# Proving the value of this integral

• bonfire09
L(f,P_n)## and ##U(f,P_n)## both converge to the integral as n goes to infinity. Thanks everyone for the help!In summary, using the preceding exercise and corollary 6.12, it can be proven that for any natural number ##p##, ##\int_0^1 x^p dx=\frac{1}{p+1}## by using a regular partition and the squeeze theorem.
bonfire09

## Homework Statement

For any natural number ##p## use the preceding exercise and corollary 6.12 to prove that ##\int_0^1 x^p dx=\frac{1}{p+1}##

## Homework Equations

Preceding Exercise: Let ##p## and ##n## be natural numbers with ##n\geq 2##. Then ##\sum_{k=1}^{n-1} k^p\leq \frac{n^{p+1}}{p+1}\leq\sum_{k=1}^{n} k^p##

Corollary 6.12: Suppose the function ##f:[a,b]→\mathbb{R}## is integrable. If ##\{P_n\}## is any sequence of partitions of ##[a,b]## such that ##\lim_{n\to\infty}||P_n||=0## then ##\lim_{n\to\infty} U(f,P_n)=\lim_{n\to\infty} L(f,P_n)=\int_a^b f##.

## The Attempt at a Solution

I am really stuck on this problem for the past hour. I proved the preceding exercise already and I moved on to this one. I tried this problem using a regular partition P but that did not work. Any help would be great. Thanks

bonfire09 said:

## Homework Statement

For any natural number ##p## use the preceding exercise and corollary 6.12 to prove that ##\int_0^1 x^p dx=\frac{1}{p+1}##

## Homework Equations

Preceding Exercise: Let ##p## and ##n## be natural numbers with ##n\geq 2##. Then ##\sum_{k=1}^{n-1} k^p\leq \frac{n^{p+1}}{p+1}\leq\sum_{k=1}^{n} k^p##

Corollary 6.12: Suppose the function ##f:[a,b]→\mathbb{R}## is integrable. If ##\{P_n\}## is any sequence of partitions of ##[a,b]## such that ##\lim_{n\to\infty}||P_n||=0## then ##\lim_{n\to\infty} U(f,P_n)=\lim_{n\to\infty} L(f,P_n)=\int_a^b f##.

## The Attempt at a Solution

I am really stuck on this problem for the past hour. I proved the preceding exercise already and I moved on to this one. I tried this problem using a regular partition P but that did not work. Any help would be great. Thanks
Should that first summation be
##\displaystyle \sum_{k=0}^{n-1} k^{\ p} \ \ ? ##​

Show what you have tried for this proof -- or at least explain it in some detail.

SammyS said:
Should that first summation be
##\displaystyle \sum_{k=0}^{n-1} k^{\ p} \ \ ? ##​

Show what you have tried for this proof -- or at least explain it in some detail.

The exact limits on the summations aren't going to matter when you take the limit n->infinity. I'm hoping the regular partition means [0/n,1/n,2/n...n/n]. If so it should have worked. As SammyS said, show what is going wrong.

No what I have for the first summation is written just correctly just like they have it in the book. What I've tried is let ##P_n## be a regular partition of ##[0,1]##. It follows that ##\lim_{n\to\infty}||P_n||=0##. Then ##L(f, P_n)=\sum_{i=1}^{n} \frac{1}{n}f(\frac{i-1}{n})##. And ##U(f,P_n)=\sum_{i=1}^{n} \frac{1}{n}f(\frac{i}{n})##. Thats all I have so far and trying to use the exercise in my proof I can't seem to figure that out. And yes the regular partition I am using is ##P_n=\{[0,\frac{1}{n}],...,[\frac{n-1}{n},1]\}##.

What's f in this case?

An increasing function on ##[0,1]## ?

You can be a lot more specific than that. What's the function you're being asked to integrate?

Oh sorry its ##f(x)=x^p## where ##p\in\mathbb{N}##. I mean yes I can integrate it using the integral properties but that would be a shortcut. I am trying to figure out how to use that inequality I am supposed to use.

So...
$$L(f, P_n)=\sum_{i=1}^{n} \frac{1}{n}f\left(\frac{i-1}{n}\right) = \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i-1}{n}\right)^p.$$ Relate that sum to one in the preceding exercise.

It looks pretty straightforward to me. Divide the interval from 0 to 1 into n equal parts. Using the left end of each subinterval to evaluate the function, we have rectangles with heights of $0^p$, $(1/n)^p$, $(2/n)^n$, ..., up to $((n-1)/n)^2$, each of width $1/n$ so the sum of their widths is $\sum_{k=0}^{n-1} (k/n)^p(1/n)= \frac{1}{n^{p+1}}\sum_{k=0}^{n-1} k^p$.
(I notice that you have the sum starting at k= 1. That is the same thing here since the k=0 term is 0.)

If instead we use the right endpoints, we will get values of $(1/n)^p$, $(2/n)^p$, ... up to $(n/n)^p= 1$. Do the same thing to get [/itex]\frac{1}{n+1}\sum_{k= 1}^n k^p[/itex].

In the particular case, the function is increasing between 0 and 1 so that the value at the left side of each sub-interval is less than the value at the right. That is, each term in the first sum is less than the corresponding term in the second sum. Further, since the left side is the lowest value of the function and the right side the highest, one rectangle lies below the curve, the other above it. So the "area under the curve" must lie between the two rectangles.

1 person
Oh would it be ##\sum_{k=1}^n \frac{1}{n}(\frac{k-1}{n})^p=L(f,P_n) \geq \frac{n^{p+1}}{p+1} \geq \sum_{k=1}^{n-1} \frac{1}{n}(\frac{k-1}{n})^p ##. Nevermind I got it. At times I don't know why I miss it when its obvious. The final result just follows from the squeeze theorem,

Last edited:

## 1. What is the purpose of proving the value of an integral?

The purpose of proving the value of an integral is to determine the exact numerical value of a mathematical expression that represents the area under a curve, or the total accumulation of a function over a given interval. This is important in various fields of science, such as physics, engineering, and economics, where accurate calculations are essential for making predictions and solving real-world problems.

## 2. How is the value of an integral determined?

The value of an integral is typically determined through the process of integration, which involves finding an antiderivative of the function being integrated and evaluating it at the upper and lower bounds of the integral. Other methods, such as numerical integration and graphical methods, can also be used to approximate the value of an integral.

## 3. Can the value of an integral be proven without using calculus?

No, the concept of integration is a fundamental part of calculus and is necessary for proving the value of an integral. However, there are alternative methods such as Riemann sums and the trapezoid rule that can be used to approximate the value of an integral without using calculus.

## 4. How is the value of a definite integral different from an indefinite integral?

A definite integral has specific limits of integration, which means it represents the exact area under a curve or accumulation of a function over a given interval. An indefinite integral, on the other hand, has no specified limits and represents a family of functions that are all antiderivatives of the original function.

## 5. What are some real-world applications of proving the value of an integral?

Proving the value of an integral has many practical applications, such as calculating the work done by a force, finding the center of mass of an object, determining the amount of medication in a patient's bloodstream, and predicting the future value of investments. It is also used in various fields of science, including physics, engineering, economics, and statistics.

Replies
3
Views
540
Replies
7
Views
694
Replies
17
Views
1K
Replies
3
Views
956
Replies
1
Views
771
Replies
8
Views
1K
Replies
2
Views
1K
Replies
13
Views
405
Replies
6
Views
781
Replies
12
Views
607