# Definitions and properties of limits (handwriting attached)

1. Feb 26, 2013

### tolove

Could someone look over this and see if I have any mistakes? I'm trying to show that
∫ y' dx = ∫ dy through definitions.

http://imgur.com/6zCHYo5

Thanks!

2. Feb 27, 2013

### joeblow

If dy and dx have been properly defined (which for some reason is a rare occurrence), then this equality is immediately apparent for differentiable functions. However, this is useless as a means to finding an antiderivative (which is what you're doing, since you have no bounds on your integral.)

Your proof is not much more than switching notations out. Even if it were a definite integral, dy is dependent on dx, so trying to evaluate the integral that way would be futile.

Last edited: Feb 27, 2013
3. Feb 27, 2013

### tolove

What you're saying is that "What you wrote has no problems, however it's a pointless exercise and proves nothing since all you're doing is changing notations around."

If that is so, then thank you! That's more or less what I was going for. These notations have been confusing me, but I think I've made sense of them.

4. Feb 27, 2013

### joeblow

Well... I'd say that it "makes sense" only if it were a definite integral.

And I wasn't complaining about the uselessness of the proof, so much as the impossibility of computing the integral on the RHS.

I hope you're not using this to "prove" FTC. If you are, you should consider the change of the area under the curve with upper bound x induced by moving delta x, then find the corresponding dA.

Last edited: Feb 27, 2013
5. Feb 27, 2013

### tolove

I think I'm only following you in pieces (I have a pretty awful understanding of the subject!)
What do you mean by properly defining dx and dy? And by apparent, do you simply mean that this is clear to see ∫ dy/dx dx = ∫ dy? I find it very unclear to simply to see that. Mathematically, just cancel out the dx, but physical equations can become hard to visualize for me sometimes, ∫ v' dt = ∫ dv.
I don't follow this either ;_;

6. Feb 27, 2013

### tolove

The physics book I'm using breaks up dx/dy in equations regularly, and I'm try to not get into the habit of just viewing dy/dx as a quotient.

7. Feb 27, 2013

### joeblow

Suppose we have the graph of a function. Let us approximate its shape with vectors whose horizontal component is dx and vertical component is dy. Then, the derivative of the function gives us the number we need so that $$dx \cdot \frac{dy}{dx}=dy.$$ (Typically, y is dependent on x, so we can assume that dx is the same at every point.) Thus, dy is dependent on dx AND the slope of the tangent line at the point. Thus, integrating dy by itself doesn't make sense unless you know something about the position.

If you keep in mind this relationship, you can detach dy and dx freely for differentiable functions.

To derive FTC (for functions that are almost everywhere continuous), you need to see what happens to the change in area (dA) of the function $$A(x)= \int_a ^x f(t)dt$$ when we move a small amount dx. Here, you will see that you can use dy by itself, keeping in mind the associated dx.

Last edited: Feb 27, 2013