# Interpretation of dx as the differential of x for Indefinite Integrals

1. Mar 13, 2012

### mbarile

Interpretation of "dx" as the differential of x for Indefinite Integrals

This question is concept-as-opposed-to-calculation based. I understand that when one sees the integral sign, followed by f(x)dx, that we can think of this as the indefinite integral, or antiderivative of f(x), with "respect to x," where the "dx" means just that ("with respect to x"). Or, we can think of the same expression as being in differential form, where "dx" no longer means "with respect to x," but now is interpreted as the "differential of x," a concept that becomes a useful tool when dealing with substitution techniques and the like. What I'm missing, in terms of understanding, is the following: Say that we integrate f(x)dx, where we think of "dx" as "with respect to x," and we get the function F(x) + C, where C is an arbitrary constant. The idea of the differential form, or at least my understanding of it, is that if we integrate f(x)dx, where "dx" is the "differential of x", we get the same antiderivative, F(x) (+ an arbitrary constant C). How is this so? If we think of "dx" as the "differential of x," is that not some quantity anywhere between (-infinity,infinity)? Are we not then multiplying f(x) by some quantity "dx" by this interpretation, as opposed to the interpretation that "dx" simply means
"with respect to x".

I understand how to manipulate the material to get the correct answer, such as using u-substitution, etc, but the concept does not sit well, so I am obviously missing something. Any help would be greatly appreciated.

2. Mar 13, 2012

### mathwonk

Re: Interpretation of "dx" as the differential of x for Indefinite Integrals

to say the antiderivative of f is F, means that dF/dx = f. To say that f(x)dx is the differential of F, means that dF = f(x) dx, which is basically the same thing.

3. Mar 13, 2012

### mbarile

Re: Interpretation of "dx" as the differential of x for Indefinite Integrals

Thank you for the response. As a follow up, so are you saying that in the same way that antidifferentiation "undoes" the process of finding the derivative of a function F(x), antidifferentiation will undo the process of finding the differential of a function F(x), and in either case will lead one back to the original function F(x) (plus constant)?

4. Mar 13, 2012

### emailanmol

Re: Interpretation of "dx" as the differential of x for Indefinite Integrals

Both interpretations are one and the same thing.

Integrals give you area.
When we multiply f(x) with dx we are infact finding the area of small rectangle of breadth dx and height f(x).
When we add all these small areas(which is why we integrate to add these up.Remember integral is a sum.Right?) we get area below the curve.

Now dx has only one meaning.It is an infinitely small change in x.

When we say f(x)d(x) the only imterpretation is that f(x) is multiplied by infinitely small value d(x).

The point which you stated with respect to x is something else.

Imagine a function y =x^2 + t^3

Here y is a function in two independent variables x and t.

Taking a derivative of y with respect to x here means that we see the change in y for an infinitely small change in variable x.

Since y =x^2+t^3

dy/dx =2x + 3t^2 dt/dx

In case t is completely independent of x , (has no relation with it at all) then dt/dx is 0 and dy/dx is just 2x.

If however you want to see how y changes with t, you take a derivative wrt to t to get dy/dt =3t^2 (remember dx/dt is 0 as they both are unrelated i.e independent variables).

This just means that if we increase x from x to dx (keeping t constant) , the effect on y depends on factor 2x.
It is called change in y wrt x.

If however we increase t from t to t+dt(x is constant)
change in y wrt to t depends upon the factor 3t^2.

While integrating its the same meaning.

In essence is you wanted to integrate the above function you had two choices.

To see its area wrt either x or t.
That is the significance of integrating wrt x.

If you wish to find integral of y wrt x, you multiply y with infinitely small change in x which is dx(and assume t to be constant as it doesnt vary with x).
This gives u small rectangle whose breadth is dx and height is y and you obtain area of y wrt x by adding all these small x's .

For indefenite integral you find a general function of area.

Last edited: Mar 13, 2012
5. Mar 13, 2012

### morphism

Re: Interpretation of "dx" as the differential of x for Indefinite Integrals

Search for "differential forms"