# Proving the value of this integral

1. Feb 22, 2014

### bonfire09

1. The problem statement, all variables and given/known data
For any natural number $p$ use the preceding exercise and corollary 6.12 to prove that $\int_0^1 x^p dx=\frac{1}{p+1}$

2. Relevant 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 $\lim_{n\to\infty} U(f,P_n)=\lim_{n\to\infty} L(f,P_n)=\int_a^b f$.

3. 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

2. Feb 22, 2014

### SammyS

Staff Emeritus
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.

3. Feb 22, 2014

### Dick

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.

4. Feb 22, 2014

### bonfire09

No what I have for the first summation is written just correctly just like they have it in the book. What ive 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 im using is $P_n=\{[0,\frac{1}{n}],...,[\frac{n-1}{n},1]\}$.

5. Feb 22, 2014

### vela

Staff Emeritus
What's f in this case?

6. Feb 22, 2014

### bonfire09

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

7. Feb 22, 2014

### vela

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

8. Feb 22, 2014

### bonfire09

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. Im trying to figure out how to use that inequality im supposed to use.

9. Feb 22, 2014

### vela

Staff Emeritus
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.

10. Feb 22, 2014

### HallsofIvy

Staff Emeritus
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.

11. Feb 22, 2014

### bonfire09

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: Feb 22, 2014