n\rightarrow} \sum_{i=1}^n \frac 2n(1+ {\frac {2i-1}n)}^\frac13=\int_0^2(1+x)^\frac13 dx##

Why it is said in the solution ##c_i=(2i-1)/n## but not ##1+c_i=(2i-1)/n##

Source: Calculus, A Complete Course by Robert A. Adams.

Thank you.

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