- #1

mcastillo356

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- TL;DR Summary
- Assumed continuity implies integrability, can´t see the Riemann sum in this exercise

Hi, PF, there goes the definition of General Riemann Sum, and later the exercise. Finally one doubt and my attempt:

(i) General Riemann Sums

Let ##P=\{x_0,x_1,x_2,\cdots,x_n\}##, where ##a=x_0<x_1<x_2<\cdots<x_n=b##, be a partition of ##[a,b]##, having norm ##||P||=\mbox{max}_{1<i<n}\Delta x_i##. In each subinterval ##[x_{i-1},x_i]## of ##P## pick a point ##c_i## (called a

$$R(f,P,c)=\sum_{i=1}^n f(c_i)\Delta x_i$$

$$\qquad{=f(c_1)\Delta x_1+f(c_2)\Delta x_2+f(c_3)\Delta x_3+\cdots +f(c_n)\Delta x_n+}$$

is called the

(...) For any choice of the tags ##c##, the Riemann sum ##R(f,P,c)## satisfies

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Therefore, if ##f## is integrable on ##[a,b]##, then its integral is the limit of such Riemann sums, where the limit is taken as the number ##n(P)## of subintervals of ##P## increases to infinity in such a way that the lengths of all the subintervals approach zero. That is,

$$\displaystyle\underset{||P||\to 0}{\lim_{n(P)\to\infty}}{R(f,P,c)}=\int_{a}^{b} f(x) \, dx$$

as a definite integral.

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

(iii) Question: how can I state the following equivalence?

$$x_{i-1}=(2i-2)/n<c_i<2i/ n=x_i$$

$$\Longleftrightarrow$$

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Attempt to solve: I can`t. ##c_i## is a tag set; any partition ##P##; but I can say: if ##f## is continous, I can assert the correspondence.

Regards!

(i) General Riemann Sums

Let ##P=\{x_0,x_1,x_2,\cdots,x_n\}##, where ##a=x_0<x_1<x_2<\cdots<x_n=b##, be a partition of ##[a,b]##, having norm ##||P||=\mbox{max}_{1<i<n}\Delta x_i##. In each subinterval ##[x_{i-1},x_i]## of ##P## pick a point ##c_i## (called a

*tag*). Let ##c=(c_1,c_2,\cdots,c_n)## denote the set of these tags. The sum$$R(f,P,c)=\sum_{i=1}^n f(c_i)\Delta x_i$$

$$\qquad{=f(c_1)\Delta x_1+f(c_2)\Delta x_2+f(c_3)\Delta x_3+\cdots +f(c_n)\Delta x_n+}$$

is called the

**Riemann sum**of ##f## on ##[a,b]## corresponding to partition ##P## and tags ##c##.(...) For any choice of the tags ##c##, the Riemann sum ##R(f,P,c)## satisfies

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Therefore, if ##f## is integrable on ##[a,b]##, then its integral is the limit of such Riemann sums, where the limit is taken as the number ##n(P)## of subintervals of ##P## increases to infinity in such a way that the lengths of all the subintervals approach zero. That is,

$$\displaystyle\underset{||P||\to 0}{\lim_{n(P)\to\infty}}{R(f,P,c)}=\int_{a}^{b} f(x) \, dx$$

**EXAMPLE 4**Express the limit $$\displaystyle\lim_{n\to\infty}{\sum_{i=1^{n}}{ \displaystyle\frac{2}{n}}\left(1+\displaystyle\frac{2i-1}{n}\right)^{1/3} }$$as a definite integral.

**We want to interpret the sum as a Riemann sum for ##f(x)=(1+x)^{1/3}##. The factor ##2/n## suggests that the interval of integration has length ##2## and is partioned into ##n## equal subintervals, each of length ##2/n##. Let ##c_i=(2i-1)/n## for ##i=1,2,3,\cdots,n##. As ##n\to\infty##, ##c_1=1/n\to 0## and ##c_n=(2i-1)/n \to 2##. Thus, the interval is ##[0,2]## and the points of the partition are ##x_i=\frac(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## is continous on that interval, it is integrable there, and***Solution*$$\displaystyle\lim_{n\to\infty}{\sum_{i=1^{n}}{ \displaystyle\frac{2}{n}}\left(1+\displaystyle\frac{2i-1}{n}\right)^{1/3} }=\int_{0}^{2}{(1+x)^{1/3}} \, dx$$

(iii) Question: how can I state the following equivalence?

$$x_{i-1}=(2i-2)/n<c_i<2i/ n=x_i$$

$$\Longleftrightarrow$$

$$L(f,P)\leq R(f,P,c)\leq U(f,P)$$

Attempt to solve: I can`t. ##c_i## is a tag set; any partition ##P##; but I can say: if ##f## is continous, I can assert the correspondence.

Regards!