What Is the Correct Evaluation of the Indefinite Integral of Zero?

[physkill]

The antideriviative of 0 is not 0x, isn't it ?
If you differentiate 1 its 0, so if you reverse that process solution a has to be 1+C and not C for its own.
And for Argument b you would destroy the original sense of C, being a constant that you add because its been removed in the process of differentiation ?
so the Argument b would be something like 0*x+C = 0+C = C
And since in Argument a, there is a constant 1 that you add to C, you could remove the 1 as well since it is included inside C ?
That would be my point of view as an 12 grader...

 

[pellikkan]

the indefinite C after integration usually has to be found from
some equation in accordance with some boundary condition, constraint, or
other condition. It can often be found to be zero that way. You might be
confusing multiplication and addition between your two paragraphs.

 

[Tim77]

I am not quite at the education level to give a rigorous answer based on Riemann integrals. But I do see that case B is invalid because you are not using linearity to pull a constant out of the integral. In this case 0 is not being treated as a constant but as a function. I'm sure we agree you can't pull the function that is being integrated over out of the integral.

Another viewpoint that might explain why the indefinite integral of 0 = C is that you can think of the output of an indefinite integral as sum of the rates of change of an unknown family of functions,which in this case is all of the constant valued functions that all differ by a constant:). A definite integral is the difference in that function evaluated at two points which in turn is the difference in total rates of change. Where C(b) -C(a) =C-C= 0 .Technically the constant of integration is always there we just omit it in the definite case because it is subtracted out.

Thinking in terms of functions being defined by sums of their rates of change (derivatives or total differentials) as opposed to the traditional Riemann picture and area underneath curves helped me to understand what is happening a little better.And it gives better intuition of why Taylor series work.The function must be analytic though:)

Sorry for a long winded post and I apologize also for not using Latec I will get there though.

 

[HallsofIvy]

I am not quite at the education level to give a rigorous answer based on Riemann integrals. But I do see that case B is invalid because you are not using linearity to pull a constant out of the integral. In this case 0 is not being treated as a constant but as a function. I'm sure we agree you can't pull the function that is being integrated over out of the integral.

I, for one, don't agree! If the function is a constant function, f(x)= a, where a is fixed number, then \int a dx= a \int dx= ax+ C<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Another viewpoint that might explain why the indefinite integral of 0 = C is that you can think of the output of an indefinite integral as sum of the rates of change of an unknown family of functions,which in this case is all of the constant valued functions that all differ by a constant:). A definite integral is the difference in that function evaluated at two points which in turn is the difference in total rates of change. Where C(b) -C(a) =C-C= 0 .Technically the constant of integration is always there we just omit it in the definite case because it is subtracted out.<br /> <br /> Thinking in terms of functions being defined by sums of their rates of change (derivatives or total differentials) as opposed to the traditional Riemann picture and area underneath curves helped me to understand what is happening a little better.And it gives better intuition of why Taylor series work.The function must be analytic though:)<br /> <br /> Sorry for a long winded post and I apologize also for not using Latec I will get there though. </div> </div> </blockquote>

 

[Tim77]

I, for one, don't agree! If the function is a constant function, f(x)= a, where a is fixed number, then \int a dx= a \int dx= ax+ C

<br /> <br /> In this case using the wrong procedure leads to the correct answer but you technically can&#039;t pull the function out of the integral.You could pull the constant 1 out though. Your example only works in this case because the function is constant and not equal to zero.In general though pulling the object treated as the function out leads to false answers.<br /> Int(e^x) =e^xInt(1) = xe^x+Ce^x??<br /> Int(x)=xInt(1)= x^2+Cx ??<br /> <br /> The form f(x)dx is required for integration to have meaning you can&#039;t separate it even if f(x)= a and especially if f(x)=0

 

[pwsnafu]

In this case using the wrong procedure leads to the correct answer but you technically can't pull the function out of the integral.

This is false. You can pull constant functions out.
Let $r \in \mathbb{R}$ then define $f_r (x) = r$ and so
$\int f_r(x) \, dx = \int f_r(x) \cdot f_1(x) \, dx = \int r \cdot f_1(x) \, dx = r \int f_1(x) \, dx = f_r(x) \int f_1(x) \, dx$
QED.

You can't do it with non-constant functions, but this thread is not about non-constants and everything you have written about non-constant functions is irrelevant to this thread.

The form f(x)dx is required for integration to have meaning you can't separate it even if f(x)= a and especially if f(x)=0

This is only true for differential forms. For other areas of analysis the dx is nothing more than notation. And even in differential forms $\int \omega$ is a common notation.

 

[Tim77]

This is false. You can pull constant functions out.
Let $r \in \mathbb{R}$ then define $f_r (x) = r$ and so
$\int f_r(x) \, dx = \int f_r(x) \cdot f_1(x) \, dx = \int r \cdot f_1(x) \, dx = r \int f_1(x) \, dx = f_r(x) \int f_1(x) \, dx$
QED.

You can't do it with non-constant functions, but this thread is not about non-constants and everything you have written about non-constant functions is irrelevant to this thread.
This is only true for differential forms. For other areas of analysis the dx is nothing more than notation. And even in differential forms $\int \omega$ is a common notation.

Yes you can hold functions "constant" and pull them out of the integral it is the key to evaluating multiple integrals, but you still have to integrate a function with respect to the variable of integration.And yes this is very connected to differential forms they are key to analysis.That aside an integral requires a function as its argument.
$\int \omega$ is actually shorthand for a general form of line integrals or Stokes theorem.

The omega encapsulates a lot of information but in order to actually evaluate the integral it has to be a differential form.I will conceed in the second case of the OP's original post it is valid to pull out the zero but then you are no longer integrating zero with respect to x you are integrating 1 or k or what ever "function plays the role of f(x) in the integral.In case 1 you have Int(0) =C in case 2 you have 0*Int(1)=0 essentially those are two different integrals. I stand corrected if you can show how you evaluate integrals of the form $\int \omega$ or how about $\int3$ without using a differential form and integrating with respect to some parameter or how about this $\int\emptyset\ dx$

 

[pwsnafu]

Yes you can hold functions "constant" and pull them out of the integral it is the key to evaluating multiple integrals, but you still have to integrate a function with respect to the variable of integration.

What does that this have to do with anything?

And yes this is very connected to differential forms they are key to analysis.

Maybe in multivariable differential geometry, but certainly not in integration of one variable which is what we are doing here.

That aside an integral requires a function as its argument.

Yes we know that. Why are you doing a brain-spew?

I will conceed in the second case of the OP's original post it is valid to pull out the zero but then you are no longer integrating zero with respect to x you are integrating 1 or k or what ever "function plays the role of f(x) in the integral.In case 1 you have Int(0) =C in case 2 you have 0*Int(1)=0 essentially those are two different integrals.

The co-domain of the indefinite integral is a quotient space. The two integrals in the OP are in the same equivalence class. This has been pointed out in this thread over and over. It doesn't matter if they are "essentially" different or not.

I stand corrected if you can show how you evaluate integrals of the form $\int \omega$ or how about $\int3$ without using a differential form and integrating with respect to some parameter

You clearly don't understand what I'm saying do you? In one dimensional calculus, there is no inherent reason to write down dx for an indefinite integral. We do it because of historical reasons. If I write $f: \mathbb{R}\to\mathbb{R}$ and $\int f$ then it is clear I am interested in the anti-derivative of some one dimensional function f. I haven't declared a variable for f, but it is clear that I am integrating with respect to it because that is the only choice. There is no need to write down the variable of a one dimensional constant function. Further, in one dimensional calculus, the "dx" is not part of a 1-form. It is part of the integral. That is, the indefinite integral is the map
$f \mapsto [\int f(x) \, dx]$
and not
$f(x) \, dx \mapsto [\int f(x) \, dx].$

In the multivariate case we need to worry about which variable to integrate with regard to, so differential forms are useful. You are arguing that differential forms are necessary in multivariable calculus therefore they are necessary in one-variable calculus. That is not true.

And I haven't even started to talk about the product measure where you see notation like:
"Let $(X_1, \Sigma_1, \mu_1)$ and $(X_2, \Sigma_2, \mu_2)$ be measure spaces and $f: X_1\times X_2 \to\mathbb{R}$. Then $\int_{X_1} f \, d\mu_1$..."
Some authors like to write down $\int_{X_1} f \, d\mu_1(x_1)$, others drop it because it is obvious that $\mu_1$ is used to integrate with respect to $x_1$.

 

[Tim77]

What does that this have to do with anything?
Maybe in multivariable differential geometry, but certainly not in integration of one variable which is what we are doing here.
Yes we know that. Why are you doing a brain-spew?
The co-domain of the indefinite integral is a quotient space. The two integrals in the OP are in the same equivalence class. This has been pointed out in this thread over and over. It doesn't matter if they are "essentially" different or not.
You clearly don't understand what I'm saying do you? In one dimensional calculus, there is no inherent reason to write down dx for an indefinite integral. We do it because of historical reasons. If I write $f: \mathbb{R}\to\mathbb{R}$ and $\int f$ then it is clear I am interested in the anti-derivative of some one dimensional function f. I haven't declared a variable for f, but it is clear that I am integrating with respect to it because that is the only choice. There is no need to write down the variable of a one dimensional constant function. Further, in one dimensional calculus, the "dx" is not part of a 1-form. It is part of the integral. That is, the indefinite integral is the map
$f \mapsto [\int f(x) \, dx]$
and not
$f(x) \, dx \mapsto [\int f(x) \, dx].$

In the multivariate case we need to worry about which variable to integrate with regard to, so differential forms are useful. You are arguing that differential forms are necessary in multivariable calculus therefore they are necessary in one-variable calculus. That is not true.

"Sheesh sorry about the "brain spew" You speak like multivariate calculus is some how a different subject when it is a generalization of "single variable calculus to higher dimensions is it not?

The first paragraph of the article you posted states:
In the mathematical fields of differential geometry and tensor calculus, differential forms are an approach to multivariable calculus that is independent of coordinates. Differential forms provide a unified approach to defining integrands over curves, surfaces, volumes, and higher-dimensional manifolds. The modern notion of differential forms was pioneered by Élie Cartan. It has many applications, especially in geometry, topology and physics.
For instance, the expression f(x) dx from one-variable calculus is called a 1-form"

By any means it seems you are twisting the use of notation with actually evaluating the integral.Arguments about differential forms aside my point was which ever "constant" you pullout you still have to integrate the function over a domain.If you pull out the zero you reduce your class from the general to the trivial case. Oh well I this is my last try to get my point across. I expect you'll pipe in with the last word and some more insults , ad hominem etc.I feel sorry for your "students" if you are as condesending to them.

 

[pwsnafu]

I expect you'll pipe in with the last word and some more insults , ad hominem etc.I feel sorry for your "students" if you are as condesending to them.

1. If I offended you I apologies. I do admit I was too rash in the post.
2. If you feel you were insulted, use the Report button.

 

[Stephen Tashi]

To return to the main topic, vela said it well. The meaning of the rule depends on how you interpret the notation of "a constant times a set". One interpretation is that it denotes a second set consisting of all things that can be formed by multiplying the constant by things in the first set. However, there is no convention that says this is the correct interpretation in all contexts.

 

[HallsofIvy]

Tim77 are you denying the existence of "constant functions"? That is, are you saying that "f(x)= 1" for all x is NOT a "function"?

 

[Ahmad Kishki]

Just an idea: Maybe 0*C will not equal zero. Because the integral of different functions will have different constants and this constant has no reason to be bounded, couldn't that constant be infinity such that 0*C=A (after taking some limit)? I don't know, its just an idea.

 

[Mark44]

"Sheesh sorry about the "brain spew" You speak like multivariate calculus is some how a different subject when it is a generalization of "single variable calculus to higher dimensions is it not?

Certainly multivariate calculus is a generalization of single variable calculus, but going there takes this thread well off topic.

The first paragraph of the article you posted states:
In the mathematical fields of differential geometry and tensor calculus, differential forms are an approach to multivariable calculus that is independent of coordinates. Differential forms provide a unified approach to defining integrands over curves, surfaces, volumes, and higher-dimensional manifolds. The modern notion of differential forms was pioneered by Élie Cartan. It has many applications, especially in geometry, topology and physics.
For instance, the expression f(x) dx from one-variable calculus is called a 1-form"

By any means it seems you are twisting the use of notation with actually evaluating the integral.Arguments about differential forms aside my point was which ever "constant" you pullout you still have to integrate the function over a domain.If you pull out the zero you reduce your class from the general to the trivial case. Oh well I this is my last try to get my point across. I expect you'll pipe in with the last word and some more insults , ad hominem etc.

What insults? What ad hominen arguments? The only things I could see that come remotely close are when pwsnafu said, "Why are you doing a brain-spew?" and "You clearly don't understand what I'm saying do you?"
Your post #37 would fit the description of "brain spew" IMO. And while his second question comes off as somewhat condescending, I fail to see anything ad hominem about it.

I feel sorry for your "students" if you are as condesending to them.

The OP's question has been asked and answered, so I am closing this thread.