# During U-substitution, is the $du$ that we use as the infinitesimal in $\int f(u) \ du$ changing size?

I am familiar and comfortable with the following definition of an integral (defined as the the Riemann Sum where n approaches infinity). I have circled in red two terms to emphasize a concept that I was taught: the $$\frac{b-a}{n}$$ term effectively becomes the $$dx$$ term. This intuitively makes sense to me.

Additionally, I am familiar with the following statement (as expressed here: How does $u$-substitution work?) However, I have boxed in red something that I find rather confusing when trying to interpret it through the understanding that $$\frac{b-a}{n}$$ sorta kinda equals $$dx$$...

Specifically, the $$du$$ value appears to be changing throughout the interval $$[ g(a), g(b)]$$. To reframe this in the "Riemann Sum" version,we would have something like: $$\text{ "value that varies depending on where you are"} * \frac{g(b)-g(a)}{n}$$

This is very confusing to me...because, if this is the correct interpretation, I'm not quite certain I understand why this is allowed.

• "the $\frac{b-a}{n}$ term effectively becomes the $dx$ term." such description in a calculus textbook is informal and should not be taken as a mathematical fact for rigorous reasoning. The expression $\int_a^bf(x)\,dx$ is a whole thing and $dx$ alone is "meaningless" unless one is talking about differential forms. – Jack Jul 2 '19 at 14:28
• The "Riemann sum" definition is about definite integrals while the u-substitution excerpt is indefinite integrals, which are not defined by Riemann sums. If one wants to see how the Riemann sum changes in the change of variable formula, then I believe one should stick to the context of definite integrals. – Jack Jul 2 '19 at 14:34

You seem to be under the impression that one has to have one fixed $$\Delta x$$, i.e. a uniform subdivision of the interval. This is a very common misconception and the full definition of what a Riemann sum can be is given in the answer by LutzL.

If you have a Riemann sum with a uniform spacing then a u-substitution (performed at the level of the sum) will lead to a non-uniform spacing so you will have a $$\Delta x$$ that varies across the interval. For example consider $$\int_0^1 2xe^{-x^2}{\rm d}x$$ and let's use a uniform division of the interval $$x_i = \frac{i}{n}$$ to get $$R_n = \sum_{i=0}^{n-1} 2x_i e^{-x_i^2} \Delta x_i$$ where $$\Delta x_i = x_{i+1}-x_i \equiv \frac{1}{n}$$ is the same across the interval. In a u-substitution we will put $$u_i = x_i^2$$ to get $$R_n = \sum_{i=0}^{n-1} e^{-u_i} \Delta u_i \cdot \frac{x_i+x_1}{x_i+x_{i+1}}$$ where now $$\Delta u_i = u_{i+1} - u_i = \frac{2i+1}{n^2}$$ which varies from point to point. The last factor above will approach unity as $$n\to\infty$$ so for large $$n$$ this sum is just $$\simeq \sum_{i=0}^{n-1} e^{-u_i} \Delta u_i$$ which is a Riemann sum for the integral $$\int_0^1 e^{-u}{\rm d}u$$.

The largest spacing is found close to $$u=1$$ where $$\Delta u_{n-1} = \frac{2n-1}{n^2} \sim \frac{2}{n}$$ compared to the spacing close to $$u=0$$ where $$\Delta u_0 = \frac{1}{n^2}$$. However as $$n\to\infty$$ note that this largest spacing does go to zero and this is all we need for the Riemann sum to converge to the value of the integral.

Perhaps it's more instructive to show a figure. Recall that a Riemann sum (before taking the limit) is nothing but a finite sum of rectangles that approximate the (signed) area below the curve. Here is one example of such a sum where the spacing varies from point to point: If we consider a finer subdivision of the interval, but still non-uniform, we generally get a much better approximation: You can from this convince yourself that for approximating the area well it doesn't matter that some regions have a larger spacing than others as long as the largest spacing is "small enough". This is the basis of the general definition of a Riemann sum.

• This is a great illustration! I do have (at least for now) one follow up question, though: for the expression du=g'(x) * dx : it seems like it would be possible for du's to take on negative values. What does that entail? I do not know what a "negative subdivision" would actually mean. I know what a negative area means (i.e. when part of the function dips below the x axis...as does the 2nd hump of your curve) but I am uncertain of this negative subdivision. – S.Cramer Jul 2 '19 at 15:21
• @S.Cramer Let's take an example. Consider $\int_0^1 1{\rm d}x = 1$. If you perform a u-subsitution $u=1-x$ then you end up with the integral $-\int_1^0 1 {\rm d}u$ which is also $1$ since $\int_1^0 1{\rm d}u = -1$. This integral is negative since we integrate "right to left" instead of "left to right". You can see this at the level of a Riemann sum as the point going "left to right" (say $x_i = \{0,0.5,1.0\}$) now become going "right to left" ($u_i=\{1,0.5,0\}$) so we have negative $\Delta u_i$. – Winther Jul 2 '19 at 15:35
• Just think of it as a convention. For the integral to denote the area we have to integrate "left to right" otherwise it's the negative of the area we get from the integral. – Winther Jul 2 '19 at 15:39
• That example makes perfect sense. However (and I’m sure you chose this for simplicity), my confusion arises for a g’(x) that is only SOMETIMES negative as we advance left to right. (i.e. the sign of g’(x) is alternating between positive and negative across the interval of interest). I feel like that would create a situation where you are “taking steps forward and then taking steps backwards” during the integration procedure. I guess I would just choose a smaller interval where this behavior does not occur? – S.Cramer Jul 2 '19 at 15:42
• @S.Cramer : Yes, exactly that. Look at the usual assumptions of the substitution rule, the function $g$ has to be monotonously increasing or falling over the integration interval in question. After that you can add areas for different monotonicity of $g$. – Lutz Lehmann Jul 2 '19 at 18:22

Actually, the (original) Riemann sum is $$\sum_{i=0}^{n-1}f(\xi_i)(x_{i+1}-x_i)$$ where $$a=x_0 is a subdivision of the integration interval and $$\xi_i\in[x_i,x_{i+1}]$$ are some midpoints (in the original text, the $$(x_{i+1}-x_i)$$ were supposed to be infinitesimals).

The Riemann integral is then the limit over these subdivisions and all selections of midpoints when the maximal length of a sub-interval in the subdivision goes to zero. Thus the inhomogeneous subdivision after application of the substitution rule is the general case, a subdivision by a strict arithmetic sequence is the exceptional case. At an elementary practical level, there is more use for a geometric sequence $$x_k=aq^k$$, $$q>1$$, $$x_n=b$$, in the subdivision.

• Not quite sure what you mean regarding your last paragraph. I’m assuming by “inhomogeneous subdivisions” you’re referring to the “changing” du size. But all comments after that don’t mean anything to me. Could you please clarify? – S.Cramer Jul 2 '19 at 14:26
• Yes. Equal sized $\Delta x$ would be homogeneous or an arithmetic sequence of subdivision points. And inhomogeneous subdivision refers to non-constant $\Delta u$. What you write as Riemann sum is not general enough, there are examples where this sum converges that are not Riemann integrable, like the Dirichlet comb. – Lutz Lehmann Jul 2 '19 at 14:34
• @LutzL: +1 for mentioning Dirichlet comb. – Paramanand Singh Jul 2 '19 at 14:43
• So is the du in the context of du=3*dx “bigger” than the du in the context of du=2*dx ( if for example we had a function whose g’(x) equaled 3 and 2, respectively) – S.Cramer Jul 2 '19 at 14:50
• That depends on what $dx$ is. And possibly on what the meaning of "is" is. It could also mean that one of the $dx$ is smaller. Standing alone, there is not much to say about infinitesimals $dx$ or $du$. But of course, $2dx<3dx$ if $dx>0$. – Lutz Lehmann Jul 2 '19 at 16:40

$$\int_a^bf(x)dx$$

Where I now introduce the variable $$u=x^2$$. Then $$du=2x dx$$ and is now explicitly a function of $$x$$. The confusion is arising because in the definition of the integral, it is correct that $$dx=\lim_{n\to\infty} (b-a)/n$$ in the integral, but if you make a variable change then this definition no longer applies. In the case that you have presented, you might pretend a variable change was made, and that is how the infinitesimal is a function of $$x$$. In my example du=\lim_{n\to\infty} 2x(b-a)/n\$

To respond to your comment below, let's consider a real example. Suppose $$f(x)=x^2$$, which is easy to integrate. Now, lets choose $$a=0$$ and $$b=1$$. I assume that you are familiar with integration, so we have

$$\int_0^1x^2dx$$

which I'm sure you know how to integrate,

$$=\left.\frac{1}{3}x^3\right|^1_0$$ $$=\frac{1}{3}$$

Now let's try the same integration, but with $$u=x^2$$. This will be the exact same integral, so we need to get the same result or we've done something wrong. Now, $$f(u)=u$$, and $$dx=2xdu=2\sqrt{u}du$$. $$f(0)=0^2=0$$ and $$f(1)=1^2=1$$ still. Now we have

$$\int_0^1u^{\frac{3}{2}}du$$

Certainly it is the case that

$$du=\lim_{p\to\infty}\frac{1-0}{p}$$

is useful in this case, but in relation to $$x$$ we have

$$du=2x\lim{n\to\infty}\frac{1-0}{n}$$

The confusion falls out because technically $$du=0$$ and $$dx=0$$, because of course they are the limit in the form

$$\frac{1}{\infty}$$

which of course, goes to zero. But in fact, we must add up each individual piece to get a non-zero value because we know the area under these curves are non-zero. It is nice, intuitively, to think of

$$dx\equiv \lim_{n\to\infty}\frac{1-0}{n}$$

but it is clear that if you try to use that definition as an axiom, you're going to encounter difficulties. The truth is that infinitesimals do not really make sense outside of integrals and derivatives. This is because infinitesimals, when outside of their natural habitat, are just zero.

The short response to your comment is that, we changed the definition, so of course the definition changed. That should be obvious. If I said $$f(x)=x^3$$, and then changed my mind and decided $$f(x)=x^{17}$$, you should understand that a plot of these two functions should be different, because the definition was changed.

• Why does the definition, “no longer apply”. I feel like it is a pretty weak definition if it is only situationally applicable. – S.Cramer Jul 2 '19 at 0:55
• I appreciate the edit, but this still seems very hand wavey to me (especially the du=g’(x)dx part...in terms of how the algebra works out and then doesn’t actually effect the Riemann Sum at all) I will write another question that hopefully clarifies some of my confusion. – S.Cramer Jul 2 '19 at 12:35