Math Is Fun Forum

Because to get 4x to be 10x, so we can substitute 10x dx = du, we need to multiply 4x by 5/2 ((5/2)*4x = 10x). But this gives us a different integral (instead of ∫f(x) dx we now have ∫(5/2)f(x) dx). So to keep the integral the same, we multiply it by the reciprocal of 5/2, which is 2/5. This will work out since (2/5)∫(5/2)f(x) dx = ∫f(x) dx, so it is indeed the same integral. Is that clear?

Yes, the image you posted with the substitution dx = du/10x works as well. It is just a matter of preference. If you understand that method better, then perhaps you should use it instead.

I am actually a student, not a teacher. I call myself a teacher to sound cooler, and perhaps I am just an unofficial teacher. I do teach other math and physics if they let me, I just don't get a salary for it .

I think I may post a thread in the Exercises section about Integration by Substitution later tonight. Keep your eyes open for it, and when it shows up, try to do the exercises to see if you get the concept of Integration by Substitution. For now, I need to go do homework . I'll be back later.

Edit: I just realized that you said "why isn't there a dx on the end" for 10x = du. There is supposed to be: du = 10x dx. I'm not sure if I made a typo somewhere and forgot the dx or what. Just in case you're wondering, we get du = 10x dx from

Mein Gott, es gibt ein gute Sammlung von Witze hier! Schöne.

Have you learned the technique of Integration by Substitution? Here's the basic idea of it:

Let's look at our integral:

In its current form, you probably don't recognize it as something you can integrate. but if it were in the form

you could solve it in a heartbeat. The goal of Integration by Substitution is to get the integral into a simple form like the one above, so you can evaluate the integral. How do we go about this? Well, I'm not sure how well I can explain this, and it may seem like I am making so leaps in logic, but this is just a thing you need to practice, and then it will become natural and you'll be able to know the correct substitution to make nearly all of the time.

Ok, now if we were to substitute some u in for x in a given integral ∫f(x) dx, by the chain rule the integral would turn into ∫f(u) du. This probably just looks confusing, doing the example will make it become clearer. So let's look at our integral here and find a suitable expression to set as u. The best way to go about this is to think of what integrals we can easily solve that look like the given integral and then what substitution x = u would be able to change the integral into that form. Looking at the integral, we may see the form

which we know we can solve easily. It turns out that we can get this form if we let u = 5x² - 1. We choose this substitution so that we can get only one term under the square root. Almost always with Integration by Substitution we wish to get only one term in places like (u)[sup]n[/sup], cos(u), e[sup]u[/sup], etc. Anyway, if we let u = 5x² - 1, then du = 10x dx (do you get what I did here?). Now we plug in u for 5x² - 1 and du for 10x dx(we multiplied 4x by 5/2 in the second step to get 4x to be 10x and to balance out that 5/2 we multiplied the entire integral by 2/5):

Now we just substitute 5x² - 1 for u to get our answer in terms of x:

I hope this was clear.

Sorry, my recent solution of a long-standing problem of functional analysis has dazed me in such a manner that I experience hallucinations.

You kind folks should check out this article, which contains information nobody here has heard about yet, especially Mikau:

http://edition.cnn.com/2006/TECH/space/ … index.html

For any anti-Firefox activists here, explain why Firefox is so bad. Also, feel free to point out factors that make particular other browsers better.

Firefox is the story of my life

I believe you are talking about perfect numbers in the first part of your post(the part where you talked about 6).

I'm not sure exactly what to make of the rest of the post.

Do you know if there is a way to type alt codes on a laptop with no number pad?

The m comes from MathIsFun.

Numen's view is great. You can also say that 3.333... = 3 + 1/3 = 10/3 and 6.666... = 6 + 2/3 = 20/3, and we have

3.333... + 6.666... = 9.999...

but equivalently

10/3 + 20/3 = 30/3 = 10,

so 9.999... = 10.

Another way to see that 9.999... = 10 is to use a geometric series like so:

Perhaps a few threads from "This is Cool" should be moved here? Examples would be ben's "Lie groups, anyone?" and Zhylliolom's "ζ(2n)". Just a suggestion.