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Topic review (newest first)

2005-08-24 19:12:52

∫udu = u/2 + c
∫u is not defined.
In integration, the dx or du component is important.
∫1/x is not defined. But, ∫(1/x)dx = logx + c or lnx + c(as denoted in some textbooks).
Similarly, ∫(x^n)dx = [(x^n+1)/(n+1)] + c.

2005-08-24 00:34:59

Thanks for the replies. Now I get where the 2 comes from. Now my question is: where does the du go? u/2 is the integral of u...what happened to the du?

Thanks again.

2005-08-23 20:55:54

Not a bad first post, Atled!

2005-08-23 20:52:39

∫ ( ln(x)/x )  dx  or  ∫ (dx/x) *ln(x)

let u= ln(x)

du/dx = 1/x or du = dx/x

replace dx/x with du

and ln(x) with u

then you are left with ∫ u du == u/2

thats where the 2 comes from.

2005-08-23 18:56:37

maybe it is the power half "brought down", as in

0.5 ln x=ln (x^0.5)

2005-08-23 14:38:37

I'm working through The Complete Idiot's Guide to Calculus. It's been great so far, but it's explanations of integration have proven inadequate to make me understand. Specifically, I've gotten quite stuck with a certain problem involving u-substitution:

[Edit: the integral symbols that I pasted from the row above came through as ?, so that will have to do. smile all the ? below are intended to be integral symbols.]

Integrate (x) = ln(x) / x  on the interval [1, 100].

So I go like this:
u = ln(x)
du = 1/x dx

?(u * du) = ln(x)dx ? ln|x|dx = u ... what?

I'm stuck; I don't get it. The solution in the book goes like this:
? [b=ln100, a=ln1] udu ->

u | ln100
2   | ln1    ->

(ln100) / 2 - 0

I understand why the limits of integration changed from [1, 100] to [ln1, ln100], and that the final expression is g(b) - g(a), which as I understand it is the final step to definite integration.

What I don't understand is, where did the /2 come from?

Thanks for the help. I hope my explanation is clear; some notation really doesn't translate to the web at all.

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