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Solve: $$\int x(2-3x)^{11} \, dx $$

The book I am following uses a weird technique to solve this. I am having trouble understanding why it works.

Let: $$ u = 2-3x$$ $$ x = \frac{2-u}{3} $$ $$ \frac{du}{dx} = -3 $$

Book Technique: $$dx = - \frac{du}{3}$$

$$ \int \frac{2-u}{3} u^{11} - \frac{du}{3}$$ $$ -\frac{1}{9} \int 2u^{11} - u^{12} du$$

My problem with this is that dy/dx is not a fraction but the limit of one as such the terms dy and dx on there own are in a sense meaningless and cannot be manipulated algebraically. For this reason, I solved it with a more formal technique:

$$ \frac{dy}{du} = (\frac{2-u}{3})(u^{11}) = \frac{2u^{11}}{3} - \frac{u^{12}}{3} $$

Given, the reverse chain rule (i.e u-substitution)

$$ u = g(x), y = f(u) $$ $$ \int \frac{du}{dx} \frac{dy}{du} dx = \int \frac{dy}{du} du$$

so:

$$ -\frac{1}{3} \int -3 x(2-3x)^{11} dx = \int \frac{dy}{dx} du $$ $$ -\frac{1}{9} \int 2u^{11} - u^{12} du $$

My question is thus:

The technique used by the book is at most an approximation, i.e by assuming that $dy/dx$ is fraction we are taking an assumption that will not always hold, but in this case it seemingly does. Is this correct way of looking at why the 'book technique' works?

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    $\begingroup$ When you do $\dfrac{\mathrm du}{\mathrm dx}\dfrac{\mathrm dy}{\mathrm du}\mathrm dx=\dfrac{\mathrm dy}{\mathrm du}\mathrm du$, aren't you too treating the $\dfrac{\mathrm du}{\mathrm dx}$ as a fraction since you cancel out the $\mathrm dx$'s ? $\endgroup$ – Prasun Biswas Oct 7 '17 at 12:35
  • $\begingroup$ This answer to a similar question has some links that may help: : math.stackexchange.com/q/2009181 $\endgroup$ – Ethan Bolker Oct 7 '17 at 12:37
  • $\begingroup$ It's actually rigorous enough to treat it as a fraction. In fact, that is one of the advantages of Leibniz's notation for derivative. $\endgroup$ – Prasun Biswas Oct 7 '17 at 12:38
  • $\begingroup$ @PrasunBiswas, 'canceling out' the dx's is not how the chain rule is derived, the fact that they cancel out is merely a coincidence of the notation. $\endgroup$ – Joe Oct 7 '17 at 12:41
  • $\begingroup$ $\int x(2-3x)^{11}\times dx$ If substituting u then =$\int \dfrac{2-u}{3}\times u^{11}\times\dfrac{du}{-3}$ $\endgroup$ – Fawad Oct 7 '17 at 12:42
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So the question is why we have $$\int f(x) dx = \int f(x(u))\frac{dx}{du}du$$ where $x(u)$ is an invertible function?

Let $$F(x)=\int f(x)dx$$ and $$G(u)=\int f(x(u))\frac{dx}{du}du$$

By definition of indefinite integral (anti-derivative), chain rule and inverse function theorem, we have

$$\frac{d}{dx}G(u(x))=\frac{dG}{du}\frac{du}{dx}=f(x(u))\frac{dx}{du}\frac{du}{dx}=f(x)$$

Hence $$G(u(x))=\int f(x)dx =F(x)$$

In other words we have $$\int f(x) dx = \int f(x(u))\frac{dx}{du}du$$ and symbolically it is as if the $du$ cancels.

From the above one can see that the origin of the "cancellation" comes from $$\frac{du}{dx}\frac{dx}{du}=1$$ But this is not due to "cancellation" but is by inverse function theorem.

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  • $\begingroup$ This is the first proof of integration by substitution I'm actually content with. I was looking for one years ago and only found overly complicated ones. Thanks :) $\endgroup$ – Jam Oct 7 '17 at 21:34
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The technique used by the book is correct and can be justified. This is done, for instance, in Spivak's Calculus at the chapter Integration in elementary terms.

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  • $\begingroup$ Thanks, I will look at this chapter, maybe I can gain some insight which I am currently lacking. Cuz if the technique is correct, this implies that dy/dx is a fraction no? $\endgroup$ – Joe Oct 7 '17 at 12:45
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    $\begingroup$ @Joe A fraction can be manipulated in certain ways, and a lot of the time (although not always) dy/dx can be manipulated in the same ways, but the justification for each is different - dy/dx is not a fraction, it just sometimes behaves like one. $\endgroup$ – acernine Oct 7 '17 at 15:17
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    $\begingroup$ @acernine ... and the fact that it often behaves as one makes many calculations intuitive that would be clumsy and hard to understand when written out rigorously. Train your intuition to use $dx$ and $dy$ separately as well as in the form $dy/dx$. $\endgroup$ – Ethan Bolker Oct 7 '17 at 21:25

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