FAQ 
Calendar 


#1




Finding an Integral Using usubstitution
Please explain finding an integral by usubstitution to me like I am a 5year old. I can do it, but I don’t remember why it works. Someone explained it to me 25 or 30 years ago but I haven’t thought about it since and now I don’t have a clue. Let me use the following example: After finding the appropriate u substitution you come up with the expression du/dx = 2x^{2}. Then you are supposed to say that therefore du = 2x^{2} dx. What the heck? I know it works, but how do you justify this? du/dx is not a fraction! Neither du nor dx is a number! What is actually going on here?
I suppose I could google it or try to figure it out myself but why waste this opportunity to be teased and ridiculed by the brilliant minds on the SDMB. If a coherent discussion develops, I am pretty sure there will some talk about limits. 
#2




First you will seldom go wrong in treating a first derivative, du/dx, as a fraction (Just don't cancel the d's). Treating du and dx as separate and separable objects is pretty much the mathematics of infinitesimals.
But without going into detail think of the operation of integrating of f(x). We write that as (integral sign) f(x) dx. The dx here simply indicates the variable you are integrating with respect to so we can distinguish int [ x^{2} y] dx = (1/3) x^{3} y int [ x^{2} y] dy = (1/2) x^{2} y^{2} so think of your problem du(x)/dx = 2x^{2} as define f(x) = du(x)/dx = x^{2} Then integrating f(x) is the same as integrating du(x)/dx nad you never have to worry about whether or not you're multiplying the dx out. Last edited by OldGuy; 05182020 at 11:21 PM. 
#3




By "usubstitution", it seems that you just mean that instead of integrating f(x) dx, you just define u = u(x) to be some function that will end up making your integral simpler; practice will help a lot in finding a correct substitution.
As for the notation, it suffers from a bit of abuse in that du/dx is not a fraction, and you should not think of it as du divided by dx, as you point out, but on the other hand recall that dx and du are not numbers but rather differential 1forms, so that you are able to integrate not f(x) but f(x) dx. The quantities with "d" in front of them are also variables you need to take account of and substitute for when changing coordinates. The formula you need to understand is that if u = g(x), then du = g'(x) dx. In the above notation, that says du = (du/dx) dx, which looks tautological, but it's not, really, at least not due to elementary arithmetic. As for your example, if du = 2x^{2} dx, what you need to do is write your original integral in terms of u and du instead of x and dx, possibly using the inverse dx = (1/2x^{2}) du if you need it. Then, once the x's are gone, you can pretend u is the independent variable and integrate as usual. Sorry for the terse explanation; please ask if anything needs elaboration. 
#4




Quote:
The problem of finding antiderivatives can be formulated without integral signs and differentials, but on the other hand one must get used to the notation everybody uses. 


#5




Quote:
The proposition that "if u = g(x), then du = g'(x) dx" is what is giving me trouble. Apparently it is true; people have been using it for hundreds of years. Can you give a simple proof that my tiny brain can comprehend? I can use the fact that "if u = g(x), then du = g'(x) dx" to evaluate an integral. That's not the problem. I just want to know how we can be certain that it is true. 
#6




Quote:
And the great 12thcentury Indian mathematician Bhaskara II has a memorable and relevant quote shown at the top of this page in Kim Plofker's book. 
#7




Basically, the changeofvariables formula boils down to the chain rule for the derivative of a composite function: if u = g(x) and y = f(u), then dy/dx = dy/du ⋅ du/dx.
Quote:
Quote:

#8




septimus, would you mind quoting that statement for us? Your Google Books link shows only a single sentence on the page you linked, in an Indian language that I can neither read nor recognize.
Quote:

#9




Quote:
"You arrive at a page that cannot be viewed. Or the restrictions on viewing this book have been reached." Neither memorable nor relevant, I'm afraid. 


#10




Quote:
The Chain Rule says, for example, that since the derivative of x^{3} (with respect to x) is 3x^{2}, the derivative of u^{3} (with respect to x) is 3u^{2} * u'. That is, the derivative of (something)^{3} is 3(something)^{2} times the derivative of the something. The derivative of (10x^{5} + 7x + 93)^{3} would be 3(10x^{5} + 7x + 93)^{2} * (50x^{4} + 7). So then the antiderivative of 3(10x^{5} + 7x + 93)^{2} * (50x^{4} + 7) would be (10x^{5} + 7x + 93)^{3}. And usubstitution is really just a way of making this obvious. There's a lot more I could say about how to actually do it, but that's the basic idea behind it. 
#11




Quote:
“ In other words, the “atthattime” Sinedifference Δ Sin for a given arc α is considered simply proportional to the Cosine of α: Δ Sin = Cos α * (225/R) It has been noted that this and related statements reveal similarities be tween Bh ̄askara’s ideas of motion and concepts in differential calculus. (In fact, perhaps these ratios of small quantities are what he was referring to in his commentary on L ̄ıla ̄vat ̄ı 47 when he spoke of calculations with factors of 0/0 being “useful in astronomy.”) This analogy should not be stretched too far: for one thing, Bha ̄skara is dealing with particular increments of partic ular trigonometric quantities, not with general functions or rates of change in the abstract. But it does bring out the conceptual boldness of the idea of an instantaneous speed, and of its derivation by means of ratios of small increments.“ 
#12




Quote:
I know how usubstitution works. I just want to know why it works. In fact, I stumbled upon with this question when preparing training on solving differential equations using the separation of variables technique. It's really the same question but in a slightly different context. I just used the usubstitution context because I thought more people would be familiar with it. I wanted to be ready with a short simple answer in case some trainee asked the same question that I asked in this thread. Who knows, maybe one of the trainees could be a pedantic ***hole like me and just want to ask that question so that they could watch me squirm. (More likely though, I would get questions like "What's a differential?" or "Couldn't you just google it?".) In looking for a simple answer, I have found several sources that resort to what I consider mere hand waving such as "using some 'informal' algebra", "treating dx as if it were a number even though it's not", etc. I have even seen some weasel words in this thread such as "you will seldom [emphasis added] go wrong ... ", "you can pretend ... ", and "... moreorless." One more thing. No one in their right mind wants to actually watch me physically squirm. 
#13




Now, the slightly more modern point of view is that if you have a smooth space, which could be Euclidean space but also something curved like a differentiable manifold, you can zoom in to an infinitesimal neighborhood of a given point by considering the tangent space at that point, one way of looking at which is that curves passing through the point represent tangent vectors. A tangent vector applied to a function will give the corresponding directional derivative of the function at that point. Usually this kind of stuff is introduced when you study multivariable calculus, not in a firstsemester course.
Furthermore, at any point you can also consider the cotangent space. Eg a smooth function equal to 0 at a given point represents a cotangent vector. So now if you start with any smooth function f, its differential df will give a cotangent vector at any point, therefore it is a differential 1form. Anyway, the advantage of this point of view is that it provides a rigorous way of the interpreting intuitive calculations with infinitesimals which is independent of a particular choice of coordinates, and can be generalized. 
#14




Quote:
Anyway, back to the beginning, when you write "dx" it is not a "number" (then again your coordinate function x is not a mere "number" either, rather at any point you may evaluate it to get a number). Like x, dx may be evaluated at any point and gives a cotangent vector, which is not a "number" but some linear map, but in terms of your fixed coordinate x (which gives a resulting coordinate dx on the cotangent space) the linear map is represented by a number, namely the derivative of the function at that point. Think of f(x) = f(a) + f'(a) (x  a) + ... df (evaluated at a) = f'(a) dx (evaluated at a) is the infinitesimal/differential version of that, and this is abbreviated by the formula df = f'(x) dx 


#15




Quote:
The short answer you can tell people seems to be that both "x" and "dx" are coordinates, so that when you change coordinates and write your integral in terms of the new coordinates, you need to substitute for both of them. Or, more basically, when finding antiderivatives you are using the Chain Rule the trainees probably remember that name. 
#16




I just found this video that does not answer my question but at least it kinda, sorta, maybe asks my question.
https://www.khanacademy.org/math/ap...algebraically 
#17




Quote:
But we must admit that the notation IS potentially confusing: dy/dx is not a fraction, and yet we have these things dx as well as d/dx defined, not to mention the curly d's, stuff like dx and (dx)^{1/2}, dx dy as well as dx ^ dy, you name it... 
#18




Quote:
We define ∫ f(x) dx to mean "the general antiderivative of the function f(x) with respect to the variable x"; and we define du to be "du/dx * dx" (that is, the derivative of u with respect to x, times dx) without ever really defining what dx by itself means. These definitions then give us a convenient way of doing what we need to do with integrals or separable differential equations. 
#19




Just calling it another name for the chain rule seems incomplete to me. Otherwise, I couldn't do more complicated substitutions, like this (admittedly contrived) example:
Say I want to compute: ∫3x^{5} dx Start with this substitution: u^{2} = x^{3} Take the derivative of both sides: 2u du = 3x^{2} dx Note that the original integral is one part of each: ∫3x^{5} dx = ∫x^{3} * 3x^{2} dx = ∫u^{2} * 2u du = ∫2u^{3} du Integrate the normal way (ignoring the +C for convenience): u^{4}/2 And substitute back to x: (x^{3})^{2}/2 = x^{6}/2 That is of course the same as what we'd get if we did it the normal way in the beginning. 


#20




Quote:
Now on to integration. Remember that when I write Int(f'(x)dx)=f(x)+c I am calculating the area under the curve f'(x) by taking little tiny slices of width dx and height f'(x) and adding their areas together. to make thing concrete I can chop the region a to b up into 10^1000 equal slices of width dx=10^1000 resulting in f(b)f(a) Now suppose I am interested in calculating Int(f(g(x))g'(x)dx) from a to b. I can take dx=10^1000 equal width bars along the x axis and add their areas together. This will be the sum of all (ba)*10^1000 regions each with area f(g(x))g'(x)x10^1000 Next I am going to consider a different graph where the height is f(g(x)) with the same set of bars, but now I am going to relabel the xaxis from x to u=g(x), so that the curve now has height f(u) and extends from u=g(a) to u=g(b) so if for example g(x)=2*x^2 then the point at x=0.1 will be replaced with the value u=0.02 and the value x=0.2 with be replaced with the value u=0.08 and the region will extend from 0 to 2. If I use the same bars as before, they will have height f'(g(x)) and width in terms of u equal to du=g'(x)dx=g'(x)x10^1000, so their total area will be the sum of the (ba)10^1000 regions with area f(g(x))g'(x)x10^1000 the same as my original graph. But it is also the graph of f'(u) vs u from the limits g(a) to g(b), and has area equal to Int(f(u)du) from g(a) to g(b). So Int(f(g(x))g'(x)dx) from a to b = Int(f(u)du) from g(a) to g(b) Last edited by Buck Godot; 05192020 at 05:02 PM. 
#21




Quote:
U^2 = X^3 => U = +/ X^ (3/2) 
#22




This is the part that I don't know how to explain. How did we get from g'(x)=du/dx to du=g'(x)dx? Is there a rigorous proof for this? Is there a simple explanation that could be used for someone that knows little or nothing about derivatives and differentials that doesn't involve "handwaving"? DPRK has had some suggestions but I'm afraid my audience would just return a blank stare. I have all summer to work on this.

#23




Are you familiar with infinitesimals? Willing to use them?
They provide a perfectly rigorous explanation for your questions, and backed the original intuition behind calculus for both Newton and Leibniz. However, the existence of the infinitesimals was not rigorously established until the 60s. This is why most calculus pedagogy uses the (ε, δ) limit form, with all the baggage that entails (such as the claim that dy/dx is not a fraction, even though Leibniz certainly thought it was). You aren't going to prove the existence of the infinitesimals in an introductory classbut you aren't going to establish the real numbers either. So that doesn't seem like a big deal to me. 
#24




I know that they are rather small. Actually they are really, really, really, really small. They may be even smaller than that.



#25




Then you are halfway there. The only other thing to remember is that they are not quite zero.

#26




Quote:
You can either define "du" to mean "the derivative of u with respect to x times dx," or you can define it to mean something else and then prove/explain why that something else is equivalent to g'(x)dx. 
#27




Quote:
I think it may be easier to think of a concrete example where there are actual physical quantities associated with x and u. Let us suppose that I am looking at an accelerating car and I want to know how far its gone after a particular time. In this case x is time in seconds and u is distance in meters and it happens that u=g(x)=x^2. Suppose I want to know how fast the car is going at a particular instant in time x. One way I can do this is to at a particular time, x, I advance the time just one little tiny tiny smidgen, dx, and see how much the distance changes, lets call this change in distance du. Well the distance at time=x was g(x)=x^2, and the distance of time (x+dx) was g(x+dx). So du defined as the change in distance was g(x+dx)g(x). Now we can work this out and find that du=g(x+dx)g(x)=(x+dx)^2x^2 = x+2xdx+dx^22x = 2x dx+dx^2 Now dx is really really small, so dx^2 is really really really small, so small that you might as well ignore it. So we compute that du=2x dx, and in particular at that instant in time the distance it travels per unit time is du/dx=2x, which by amazing coincidence is g'(x). Now suppose I have a wonky car that is doing all sorts of wonky things. So that its position over time is. u=g(x) = x^35/x+ sin(x) (e^x). I start out the same and get as far as, du=g(x+dx)g(x), and then divide both side by dx. So we get du/dx=(g(x+dx)g(x))/dx, now computing all those differences in a pain in the ass so I'm not going to do that right now. Instead I'm going to just keep things in terms of functions. Now remember that I wanted dx to be really really small, in fact I am going to take the limit as dx goes to zero. du/dx = limit (g(x+dx)g(x))/dx) now to make thigs possibly more familiar I have going to change notation and say h=dx so du/dx=limit (g(x+h)g(x)/h) this right side is the definition of g'(x). So we see that for dx really really small du/dx=g'(x). Note you can't argue with this step since without agreeing that its true we can't discuss g'(x) at all. now I use what I know about differentiation to compute that g'(x)=3x^25ln(x)+cos(x)e^x so if I move time ahead a little tiny smidgen dx the distance will change by du=(3x^25ln(x)+cos(x)e^x)dx 
#28




Quote:
Not sure if the limit even exists as x—>0 and not sure if you can integrate it. Are you sure this is a concrete example ? 
#29




How does Google Books decide which pages to let some user see and which pages produce the "That page is not in this preview" message?
Quote:
Quote:
Quote:
Blame me for the Thai error message (via the 'co.th' URL). Sometimes I manually edit those URLs to be '.com' but Google changes them back to 'co.th' in the browser's locator bar (unless I stay logged in to Google). I guess I figured that since Google changes English to Thai automatically for me, it would do the vice versa for you. (But that's not how Google works. For me it leaves the English 'co.uk' intact; it just changes '.com'.) 


#30




Something to consider:
To the real numbers ℝ you may adjoin an infinitesimal element, let's call it dx, which satisfies (dx)² = 0. This is a function on a formal infinitesimal line segment, a fattened point if you will, just as functions at a single point are given by a real number. dx is a coordinate function, just like x is on the real line. Differential 1forms on the real line now take the form g(x) dx, where g(x) is an ordinary smooth function. g(x) dx is not a function on the real line, rather a function on the space of infinitesimal paths in ℝ, i.e., something of the form f(x) + g(x) dx, with the f(x) = 0 so you just have g(x) dx. Now let f(x) be an ordinary smooth function we wish to differentiate. Define a new function h(x,ε) = f(x+ε) (on the infinitesimal path space, so ε² = 0). The derivative of f will now be obtained via taking h(x,ε)  f(x) = f(x+ε)  f(x). This will be a linear function of ε, namely εf'(x). The corresponding 1form is (df)(x) = f(x + dx)  f(x) = f'(x) dx. In summary, nothing that has not been said already in this thread, including that df = f'(x) dx very easily and that 1forms are functions on an appropriate space (nothing nonrigorous about it). It still remains to make the connection to antidifferentiation and to definite integrals, but the differential forms that appear should not be confusing anybody. Last edited by DPRK; 05202020 at 01:19 AM. 
Reply 
Thread Tools  
Display Modes  

