1-dimensional substitution rule for integrals

[scorpion990]

This is kind of an awkward post, but:
The only topics that have always bugged me in calculus I-III are those which deal with differentials... I've convinced myself (and proven) that if a function can be written as f(x)dx = g(y)dy, then it is possible to "separate" the variables and solve the differential equation. However, I haven't even been able to convince myself of the 1-dimensional substitution rule for integrals.

For example:
u = 2x+6
du = 2xdx

I have a hard time seeing why I can just "solve for dx", replace that in the integral, and go on my way integrating with respect to u. I just think that there is a lot hidden in the method that is not documented well in calculus textbooks.

Another problem: I'm reviewing a little bit of differential equations, and I've stumbled across equations with homogeneous coefficients. The details aren't necessary, but as is easily proven, the substitution y = ux eases the problem. However, with this substitution, it "follows" that dy = udx + xdu. This clearly follows from the product rule, but I'm confused as to how it's possible to replace the dependent variable (y) with two independent variables (u and x). And of course, the nature of differentials is still a mystery to me.

Anybody willing to put me out of my misery and explain?
Thanks!

 

[HallsofIvy]

This is kind of an awkward post, but:
The only topics that have always bugged me in calculus I-III are those which deal with differentials... I've convinced myself (and proven) that if a function can be written as f(x)dx = g(y)dy, then it is possible to "separate" the variables and solve the differential equation. However, I haven't even been able to convince myself of the 1-dimensional substitution rule for integrals.

For example:
u = 2x+6
du = 2xdx

I hate to tell you this but that is incorrect. If u= 2x+ 6, then du/dx= 2 (the "slope" of the straight line) so du= 2dx.

If u= x², then du/dx= 2x so du= 2xdx.

I have a hard time seeing why I can just "solve for dx", replace that in the integral, and go on my way integrating with respect to u. I just think that there is a lot hidden in the method that is not documented well in calculus textbooks.

Well, what is the definition of "differential" in your textbook?

Another problem: I'm reviewing a little bit of differential equations, and I've stumbled across equations with homogeneous coefficients. The details aren't necessary, but as is easily proven, the substitution y = ux eases the problem. However, with this substitution, it "follows" that dy = udx + xdu. This clearly follows from the product rule, but I'm confused as to how it's possible to replace the dependent variable (y) with two independent variables (u and x). And of course, the nature of differentials is still a mystery to me.

If y= x cos(x) then I can define u to be cos(x): y= xu. you can define u or any other new variable any way you like.

Anybody willing to put me out of my misery and explain?
Thanks!

Your problems seem so fundamental that I not sure what I said will answer your questions.

 

[LukeD]

Unfortunately, the problems that you have are actually fairly difficult ones that really can't be explained easily in a single post (or even, probably, an entire thread). To understand what's going on here, you really should take a course in undergraduate Real Analysis.

If you think you're up to the challenge, try looking for a copy of Tom Apostol's "Calculus", which will get you started and explain about 95% of what you've asked. You should be able to attack this after a fair background in Calculus I and II.

For the other 5%, you'll need to spend another semester or two studying some harder analysis. I suggest C. C. Pugh's book "Real Mathematical Analysis" (some other people will suggest Walter Rudin's "Principles of Real Analysis". This will be much more difficult to understand than the first 95% though and will take a good deal of effort, but it will eventually answer all of your questions and much more.

 

[scorpion990]

HallsofIvy:
Sorry about the first part... I don't know how I managed to mess up differentiating a polynomial... Anyway... My calculus book defines the differential dx as an independent variable. It then defines the differential dy as the depedent variable such that: dy = f'(x)dx. I understand the definitions... After all, I'm allowed to study any system I wish to define.

I'm not having trouble with the definitions. I'm having trouble applying the definitions to solving practical problems, such as changing the variables on a single integral or solving differential equations. These methods are NOT definitions, and they do NOT follow from the definitions directly. They require theorems. After all, I can define the differential dy as dy = f''(x)dx, but the substitution rule for single integration will not follow from this definition. In my experience, calculus notation was created to make it seem like the entire subject is simpler than it really is. At the end of the day, you cannot rely on clever notation to prove theorems.

 

[scorpion990]

Unfortunately, the problems that you have are actually fairly difficult ones that really can't be explained easily in a single post (or even, probably, an entire thread). To understand what's going on here, you really should take a course in undergraduate Real Analysis.

If you think you're up to the challenge, try looking for a copy of Tom Apostol's "Calculus", which will get you started and explain about 95% of what you've asked. You should be able to attack this after a fair background in Calculus I and II.

For the other 5%, you'll need to spend another semester or two studying some harder analysis. I suggest C. C. Pugh's book "Real Mathematical Analysis" (some other people will suggest Walter Rudin's "Principles of Real Analysis". This will be much more difficult to understand than the first 95% though and will take a good deal of effort, but it will eventually answer all of your questions and much more.

I figured that I'd need to look into Real Analysis. Unfortunately, I don't have enough time this summer to look into it. I'll probably take a look at those books and work through one of them during winter break. Thanks! :)

 

[sour_kremlin]

Have you looked into Hyperreals? They provide an extremely intuitive (and correct!) way of working with differentials. With some calculus background it shouldn't take you more than a day to get the hang of either.

http://www.math.wisc.edu/~keisler/calc.html

 

[john456]

Hi !

If your having understanding differentials it is because they really are quite
difficult to understand and quite counter-intutive and they involve infintesimals
which are not rigourously defined in introductory calculus texts.

Let's go back to basics say we have a straight line

y = 2x

x1 = 1, y1 = 2
x2 = 4, y2 = 8

gradient m = y2 - y1 / x2 - x1

8 - 2 / 4 - 1 = 6/3 = 2

This is simple. The gradient is simply the slope of a straight line.

But what about finding the gradient of a curve. Now the gradient
of a curve is not constant but is changing. That is why the gradient is
also called a rate of change.

In order to calculate a function called the derivative which will tell
us the gradient of a curve at any point with have to start using
infintesimals and take a limit as change_in_x goes to -> 0

Now in a straight line


change_in_y = dy
change_in_x = dx

in a curve

change_in_y NOT = dy
change_in_x NOT = dx

Why ?

Because for a curve you are talking about an infinitely small quantity which
are called infintesimals and so ordinary arithmetic breaks down strictly
speaking but don't worry we can still calculate the answer it just needs
some adjusting on our part about how we think about things.

change_in_y / change in x - > 0 = ?


So now I will tell you what this all means


The derivative of x^2 = 2x

and so the equation dy/dx = 2x is a short hand expression for the complete
limiting process of taking the tangent line of the curve at that point and
shrinking the secant line. It is quite abstract and so you should take
a lot of time thinking about it because it is the key to calculus.


John

 

[john456]

You say that you don't need to use subsitution in many cases you can
just integrate by figuring out how the integrand fits in with the chain rule.

This is very true you can do that.

Then why use subsitution ?

Because integration is a pain in the *** and it is simply easier to do that.
Sometimes people don't want to think too hard.

John

 

[john456]

Here is something ripped from the web which talks about the problem

" One of the most puzzling ideas in elementary calculus is that
of the of the differential. In the usual definition, the differential
of a dependent variable y = f(x) is given in terms of the differential of
the independent variable by dy = f0(x)dx. The problem is with the
quantity dx. What does ”dx“ mean? What is the difference between

x and dx? How much ”smaller“ than
x does dx have to
be? There is no trivial resolution to this question. Most introductory
calculus texts evade the issue by treating dx as an arbitrarily small
quantity (lacking mathematical rigor) or by simply referring to
dx as an infinitesimal (a term introduced by Newton for an idea that
could not otherwise be clearly defined at the time.)
In this section we introduce linear algebraic tools that will allow us to
interpret the differential in terms of an linear operator."


John

 

[snipez90]

Well as the above posters have mentioned, you probably won't get a rigorous treatment of differentials until you take delve into theoretical calculus/analysis. But having said that, there is a very good book that discusses at length the intuitive notion of the differential and builds calculus up through differentials entirely (whereas most books use the limit approach)

It's called Calculus for the Practical Man but it's fairly old. This link should work:

http://books.google.com/books?id=1L...a=X&oi=book_result&resnum=4&ct=result#PPA5,M1

 

[john456]

If your still having trouble with differentials don't feel bad about it. I know quite a bit about calculus but differentials still puzzle me. They are highly mysterious.


John