# Expressing a limit as a definite integral

#### mech-eng

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.

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#### Mark44

Mentor
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.

#### Ray Vickson

Science Advisor
Homework Helper
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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 $(1+c_i)^{1/3}.$

#### mech-eng

$c_i$ 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.

#### Mark44

Mentor
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.

• mech-eng

#### mech-eng

I corrected the issue. Thank you very much.

#### mech-eng

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.

#### Mark44

Mentor
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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