# Integration of f(dx)

1. Feb 13, 2012

### Shootertrex

While I was attempting to prove the area of a circle formula, I ran into a snag. This is the work that I had (note that I am using the area of an infinite amount of triangles within the circle):

a=1/2bh
=1/2br
da=1/2r db
---(splitting the triangle into 2 right triangles produces b=2r*tan($\frac{θ}{2}$))
=1/2r (2r tan($\frac{dθ}{2}$))
=r2 tan($\frac{dθ}{2}$)
a=r2$\int^{2π}_{0}(tan(\frac{dθ}{2}))$

I know that the integration will equal π not only due to inspection, but also because I proved it empirically using Excel. If anyone knows how to do this type of integration, if it is possible at all, or if you see any other problems in my work I would appreciate your help.

Last edited: Feb 13, 2012
2. Feb 13, 2012

### lugita15

tan(dx/2)=dx/2 is a formula that's very easy to get. You can find it using the fact that limit of sin(x) /x as x goes to zero is 1 and the fact that the limit of (cos(x)-1)/x as x goes to zero is 0, from which with a little work you can get that the limit of tan(x)/x as x goes to zero is 1. Or you can use Taylor series. There should also be a geometric reason why tan(x) is approximately equal to x (measured in radians) if x is really small, and thus tan(dx)=dx where dx is infinitely small.

Last edited: Feb 14, 2012
3. Feb 14, 2012

### JJacquelin

Since dx is a very small variation of x, an infinitesimal in fact, then tan(dx)=dx close to =0.
I suppose than you mean tan(x+dx)= tan(x) + d(tan(x))
where d means "a very small variation of something". So, d(tan(x) is the very small variation of tan(x) caused by the very small variation of x :
d(tan(x)) = (1 / cos²(x))*dx

4. Feb 14, 2012

### lugita15

No, that's not what I mean, but you're on the right track. Let's use the formula you stated, $tan(x+dx)=tan(x)+sec^{2}(x)dx$. If we plug in x=0, we get $tan(dx)=tan(0)+sec^{2}(0)dx=0+1dx=dx$. Another way to see this is that sin(dx)=dx (because the limit of sin(x)/x as x goes to zero is 1) and cos(dx)=1 (because the limit of (cos(x)-1)/x as x goes to zero is 0), therefore tan(dx)=dx. (If you want to be fancy, I am expressing the Taylor expansion to first order in dx.)

5. Feb 14, 2012

### JJacquelin

Of course, my answer was not adressed to lugita15, but to Shootertrex.

Shootertrex wrote :
which is nonsens
.
The title of the topic : " Integration of f(dx) " is also the result of a missunderstanding. Shootertrex should understand that f(dx) must not be confused with f(x)*dx

Last edited: Feb 14, 2012
6. Feb 14, 2012

### Shootertrex

No, I actually do mean $tan(\frac{dθ}{2})$. This is because within in the circle I created an indefinite amount of triangles with $θ$ that approaches 0. The actual problem that I am having is not showing that this formula is correct (I am fairly certain that it is), but how to integrate a function where $dx$ is within a function, in this case $tanx$

I am not sure what you mean here. Are you saying that I do not understand that f(dx) should not be confused with f(x)dx or that I already know the difference? I do know the difference, and that is why I am asking the question.

After asking two of the math teachers at my school, I have not progressed far. Neither knew of a way to integrate the equation as is, but both suggested substitution. An alternate substitution that was proposed was to do this:

$b=2r tan\frac{θ}{2}$
$\frac{db}{dθ}=2r*\frac{1}{2}$$sec$2$\frac{θ}{2}$
$db=r*$$sec$2$\frac{θ}{2}dθ$
$da=\frac{1}{2}r$2$sec$2$\frac{θ}{2}dθ$

I see two problems with this equation though. First, it has a 1/2. I do not know if this will work in compliment with the secant. Second, the integration again. Since secant is a non-continuous function, it will be hard to do a definite integral with it. I was thinking that maybe an improper integral would work. Please take note that I am still in Calc 1. I am willing to learn things outside of Calc 1 to finish this proof, but please use these techniques sparingly.
If this substitution works and improper integrals will solve it, I will research how to actually do this kind of integration.

I agree with both of the limits that were proposed, but I am having trouble seeing how to get to the tangent limit. This is what I thought to do:

$lim_{x→0}\frac{xcosx}{xsinx}=\frac{0}{0}$
$lim_{x→0}\frac{cosx}{sinx}=\frac{0}{0}$ (cancel out x's, then apply L'Hopital's Rule)
$lim_{x→0}\frac{sinx}{cosx} \Rightarrow lim_{x→0}-tanx=0$

I was able to get tanx, but not tan(x)/x

7. Feb 14, 2012

### lugita15

Shootertrex, one simple way to see that the limit of tan(x)/x as x goes to zero is 1 is L'hopital's rule.

8. Feb 14, 2012

### Shootertrex

I agree with the limit, I just don't know how you came about that limit. You had said to use the sin(x)/x limit and the (cos(x)-1)/x limit to get it.

9. Feb 14, 2012

### Shootertrex

Okay, so I have thought it over some more. Lugita15, you had stated that $tan\frac{dx}{2}=\frac{dx}{2}$. I would happily make this substitution because it allows me to easily integrate the equation and would get me $a=r$2$π$, but I don't know if I can agree with it or not. I know that $lim_{dx→0}tan\frac{dx}{2}=0$ and $lim_{dx→0}\frac{dx}{2}=0$. So both sides would be equal but it seems like one of those "2+2=5" moments. Am I just over analyzing it or what? I will take your word for it either way, but I would like help understanding it.

10. Feb 14, 2012

### lugita15

There are two different languages we use to talk about calculus, the language of infinitesimals and the language of limits. Infinitesimals (things like dx and dy) are the way calculus was originally understood by Newton and Leibniz, but they were thought to be vague and nonrigorous, so in the 1800's the notion of limits was invented. But people still kept using the old infinitesimal notation, like writing the derivative as df/dx and writing the integral as ∫f(x)dx. It was only very recently, in the twentieth century, that Abraham Robinson and others showed that manipulating infinitesimals is mathematically valid. Infinitesimals are for many purposes easier to work with than limits: for instance the chain rule is quite difficult to prove using limits, but in infinitesimal language it's just canceling fractions, dy/dx=(dy/du)*(du/dx).

To illustrate the differences between infinitesimals and limits, the definition of the derivative in terms of infinitesimals is (f(x+dx)-f(x))/dx, whereas the limit definition is the limit of ((f(x+Δx)-f(x))/Δx as Δx→0. And a good tip in translating between the two languages is that the infinitesimal statement f(dx)=g(dx) translates into the statement that the limit of f(x)/g(x) as x goes to zero is 1. So sin(dx)=dx corresponds to the statement that the limit of sin(x)/x as x goes to zero is 1, which is true.

Getting back to your case, it's wrong to say tan(dx)=0. tan(dx)=dx, which is an infinitely small number, not zero. The reason that tan(dx)=dx is not that the limits of both tan(x) and x as x goes to zero are equal to 0. If that was true, then cos(dx)-1 would equal dx, because the limits of both cos(x)-1 and x as x goes to zero are equal to 0. But in fact, cos(dx)-1 equals zero, not dx (because the limit of (cos(x)-1)/x as x goes to 0 is 0). Bottom line, just because the limit of f(x) as x goes to 0 is 0 does not mean f(dx)=dx. The correct formula is that f(dx)=f(0)+f'(0)*dx. Can you see how to get this formula?

11. Feb 14, 2012

### Shootertrex

I think I understand the last paragraph, and the history lesson was quite interesting. I'm sorry though, but I do not think I fully understand the f(dx) equation you stated at the end. I will explain what I am thinking and you can tell me if I am right, on the right track or just wrong.

f(dx) will be whatever the function is at almost zero. f(0) will be what the function is at zero. Now, is adding f'(0)*dx sort of like a correction factor to get f(0) to be f(dx)?

Am I at least close? Even if I understand how the formula works, I would not be able to derive the formula. Thank you ahead of time for being patient with me.

12. Feb 14, 2012

### lugita15

That's the general idea. But the more precise derivation is as follows: as I told you in my previous post, the definition of the derivative in terms of infinitesimals is f'(x)=(f(x+dx)-f(x))/dx. Plugging in x=0, we get f'(0)=(f(dx)-f(0))/dx. If you solve this equation for f(dx), you get f(dx)=f(0)+f'(0)*dx. Does that make sense?

13. Feb 14, 2012

### Shootertrex

Yes, I now see how you got the equation. It makes sense. But will this help me with my original dilemma?
Using this formula, if I assume dx=1/250, then this is what I get:

$tan\frac{dx}{2}=4.44*10$-16
$tan(0)+sec$2$(0)*dx=8.88*10$-16

but I'm guessing that the dx in the second formula needs to be dx/2, which would then give:4.44*10-16 which would be equal, correct?

Now if I were to plug this new formula into my integral then:

$a=r$2$\int^{2π}_{0}(tan(0)+sec$2$(0)*dx)$
$=r$2$*(tan(0)x+sec$2$(0)x)$|$^{2π}_{0}$

Now, I am unsure if tan(0) would get an x or not, but it wouldn't affect the answer because it is equal to zero. So the final equation would be:

$a=r$2$2π$

Which is not correct... Should the interval for the integral be instead just to π? If so, why? The triangles would line the inside of the circle all the way around, being 2π.

14. Feb 14, 2012

### lugita15

Remember, dx is an infinitesimal, meaning it is an infinitely small number. It is neither equal to zero nor is it equal to a finite number like 1/250. Note that tan(dx)=tan(0)+sec^2(0)*dx is true, but tan(Δx)=tan(0)+sec^2(0)*Δx where Δx is a finite number is false, although if Δx is small then it is approximately true. It only is exactly true for infinitely small numbers.

15. Feb 14, 2012

### Shootertrex

Wait! I think I found an error in my substitution. I put $(tan(0)+sec2(0)∗dx)$ instead of $(tan(0)+sec2(0)∗dx/2)$. If I use this new substitution, then I would get

$a=r$2$π$
edit2: Q.E.D.

Am I correct?

Edit: the dx=1/2^50 was just so I could actually see if the two equations were equal, not to prove anything.

Last edited: Feb 15, 2012
16. Feb 14, 2012

### lugita15

Yes, tan(dx/2)=tan(0)+sec^2(0)*dx/2, and tan(dx)=tan(0)+sec^2(0)*dx.

17. Feb 15, 2012

### Shootertrex

Thank you very much for all of your help! You are truly a god among men! I can finally lay this problem to rest.