## Oh me yarm I GET IT!

For the discussion of math. Duh.

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superglucose
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### Oh me yarm I GET IT!

So I took a calculus class... twice. Once in high school, again in college. I did fine, you know 4 on the AP, was getting decent grades on the tests, etc. etc. but I never really felt like I got the point behind integrals. It just never... snapped into place.

Then today in Physics discussion I was looking at a spring, and thinking "man, I wish there was an easy way of summing up those infinitely small pieces of force during the moments I compressed the spring, then I could easily determine the energy."

So after two years of that integral crap being force fed down my throat, looking at a physics problem for five seconds I suddenly saw the magic and power of integrals. Man... when something clicks, it clicks don't it.

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### Re: Oh me yarm I GET IT!

Now you can get differential equations...
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imMAW
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### Re: Oh me yarm I GET IT!

Yea, I took calculus and calculus based physics at the same time, and it was great. I liked seeing the interconnections between the two.

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### Re: Oh me yarm I GET IT!

imMAW wrote:Yea, I took calculus and calculus based physics at the same time, and it was great. I liked seeing the interconnections between the two.
Same here, it was marvelous. Now I'm taking a new Calc class that starts out with a lot of stuff I already know, so it's a bit boring but I'm also getting a somewhat better grasp on the underlying concepts this time around. Now if only calculus would start explaining more physics for me...
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### Re: Oh me yarm I GET IT!

superglucose wrote:but I never really felt like I got the point behind integrals.

To get good grades without an understanding of the subject. That is why I never worked hard at any class I ever took.

Integrals are the ying to the derivative's yang. You wouldn't want the universe to fall out of balance, would you?

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### Re: Oh me yarm I GET IT!

Speaking of which, I got the same sort of feeling when I got why the dx is at the end of the integral.
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### Re: Oh me yarm I GET IT!

The "Ohh is that how it works..." feeling is the best thing about science. Enjoy.
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### Re: Oh me yarm I GET IT!

headprogrammingczar wrote:Speaking of which, I got the same sort of feeling when I got why the dx is at the end of the integral.

The integral symbol is an S, and stands for SUM.
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### Re: Oh me yarm I GET IT!

Technically it stands for summa, just like the square root symbol was originally a fancy r which stood for radix.

Matterwave1
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### Re: Oh me yarm I GET IT!

I learned it at first as an anti-derivative, and then as an infinite Reimann sum. I never got why an anti-derivative should be the sum, but I just take the for granted.

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### Re: Oh me yarm I GET IT!

Matterwave1 wrote:I learned it at first as an anti-derivative, and then as an infinite Reimann sum. I never got why an anti-derivative should be the sum, but I just take the for granted.

That's what the Fundamental Theorem of Calculus[1] is all about.

[1] So Fundamental it's got Capital Letters! This thing is more Fundamental than Young Earth Creationists!
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### Re: Oh me yarm I GET IT!

headprogrammingczar wrote:Speaking of which, I got the same sort of feeling when I got why the dx is at the end of the integral.

Enlighten me! I still don't quite get what dx or dy mean on their own. dy/dx isn't really a quotient of infinitesimals, the Chain Rule doesn't work because of du cancelling, and S[thing]dx is either all one operator that surrounds the integrand or dx is being multiplied... not to mention how you can split up the dy/dx for implicit differentiation...

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### Re: Oh me yarm I GET IT!

headprogrammingczar wrote:Speaking of which, I got the same sort of feeling when I got why the dx is at the end of the integral.

Nobody's cared to explain it to me.
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### Re: Oh me yarm I GET IT!

davetp425 wrote:
headprogrammingczar wrote:Speaking of which, I got the same sort of feeling when I got why the dx is at the end of the integral.

Enlighten me! I still don't quite get what dx or dy mean on their own. dy/dx isn't really a quotient of infinitesimals, the Chain Rule doesn't work because of du cancelling, and S[thing]dx is either all one operator that surrounds the integrand or dx is being multiplied... not to mention how you can split up the dy/dx for implicit differentiation...

Did you ever approximate the area of a curve by rectangles? Look at this image for an example. http://en.wikipedia.org/wiki/Image:MidRiemann.png

You can probably see that the integral of the curve should be approximately the sum over all rectangles of rectangle width times the value of f(x) at the top of the rectangle.

As you shrink the rectangles and make them more numerous, your approximation gets better and better. In the limit, your sum is over all values f(x) takes on and you are multiplying each of these values by an infinitesimally small rectangle width, dx.

That's how the integral was originally thought of, as I understand it. Similarly, you say that dy/dx isn't really a quotient of infinitesimals--and that's true with the modern definition of a derivative. But the modern definition involving limits was originally designed to serve the same purpose as the quotient of infinitesimals! The notation dy/dx dates back to the day when it really was thought of as dy divided by dx.

And you'll find that physicists still love to treat a derivative as a ratio of infinitesimals.
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superglucose
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### Re: Oh me yarm I GET IT!

isn't dx just using the lowercase greek letter Delta? And delta is, more or less, used to signify 'change.' So I always think of dx as the change in X. Hence why dx/dy is the change in x with respect to the change in y, thus, the slope. And the integral of xdx, for instance, is the sum of all the portions of x as x changes.

At least that's my noob interpretation of it.

davetp425
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### Re: Oh me yarm I GET IT!

From what I gather, that is how they thought about it when they invented calculus, but that's not technically rigorous--it's the $.1^\infty > 0$ fallacy or something. I guess I'm looking for a rigorous and intuitive explanation for it.

Matterwave1
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### Re: Oh me yarm I GET IT!

Well in physics, dx or dt is usually just said to be "a small change". So it's no different than a very small piece of it, and therefore, we cancel them and move them around and stuff. I'm sure a mathematician would have a heart attack.

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### Re: Oh me yarm I GET IT!

Matterwave1 wrote:Well in physics, dx or dt is usually just said to be "a small change". So it's no different than a very small piece of it, and therefore, we cancel them and move them around and stuff. I'm sure a mathematician would have a heart attack.

Nah, they are doing it too. This is just an example why our notation is good - it is robust to taking it literally .
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### Re: Oh me yarm I GET IT!

davetp425 wrote:Enlighten me! I still don't quite get what dx or dy mean on their own. dy/dx isn't really a quotient of infinitesimals, the Chain Rule doesn't work because of du cancelling, and S[thing]dx is either all one operator that surrounds the integrand or dx is being multiplied... not to mention how you can split up the dy/dx for implicit differentiation...

It is the quotient of infinitesimals though. The derivative is the slope formula (the quotient of an infinitesimal change in y and an infinitesimal change of x). The integral is a Riemann sum, which is:
Since we know from the Fundamental Theorem of Calculus that the integral is actually the anti-derivative, we can use the same notation. So from [imath]\frac{dy}{dx}=f(x)[/imath] we can see that [imath]Δx=dx[/imath], and so, changing the sigma to an integral, we get [imath]\int f(x)dx[/imath]. It is not perfect, but it is how I think of it.
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### Re: Oh me yarm I GET IT!

Regarding infinitesimals: in nonstandard analysis, the derivative is given as an actual quotient of infinitesimal differences. Specifically, it is the "standard" (real) part of the quotient
$\frac{f(x+\epsilon)-f(x)}{\epsilon},$
if the standard part is the same for any nonzero infinitesimal [imath]\epsilon[/imath]. (If it's different for different choices of [imath]\epsilon[/imath], the function isn't differentiable.) The change in [imath]x[/imath] is [imath]dx = \epsilon[/imath], and the change in [imath]f(x)[/imath] is [imath]df(x) = f(x+\epsilon) - f(x)[/imath].

Likewise, the integral is given as a sum of an infinite number of infinitesimally small areas, where each area looks like [imath]f(x)\epsilon = f(x)dx[/imath].
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