What Does ∫ (dy)^2 Mean in Calculus?

  • Thread starter kingwinner
  • Start date
In summary: I don't know if that's related, but I think it's worth mentioning that the limit of a sum is always zero.
  • #1
kingwinner
1,270
0
What is ∫ (dy)^2 ?

Fact:
Let a<b.
b
∫ (dy)2 = 0
a

=====================

1) First of all, I'm having trouble understanding the meaning of the above integral. I tried searching my elementary calculus textbooks, but I really couldn't find ANY integral like this. I have only seen integrals like ∫ (dy) in my calculus books and so I have no idea how they can have (dy)2 which makes no sense to me...

2) Why is the above fact true? How can we rigorously prove it?


Hopefully someone can clarify this.
Thank you!
 
Last edited:
Physics news on Phys.org
  • #2


I guess a,b are real numbers. Where did you see this integral?
 
  • #3


Yes, a and b are real numbers, a<b.

I saw this integral in my first study of stochastic calculus and brownian motions.
It says...from ordinary integration, we have
b
∫ (dy)2 = 0
a

By ordinary, I'm assuming it's referring to standard elementary calculus, but I just can't recall anything similar to integrals like this in my calculus courses. Does such a thing exist in math? What is the precise definition of the above integral?

Thanks!
 
  • #4


One work around on this is to consider the fact that dy is independent of y thus it is like a constant when integrating with respect to y. So it is [tex] \int (dy)^2=\int cdy[/tex]

Integrating cdy from a to b will give c(b-a)=dy(b-a). But dy is an infinitesimally small quantity therefore dy(b-a)=0.
 
  • #5


dy is an infinitesimally small quantity
But then what does squaring an infinitesimally small quantity (i.e. (dy)2) mean? Squaring a differential just doesn't make much sense to me...

Also, I believe integral is defined as the limit of some kind of sum.
Is there an analogous sum being used to define the integral
b
∫ (dy)2 ?
a
 
  • #6


dy is something like (b-a)/n and the integral is defined as the limit of a sum.
[tex] \int_{a}^{b}(dy)^2=lim_{n→∞}\sum _{i=1}^{n}(\frac{b-a}{n})^2[/tex]

You can see that this limit is zero as n goes to infinity. But all i did here is not rigorously mathematically defined, i mean in math textbooks there is no official definition for the integral of (dy)^2 but an attempt to define such a thing in the most straightforward way is via the limit of sum i gave above.
 
Last edited:
  • #7


I don't mean to be pedantic about this, by bringing-in the big machinery, but I can't
come up at this moment with something simpler; tho I will try: using
Green's theorem, one integrates 1-forms over surfaces, so there seems to be
a dimension problem. Maybe it is understood/implied that when using ∫ and (dy)2, that we are actually using a double integral, and using dy repeatedly as the dummy variable
so that ∫(dy)2 can maybe be understood as the integral over ℝ2
of the function f==1 along the x-axis , i.e., a line has measure zero as a subset of ℝ2 . Or, more simply, a line has area=0 .

Maybe someone else can double-check.
 

FAQ: What Does ∫ (dy)^2 Mean in Calculus?

What is ∫ (dy)^2?

The integral of (dy)^2 is the integral of the square of the differential of a variable y. This means that we are finding the area under the curve of (dy)^2 with respect to y.

Why is it important to know the integral of (dy)^2?

Knowing the integral of (dy)^2 is important because it allows us to calculate the area under the curve of (dy)^2, which has many applications in physics, engineering, and other scientific fields.

How do you solve for ∫ (dy)^2?

In order to solve for ∫ (dy)^2, you must first use integration techniques, such as substitution or integration by parts, to simplify the integrand. Then, you can apply the fundamental theorem of calculus to evaluate the integral.

What is the result of ∫ (dy)^2?

The result of ∫ (dy)^2 is a function of y, since the integral is with respect to y. This means that the result will be a curve on a graph, and not a single numerical value.

Can the integral of (dy)^2 be negative?

Yes, the integral of (dy)^2 can be negative. This means that the area under the curve of (dy)^2 is below the x-axis on a graph. The sign of the integral will depend on the bounds of integration and the shape of the curve.

Similar threads

Back
Top