# In integration notation - why can you multiply du by a number?

When integrating sin(x/2) for example using substitution as in:

$\int \sin(\frac x2)$ $dx$

$\frac x2 = u$, $\frac{du}{dx} = \frac{1}{2} \Rightarrow 2du = dx$

$\int \sin(u)$ $2du$

etc

How is it that the du can be multiplied by 2 here, I thought that the dx represented a 'with respect to x' or a very small value of x. So does this mean that $2du = dx$ means 2 * a really small change in u is a really small change in x? And how come it can be separated from the main part of the integral so you just multiply by two at the end with the du 'disappearing'?

$-\cos(u) * 2$

$-2\cos(\frac{x}{2})$

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This is a good question.

In general we take

$$\int f(x)\; dx$$ as a symbol on its own. In general we can't break down the symbol and assignment meaning to each of the component. So $\int$ isn't a defined quantity. (Of course $f(x)$ makes sense). Likewise, $dx$ on its own isn't defined.

Now, we do have integration by substitution. According to this, as a matter or pure notation we allow for writing $dx$ on its own. It is a result that when we do that, like you have done in your example, then it actually "works".

So we are allowed to treat $du$ and $dx$ as quantities that are defined, and we are allowed to multiply and divide by them when we do integration by substitution. So if $$\frac{du}{dx} = \frac{1}{2}$$ we move ("multiply") by $dx$ on both sides to get $$du = \frac{1}{2}dx.$$ Then we multiply by $2$ on both sides and get $$2du = dx.$$ All this means is that you are allowed to replace the $dx$ in the orignial integral by $2du$. And so you get $$\int \sin(u)\; du.$$ So in this sense we don't in general consider $dx$ and $du$ to have a life outside of the use in integration by parts. (Ok this is not true. There are other places where we write $dx$ on its own).

In all this we don't assign an precise meaning to $dx$. So $dx$ doesn't mean "a small change". Sure we can think about it as such, but to be precise we would have to define what "small change" means.

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I think I understand, du & dx are 'made up' for want of a better term quantities that we use to manipulate. So $\int \sin(u)$ $2du$ is actually ($\int \sin(u)) * (2) * (du)?$ or rather $(\int \sin(u) du) * 2$ –  Jake Burton May 2 '13 at 19:15
Out of interest because I haven't come across it yet, where else is $dx$ written on its own? –  Jake Burton May 2 '13 at 19:17
So you might think of $\int \sin(u)2du$ as $\int\sin(u)$ times $2$ times $du$. But technically it isn't. As I mentioned, $\int \sin(u)2du$ is taking as one symbol. The only thing that you can manipulate is the function. –  Thomas May 2 '13 at 19:17
@JakeBurton: One other place that I can think of where $dx$ shows up on its own is with differentials. Here if $y = f(x)$, then we define $dy$ as $f'(x) dx$. –  Thomas May 2 '13 at 19:18
Makes sense, thanks! –  Jake Burton May 2 '13 at 19:18

This is a case where we are not told the whole story in a first semester calculus course. We are taught that the derivative $f^{1}(a)$ represents the slope of the curve $y=f(x)$ at the point $x=a$. While this is true, another way to look at the derivative is as a linear approximation to $f$ at $a$. This means that if we look at $a$ as the origin in a space tangent to our curve, in this case a line, $f^{1}(a)$ is the slope of the function that takes an input $h$ to $f^{1}(a)h$. In this tangent space, the input variable is not $x$ but $dx$. Essentially, we get a function $dy = f^{1}(a)dx$ in this tangent space.

There are many details that have to be filled in to make this work but this is essentially why we can say that if $u=x/2$, $du = \frac{1}{2}dx$. Since this is just a function like $y=mx$, we are free to algebraically manipulate it.

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So the derivative $f'(x)$ is a function that describes the gradient of a function $f(x)$ but for a given point $a$ it is also the equation/function of a tangent - I'm assuming thats roughly what is meant by linear approximation. But this tangent $f'(a)$ also takes a variable $dx$ which makes it describe $dy$, how does that work? –  Jake Burton May 2 '13 at 19:40
The deriviative is $f^{'}$. $f^{'}(x)$ is the derivative at a point $x$. I used $a$ to try to avoid confusion. If you fix $x$, $f^{'}(x)$ is a number, say $m$. The graph of the function $y=mx$ is the line tangent to the graph of $f$ at $x$. –  John Douma May 2 '13 at 19:42
Say then you have the function $f(x) = x^{2}$ for which $f'(x) = 2x$. So the equation of the tangent at $x = 1 e.g. f'(1)$ is 2 but how does that relate to $dy = f'(1)dx$? Oh I think I see it now $dy = 2 dx$ is a function just like $y=mx+c$ which you can manipulate? Is that correct? –  Jake Burton May 2 '13 at 19:52
You would then get that $dy=2dx$. This describes the line tangent to the graph of $f$ at $1$. The point $(1,1)$ is the origin of the space where this occurs. –  John Douma May 2 '13 at 19:53
We are saying that $\sin (\frac{x}{2})=\sin (u)$ where $u$ is a function of $x$. If that's what you mean by 'u space', then yes. –  John Douma May 2 '13 at 20:01