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Expressing a limit as a definite integral

  • Thread starter mech-eng
  • Start date
753
11
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.
 
Last edited:
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Express the limit ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac2n\ (1+\frac {2-1}{n})^\frac13##
You omitted i in the fraction.
It should be ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac 2 n\ (1+\frac {2i-1}{n})^\frac 1 3##. Same in your problem statement.
mech-eng said:
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##
##c_i## is a point in the i-th subinterval, with the interval being [0, 2] (as you know). So ##c_1 = \frac 1 n, c_2 = \frac 3 n, c_3 = \frac 5 n, \dots c_n = \frac {2n - 1} n = 2 - \frac 1 n##.
Does that make sense?
mech-eng said:
Source: Calculus, A Complete Course by Robert A. Adams.

Thank you.
 

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Express the limit ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac2n\ (1+\frac {2-1}{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.
The point is ##x_i = c_i##, but the integrand (the function being integrated) is ##(1+x)^{1/3}.## Therefore, you need ##(1+c_i)^{1/3}.##
 
753
11
##c_i## is a point in the i-th subinterval, with the interval being [0, 2] (as you know). So ##c_1 = \frac 1 n, c_2 = \frac 3 n, c_3 = \frac 5 n, \dots c_n = \frac {2n - 1} n = 2 - \frac 1 n##.
Does that make sense?
Loosely yes but strictly speaking no. If the length of the each subinterval is 2/n and the the first point in the interval is 0, then should ##c_1## not be equal to 2/n since 0+2/n=2/n?

Thank you.
 
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Loosely yes but strictly speaking no. If the length of the each subinterval is 2/n and the the first point in the interval is 0, then should ##c_1## not be equal to 2/n since 0+2/n=2/n?
##c_i## could be any point in the i-th subinterval, but it looks to me like the author picked ##c_i## to be the midpoint of the i-th subinterval.
 
753
11
I corrected the issue. Thank you very much.
 
753
11
How do conclude from this equation ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac2n\ (1+\frac {2i-1}{n})^\frac13## that

cicic_i is a point in the i-th subinterval, with the interval being [0, 2] (as you know). So c1=1n,c2=3n,c3=5n,…cn=2n−1n=2−1nc1=1n,c2=3n,c3=5n,…cn=2n−1n=2−1nc_1 = \frac 1 n, c_2 = \frac 3 n, c_3 = \frac 5 n, \dots c_n = \frac {2n - 1} n = 2 - \frac 1 n.
the general term should be 2n-1/n? Is this about series?

Thank you.
 
32,344
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From post #1:
"We want to intepret the sum as a Riemann sum for ##f(x)=(1+x)^3##
How do conclude from this equation ##lim_{n\rightarrow\infty} \sum_{i=1}^n \frac2n\ (1+\frac {2i-1}{n})^\frac13## that the general term should be 2n-1/n? Is this about series?
2n - 1/n isn't the general term -- this is ##c_n##, the midpoint of the n-th subinterval.

The integral is ##\int_0^2 (1 + x)^{1/3}~dx## -- in post #1 you had ##f(x) = (1 + x)^3##, but I'm pretty sure the exponent 3 should be 1/3.

In the Riemann sum, you have ##dx## or ##\Delta x = \frac 2 n##. The x-value in the subinterval is ##c_i = \frac{2i - 1}n##, and the integrand ##(1 + x)^{1/3}## becomes ##(1 + c_i)^{1/3} = (1 + \frac{2i - 1}n)^{1/3}##

Putting these pieces together, you have ##\int_0^2 dx~(1 + x)^{1/3} = lim_{n \to \infty} \sum_{i=1}^n \frac 2 n (1+\frac {2i-1}{n})^\frac 1 3##
 

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