# Integration of dx^2

1. Jul 20, 2012

### drwajih

Hello, I am a medical doctor but interested in mathematics and physics. Consider a parabola with its equation y = x2
dy/dx = 2x
and dy = 2x.dx
If we make small right angled triangles under the parabola with the base dx then the height of the triangles will be dy. The hypotenuse (dh) of the triangles will be
dh2 = (dx)2 + (dy)2
dh2 = (dx)2 + (2x.dx)2
dh2 = (dx)2 (1 + 4x2)
if we integrate dh it will give us the sum of squares of the small pieces of the curve, which is parabola here (Sh2).

Sh2 = ∫ 1 + 4x2.dx2

But I have been told that it is not allowed in calculus to integrate dx squared. Can some one please shed some light on it? Thank you.

Last edited: Jul 20, 2012
2. Jul 20, 2012

### Curious3141

Well, I'm a medical doctor too. So, hi there!

Well, that's wrong. $\frac{dy}{dx} = 2x$

Up to here, there's nothing fundamentally wrong, except that the expression should read:

$${(dh)}^2 = {(dx)}^2 + {(2xdx)}^2 = {(dx)}^2 + 4x^2{(dx)}^2 = (1 + 4x^2){(dx)}^2$$

Big problem at this point. *What* are you integrating with respect to? The left hand side (LHS) is (dh)2. You can only integrate it if it's simply dh.

What you can do is take square roots of both sides, like so:

$$dh = \sqrt{1 + 4x^2}dx$$

*Then* integrate:

$$H = \int_0^H dh = \int_{x_1}^{x_2} \sqrt{1 + 4x^2}dx$$

which will give you the length of the curve (H) from point x1 to x2.

In fact, this is based on the well-known formula for arc length, often expressed as:

$$\int_0^S ds = \int_a^b \sqrt{1 + {(\frac{dy}{dx})}^2}dx$$

Hope this clears things up.

3. Jul 20, 2012

### drwajih

Thanks and hello
I did make the corrections. It was a hasty post!
But why can we not integrate (dh)2?
It makes sense. We will get the sum of the squares of the small pieces of the curve.
Actually I want to calculate the circumference of an ellipse. The given formula is very ugly. I have an idea to calculate the circumference from a new method for which I need to integrate dx squared.

4. Jul 20, 2012

### chiro

Hey drwajih and welcome to the forums.

You could integrate with respect to the square, but you are going to have to use a different theory of integration and some measure theory.

The normal integration that you are familiar with is the Riemann integration which is defined with respect to an infinitesimal, and it also requires that the functions have specific conditions (like have the proper analytic anti-derivative).

The anti-derivatives that you are used to work for the Riemann Integral but typically not other ones. If you want to use a different measure other than dx like say (dx)^2 you will need to use a different kind of integral, but this means you lose the benefits of the Riemann integral which allow you to do things like calculate the integral as F(b) - F(a).

That's the short answer to your question and if you want to find out why the Riemann integral works with the whole anti-derivative stuff you will need to read an analysis book.

5. Jul 20, 2012

### drwajih

Thank you very much.
So it is a complex problem!
What if we integrate it twice?
Will that not get rid of the square over dx?

6. Jul 20, 2012

### chiro

No it doesn't work like that.

Basically if you think about the Riemann integral, we add up all the rectangles and we take the limit as the width of the rectangle goes to zero, where the width of the rectangle is dx.

But in this situation, the width of the rectangle is dx^2 which complicates things a little and the reason we can't just use the normal anti-derivative is because the derivative is defined in terms of the linear differential dy, dx (and so on) and not (dx)^2 so if you wanted to use some special kind of integral, you need an anti-derivative defined for (dx)^2 as opposed to dx and this is why you can't use the normal calculus for this situation.

7. Jul 20, 2012

### drwajih

Thanks I have got it now.
So then how will I solve the problem of the circumference of an ellipse?
This means that the formula will remain very ugly!

8. Jul 20, 2012

### Curious3141

The formula for the circumference of an ellipse cannot be exactly expressed in terms of elementary functions. You need the elliptic integral of the second kind. In other words, you are left with a definite integral in your expression.

Of course, you could express it even more messily in terms of power series, etc. Either way, you're stuck with "ugly".

9. Jul 20, 2012

### drwajih

OK
Thank you. Both of u are very talented.