Why can "dx" in integration be multiplied?

[Kelly333]

Hi,
This is an example in "Barron AP calculus"

I learned from some past threads that "dx" in integration either means △x which is a infinite number or indicates the variable with respect to which you're integrating.
In the equation above, it seems that dx is multiplied by (1-3x)^2. Isn't dx just a notation?

I haven't learned definite integrals so far so please explain in detail if your answer is related to that.

Thanks.

 

[cnh1995]

△x which is a infinite number

I don't understand what you mean here by infinite number. It is an infinitesimally small change in x. If you refer definite integration expressed as a limit of sum, you'll see that dx is actually the infinitesimal width of the rectangular strip(area element) and y is the height of of the strip. Hence, it's area becomes y*dx and when you sum the areas of all such strips having width dx and height y over an interval, you get the total area under the curve. This sum is nothing but integration of y w.r.t x in that interval i.e.∫y⋅dx.

 

[HallsofIvy]

In this problem, "dx" is simply a symbol telling us which variable this is to be integrated with respect to. It never "means \Delta x which is a infinite number"- I don't know where you got that! But it is not "multiplied by (1- 3x)^2". The notation here is just asking you to integrate the function 2(1- 3x)^2 with respect to x.

 

[Kelly333]

In this problem, "dx" is simply a symbol telling us which variable this is to be integrated with respect to. It never "means \Delta x which is a infinite number"- I don't know where you got that! But it is not "multiplied by (1- 3x)^2". The notation here is just asking you to integrate the function 2(1- 3x)^2 with respect to x.

Thanks for your answer! If dx is just a symbol telling us which variable this is to be integrated with respect to, what does (-3*dx) actually means?

 

[Kelly333]

I don't understand what you mean here by infinite number. It is an infinitesimally small change in x. If you refer definite integration expressed as a limit of sum, you'll see that dx is actually the infinitesimal width of the rectangular strip(area element) and y is the height of of the strip. Hence, it's area becomes y*dx and when you sum the areas of all such strips having width dx and height y over an interval, you get the total area under the curve. This sum is nothing but integration of y w.r.t x in that interval i.e.∫y⋅dx.

Thanks for correcting that error! But in antidifferentiation, does dx also mean delta x?

 

[Ray Vickson]

Thanks for correcting that error! But in antidifferentiation, does dx also mean delta x?

No. Typically, we use the notation $\Delta x$ when writing finite sums such as $\sum_{i=1}^n f(x_i) \Delta x_i$, and $dx$ in the limit as we take $n \to \infty$ and each $\Delta x_i \to 0$. Of course, "$dx$" is just notation; it is not really "zero".

Now let's look in more detail at your little example. First, do you agree that initially we want to integrate $(1-3x)^2$ with respect to $x$? We can indicate this fact by introducing the $dx$ notation, but that is not really necessary. In some computer-algebra systems we drop the $dx$ altogether; for example, in Maple we can write "int((1-3*x)^2,x)" to indicate the integrand (before the ",") and the integration variable (after the ","). In Wolfram Alpha you can just enter "integrate (1-3*x)^2 with respect to x".

Anyway, we want to change variables to $y = 3x$, so the integrand involves the simpler-looking expression $(1-y^2)$. However, if we write $I_1 = \text{x-integral}\; (1-3x)^2$ and $I_2 = \text{y-integral}\; (1-y)^2$, we DO NOT have $I_1 = I_2$; in fact, we have $I_1 = \frac{1}{3} I_2$. Understanding this last equality is where the $dx, dy$ notation becomes helpful: if $y = 3 x$ then $dy = 3 dx$ and so $dx = \frac{1}{3} dy$. Now when we erase the $dx$ and $dy$ we are still left with the factor 1/3.

 

[Ssnow]

No, $\Delta x$ is not $d\,x$.

 

[Kelly333]

No. Typically, we use the notation $\Delta x$ when writing finite sums such as $\sum_{i=1}^n f(x_i) \Delta x_i$, and $dx$ in the limit as we take $n \to \infty$ and each $\Delta x_i \to 0$. Of course, "$dx$" is just notation; it is not really "zero".

Now let's look in more detail at your little example. First, do you agree that initially we want to integrate $(1-3x)^2$ with respect to $x$? We can indicate this fact by introducing the $dx$ notation, but that is not really necessary. In some computer-algebra systems we drop the $dx$ altogether; for example, in Maple we can write "int((1-3*x)^2,x)" to indicate the integrand (before the ",") and the integration variable (after the ","). In Wolfram Alpha you can just enter "integrate (1-3*x)^2 with respect to x".

Anyway, we want to change variables to $y = 3x$, so the integrand involves the simpler-looking expression $(1-y^2)$. However, if we write $I_1 = \text{x-integral}\; (1-3x)^2$ and $I_2 = \text{y-integral}\; (1-y)^2$, we DO NOT have $I_1 = I_2$; in fact, we have $I_1 = \frac{1}{3} I_2$. Understanding this last equality is where the $dx, dy$ notation becomes helpful: if $y = 3 x$ then $dy = 3 dx$ and so $dx = \frac{1}{3} dy$. Now when we erase the $dx$ and $dy$ we are still left with the factor 1/3.

Thanks!Your points are clear and easy to understand. Now I grasp the idea that dx is just a symbol but is also crucial in some equations.
But I don't believe it makes sense to multiply dx by (-3) in this equation in my example. Can you explain that?

 

[HallsofIvy]

Thanks for your answer! If dx is just a symbol telling us which variable this is to be integrated with respect to, what does (-3*dx) actually means?

By itself it doesn't mean anything! With an \int, \int -3dx it would mean "integrate -3 x". Or, it could be a "differential". "dy= -3 dx" means that however x increases, y will decrease by three times that much. You should never see something like "f(x)dx" by itself. It should either be inside an integral, as \int f(x) dx, or in connection with some other differential, as dy= f(x)dx.

 

[Ray Vickson]

Thanks!Your points are clear and easy to understand. Now I grasp the idea that dx is just a symbol but is also crucial in some equations.
But I don't believe it makes sense to multiply dx by (-3) in this equation in my example. Can you explain that?

If you change variables to $y = 3x$ you get $\frac{1}{3} \int (1-y)^2 \; dy$, but if you change to $y = -3x$ you get $-\frac{1}{3} \int (1+y)^2 \, dy$.

You say " I don't believe it makes sense to multiply dx by (-3) in this equation in my example". Belief has nothing to do with it; there are standard change-of-variable formulas for integration, and they give you the 1/3 or -1/3 factor. What I mean is this: if we want to evaluate
I = \text{x-integral} \; f(x)
we may find it easier to change variables from $x$ to $y$, where $x = g(y)$, so that in the integral we will have $h(y) = f(g(y))$ instead of $f(x)$. However, that is not the end of the story: we also need to "transform the $dx$", giving
I = \text{y-integral} \; h(y) g'(y)
This is a provable theorem, not something subject to belief or dis-belief.