Help with dx

1. Nov 28, 2004

QuantumTheory

Well, I love calculus. But sometimes, its confusing. Ever been though a time while you haven't been to a single calc class, and just learn it all though reading a textbook not being able to ask questions?

It's hard.

I'm having a problem.

f (x) = (integral sign) F(x) dx

I don't understand the constant of integration, and why it can be anything.

but the main problem here is whast the difference between an indefinite integral and a definite integral with dx?

I know the integral sign from a to b dx means all the infinitely small pieces of x added together.

But I don't understand the integral sign and dx together without area, like the above example.

What does x stand for?

Since this is a function of f(x), if it was f(t) would it be dt?

Would this just mean the "change in" aka delta t?

And i asked my teacher at my school what the difference between dx and dx/x is, but I forgot already :(

Thanks for the help, I wish I would have known of this forum earlier.

2. Nov 29, 2004

James R

Integration is the reverse operation of differentiation. But lots of different functions have the same derivative. For example, consider:

f(x) = x^2 + 7
g(x) = x^2 - 13
h(x) = x^2 + pi/2

When you differentiate f,g or h with respect to x, you get 2x in all cases. So, when you integrate 2x, how do you know whether it should give you f,g or h, or something else? You don't. That's why you add an arbitrary constant.

A definite integral is evaluated between definite limits, and so requires no arbitrary constant. An indefinite integral leaves in the uncertainty of the undetermined constant.

For example, here's a definite integral (no arbitrary constant needed):

$$\int\limits_0^x 2t dt = t^2|^x_0 = x^2$$

Here's an indefinite integral:

$$\int 2t dt = t^2 + c$$

where c is an undetermined (indefinite) constant.

You can think of $\int\limits_a^b f(x) dx$ as the (signed) area between the function f(x), the x axis, and the lines x=a and x=b, if you like.

Yes. The dt or dx just tells you which variable to integrate. That is particularly important when you have multiple variables. For example:

$$\int 2x + t dt = 2xt + t^2/2 + c$$

but

$$\int 2x + t dx = x^2 + tx + c$$

These are equivalent:

$$\int \frac{1}{x} dx \equiv \int \frac{dx}{x}$$

3. Nov 29, 2004

arunma

Good question. Now that you mention it, it does seem a bit confusing at first. After all, the definite integral is defined as a sum of infinitely many, small numbers. So the dx makes some sense there. But how does it fit in with the notion of antidifferentiation?

It turns out that the dx isn't just a notational issue. It's actually an extremely helpful notation when you start doing direct and inverse substitutions (but more on this next semester). So if you figure out what it means right now, that's very helpful.

Now go back to derivatives. When y = f(x), we can write dy/dx = f'(x), so long as f'(x) exists. But if we think of the symbol dy/dx as a division, we can write dy = f'(x) dx (this is called differential form).

Think of the integral sign as the inverse of the "d" operator. If y = f(x), then dy = f'(x)dx. By this logic, the dx at the end of an integral is what is left over from the derivative.

On to your next question. Why can the constant of integration be anything?

Well, if we again write y = f(x), then we can write dy/dx = f'(x). As you know, the integral of dy/dx is not f(x). It is f(x) + C. But if we know that y = f(x), then why do we have to add the constant anyway?

Here's why. There is a difference between taking an integral, and solving a differential equation. If we had the equation dy/dx = f'(x), we would be trying to "solve" for y. In that case, writing down "f(x)" as a solution would suffice. But when we evaluate the integral of dy/dx, we are not trying to find y. We are trying to find any function which, when differentiated, will give dy/dx = f'(x). The function f(x) certainly satisfies this, but so does f(x) + 1 or f(x) - 6. Because any constant will differentiate to 0, we need to add an arbitrary constant.

I know this sounds rather pointless. But the constant of integration is actually very important in applications to physics, as well as in the study of differential equations. But then, we physics people are known for using all kinds of smarmy tricks to get out of writing down the constant (physicists are lazy people, you know), so I could get in trouble for saying all of this.

Well, I hope this helped. Let me know if this post didn't make any sense, and I'll see if I can explain more clearly.

4. Nov 29, 2004

arunma

Hey James, quick question for you. How do you format mathematical symbols (such as the integral sign and all those itallic characters) into your posts?

5. Nov 29, 2004

WaR

He uses LateX syntax and the forums use a combination of that, dvi2ps, and pstoimg. Hover your mouse over his formulas and you will see the code.

Edit: Woohoo, my first post :)

Last edited: Nov 29, 2004
6. Nov 29, 2004

QuantumTheory

Hehe, thank you for the lengthy response. I guess in order to understand the constant of integration better I must understand differenation, I do not fully understand what it is.

But I do understand (according to the book Calculus For Dummies) that if dy/dx is the change of dy/dx of a curve, then the differivative (or the slope) of dy/dx = f'(x) if and only if f(x) = dy/dx

7. Nov 29, 2004

Ethereal

How does Lebesgue integration differ from Riemannian integration?

8. Nov 30, 2004

matt grime

It is a device that allows you to integrate more functions: some of the ones where the (upper and lower) Riemann sums do not converge (to the same thing).

For exanple: f:R->R, f(x)=0 if x is rational 1 otherwise is not Riemann integrable, but we consider there to be "far more irrationals than rationals" uncountable v countable, and that they should carry more "weight" (white lie, please forgive me) and hence the lesbesgue integral over the unit interval is actually 1.
So we try and find a way to assign "weightings" to subsets of the reals. countable subsets carry, on their own, no "weight" - we say they have measure zero (even if, like the rationals they are dense).