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Integration by substitution is defined as something like

$\displaystyle\int_a^b f(\phi(t))\phi'(t)dt = \int_{\phi(a)}^{\phi(b)} f(x)dx$

But for my taste, the variables $x$ and $t$ are exactly switched! It's still correct since it's just a name, but why is this written in the exact counter intuitive way? I Feel like maybe I'm missing something. To illustrate, an example:

$\displaystyle\int_{x=0}^{x=2} x \cos(x^2+1)dx$

Then we have $t=ϕ(x)=x^2+1$ and therefore $\frac{dt}{dx}=2x$ or $dx=\frac{dt}{2x}$. So we have

$\displaystyle\int_{\phi(0)}^{\phi(2)} \frac 1 2 \cos(t)dt = \int_{t=1}^{t=5} \frac 1 2 \cos(t)dt = \frac 1 2 (\sin 5 - \sin 1)$

Which is all correct and wonderful. But why on earth would you name the variables $t$ and $x$ exactly the opposite way of how they would appear in almost every situation?

MJD
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Basti
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    You seem to bring with you some peculiar conceptions about the uses of some specific dummy variables. Be aware that these are far from being universally shared. – Did Aug 14 '14 at 14:03
  • Not universal, but it is defined in this manner in 3 different books I looked at and Wikipedia (both English and German editions) – Basti Aug 14 '14 at 14:12
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    I don't really understand why you feel it's backwards. Can you explain? Also, I see $x$ written as a function of $t$ somewhat more often than I see $t$ written as a function of $x$. –  Aug 14 '14 at 14:20
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    The typical use would be like my example, where you have a function f(x) in which you want to substitute a part. When you go through, you end up with an integral of t (or u). Yet in the "definition", the original integral is over t and the result is over x - the exact opposite of what would be intuitive – Basti Aug 14 '14 at 14:28
  • As far as I can tell, you would find this whole calculation easier to understand if the roles of the variables $x$ and $t$ were interchanged. Can you indicate why the choice of variables makes a difference for you? For myself, I see no difference; it wouldn't matter to me if the variables were Chinese characters (as long as I can easily distinguish them visually). – Andreas Blass Aug 14 '14 at 16:20
  • It doesn't, it's completely the same whatever you call them. But of all the characters (chinese or others) in the world, why would you chose the exact two which actually appear in the calculation, but use them in the exact opposite way? – Basti Aug 14 '14 at 17:36
  • See my answer at http://math.stackexchange.com/questions/627926/understanding-integration-techniques. – Martín-Blas Pérez Pinilla Aug 16 '14 at 14:24

3 Answers3

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Since forever we have studied functions with the idea of the variable $x$ and write $y=f(x)$ for the graph of $f$ in the coordinate plane. The next common letter for a dummy variable is $t$, possibly from physic for time.

Similarly, when study integration of a function, it's common that we write the integration as $\int f(x)dx$. And most formulas/exercises are given in this form (power law, exponential law,...). It's up to the writer's taste to choose the notations.

In this case my guess is that when he wrote the Substitution Law, he had the picture of the final integral in mind. That is, the last integral that the students/readers would encounter in the computation process.

I would like to add that many other books choose different notations. For example $$\int^b_a f(u(x))u'(x)dx= \int^{u(b)}_{u(a)}f(u)du.$$ This choice puts the original integral first which in some cases helps the students recognize the substitution patterns faster. Again, it's totally personal taste and I wouldn't take that too seriously.

Quang Hoang
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The variables $x$ and $t$ are dummy variables. This means, essentially, that we can rename $x$ and $t$ without changing anything of substance. You might as well write the change of variables formula as $$ \int_a^b f(\phi(x))\,\phi'(x)\,dx = \int_{\phi(a)}^{\phi(b)} f(t)\,dt. $$ (Does this answer your question? I'm not sure that I know what you're asking.)

user134824
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  • I know! Sure they are dummies. But of all the letters of the world, why does everyone chose the exact two names to maximize confusion? I just thought that I might have misunderstood something, hence my question. – Basti Aug 14 '14 at 13:59
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If $\int_a^b f(x)dx$, then you set $x=g(y)$ then if $x$ move from $a\to b$ then $y$ move from $g^{-1}(a)\to g^{-1}(b)$ and $dx=dg(y)= \frac{dg(y)}{dy}\cdot dy=g'(y)dy$ then $$\int_a^b f(x)dx=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(y))g'(y)dy$$

I hope it answer at your question.

idm
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