# I Proofs of various integral properties

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1. Oct 5, 2016

### Frank Castle

I've been trying to prove a couple of properties of integrals using the Riemann sum definition: $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\Delta x$$ where the interval $[a,b]$ has been partitioned (such that $a=x_{1}<x_{2}<\cdots <x_{i-1}<x_{i}<\cdots <x_{n-1}<x_{n}$) into $n$ sub-intervals of equal width $\Delta x=\frac{b-a}{n}$, where $x^{\ast}_{i}\in [x_{i-1},x_{i}]$ is an arbitrary point within the $i^{th}$ sub-interval $[x_{i-1},x_{i}]$.

Given this, I'm trying to prove that

1. $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

2. $\int_{a}^{a}f(x)dx=0$.

Here are my attempts:

1. Using the definition of the Riemann integral, we have that
$$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\Delta x=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{b-a}{n}=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{-(a-b)}{n}\\ =-\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{a-b}{n}=-\int_{b}^{a}f(x)dx\qquad\qquad\qquad\quad$$

2. This follows from 1. for $a=b$, we have $$\int_{a}^{a}f(x)dx=-\int_{a}^{a}f(x)dx\Rightarrow \int_{a}^{a}f(x)dx=0$$ or, from the definition of the Riemann integral $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\Delta x=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}f(x^{\ast}_{i})\frac{a-a}{n}=\lim_{n\rightarrow\infty}0=0$$

Are these valid proofs?

Furthermore, how can one prove these using the alternative definition of the Riemann integral: $$\int_{a}^{b}f(x)dx:=\lim_{n\rightarrow\infty}\lim_{\text{Max}\,\Delta x_{i}\rightarrow 0}\sum_{i=1}^{n}f(\zeta_{i})\Delta x_{i}$$ where $\Delta x_{i}=x_{i}-x_{i-1}$ is the width of the $i^{th}$ sub-interval and $\zeta_{i}\in [x_{i},x_{i-1}]$ is an arbitrary point within this interval, with $\text{Max}\,\Delta x_{i}=\lbrace x_{2}-x_{1},\ldots , x_{i}-x_{i-1},\ldots , x_{n}-x_{n-1}\rbrace$?!

Finally, is there an easy way to prove the additivity property of integrals, i.e. $$\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx\;?$$

2. Oct 5, 2016

### Erland

These two properties cannot be proved, they must be taken as definitions. In your definition, and in all alternative definitions of the Riemann Integral, it is assumed that $a<b$ when we write $\int_a^b f(x)dx$, so using this definition alone, the expression $\int_b^a f(x)dx$ is not defined for $b<a$. In this case, we define $\int_b^a f(x)dx=-\int_a^b f(x)dx$, and we also define $\int_a^a f(x)dx =0$.

Edit: The additivity property should be quite easy to prove if you use your "alternative definition". Personally, I prefer the definitions based upon upper and lower sums.

3. Oct 6, 2016

### Frank Castle

This leaves me confused as in some notes I've read the authors all proceed to prove these properties (for example http://tutorial.math.lamar.edu/Classes/CalcI/ProofIntProp.aspx), whereas others confirm what you've said and simply give them as definitions (for example https://www.math.ucdavis.edu/~hunter/intro_analysis_pdf/ch11.pdf)? [Broken]?!

Last edited by a moderator: May 8, 2017
4. Oct 6, 2016

### Erland

If $\int_b^a f(x)dx$ is not defined as $-\int_a^b f(x)dx$ when $a>b$, then how is it defined? In your definition, you presuppose that $a<b$ since otherwise you cannot write the partition as $a=x_0<x_1<\dots< x_n=b$, and the link you gave doesn't resolve this either. Of course one can generalize the definition to include "backwards" partitions $a=x_0>x_1>...> x_n=b$ and talk about subintervals $[x_i,x_{i-1}]$ instead of $[x_{i-1},x_i]$, but then we must handle several cases in the definition. This complicates things unnecessarily, in my opinion. And then, how do we generalize this to double and triple integrals? It's just simpler to give the definition for the case $a<b$ only and define the integral from $b$ to $a$ in the case $a<b$ as $-\int_a^b f(x)dx$.

The best way to define the Riemann integral, in my opinion, is to use upper and lower sums as here: https://en.wikipedia.org/wiki/Darboux_integral
What they call "Darboux integral" is the same thing as Riemann integral.

5. Oct 6, 2016

### Frank Castle

Is it fairly standard to take $\int_{a}^{a}f(x)dx=0$ and $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$ as definitions then, and then prove all other properties of the definite integral, for example $\int_{a}^{c}f(x)dx=\int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx$, using these definitions as starting points?

I have to admit, I was never taught about definite integrals in terms of upper and lower sums, but I'd like to learn. Would you happen to know of any accessible introductions to definite integrals using this approach?

6. Oct 6, 2016

### Erland

It's definitely standard. In fact, I never saw anything else before.
Well, you could look up the Robert A. Adams & Christopher Essex: Calculus, a complete course, Chapter 5. which I taught engireering students for many years.
A drawback is that it is not very general, since it assumes continuous functions. This is because otherwise we cannot ensure that the functions have a maximum and minimum on every closed interval. Adams don't want to get into the more complicated issue with supremum and infimum.

Walter Rudin's Principles of Mathematical Analyis, Chapter 6, fills this gap, but it is on the other hand a bit too general, since it introduces a more general type of integral, the Riemann-Stieltjes integral, but you can disregard that and think of the Riemann integral.

Last edited: Oct 7, 2016
7. Oct 7, 2016

### micromass

I recommend Bloch, which actually treats both definitions, i.e. the definition with lower/upper sums, and the definition in the OP. It also proves Lebesgue's theorem which gives us information about exactly which functions are integrable. https://www.amazon.com/Real-Numbers-Analysis/dp/0387721762

Note that while lower and upper sums are very intuitive and definitely serve a useful purpose, they don't exactly generalize well to more advanced settings.

Last edited by a moderator: May 8, 2017
8. Oct 7, 2016

### Frank Castle

Thanks for the recommendations, I'll take a look at them.

What would a better definition to use in more advanced settings?

Last edited by a moderator: May 8, 2017