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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

ryos
2006-02-18 12:15:10

LOL. I guess the tutorial I found was a good one...

This page talks about mbox{}.

Ricky
2006-02-18 07:24:57

Theres a command I never heard of...

Thanks ryos.

ryos
2006-02-18 04:21:52

\mbox{The first post, mathsy. 8O)}

mathsyperson
2006-02-18 00:49:42

ryos wrote:

Edit: on the plus side, this post made me finally learn LaTeX, or at least enough to write it.



[Please ignore this post as it is made of 100% pure stupidity. x_x]

ryos
2006-02-17 15:44:54

Stupid typos. Again. It's fixed now.

As to where I got it, umm...read my posts again.

mikau
2006-02-17 15:21:35

(3e - 2e) - (e - e)

What the math? Where did you come up with that? It should be 3e^3 - e^3 - (2e^2 - e^2). Where the heck did you come up with such a visually repelling expression?

Well I'm just about finishing calculus now. So I'll probably be getting into calculus 2 soon.

ryos
2006-02-17 14:06:05

I'm advanced eh? I guess I'm good at sounding smart or something. I'm pretty sure I'm in the same place (Calculus 2) as you, mikau.

Anyway mikau, my question (which both you and irspow answered the same way) is this, using your example:

xe^x - e^x

Is it:
(3e³ - 2e²) - (e³ - e²)

...or is it:
(3e³ - e³) - (2e² - e²)

...? Actually, working through the two, they both come out the same. Is this always the case, though? Let's see:

Let h(x) = uv and g(x) = ∫u′v, then assert:
[h(b) - h(a)] - [g(b) - g(a)] = [h(b) - g(b)] - [h(a) - g(a)]

h(b) - h(a) - g(b) + g(a) = h(b) - g(b) - h(a) + g(a)

So it is. Ok, dumb question. *smacks self*

Edit: on the plus side, this post made me finally learn LaTeX, or at least enough to write it.

mikau
2006-02-17 12:03:01

pardon me if I'm not interpreting this correctly, I think Ryos is far more advanced then me in math knowledge and I find it suprising he would ask such a simple question.  Which is why I doubt I'm interpreting this question correctly.

But if I am...
Evaluating a definite integral that requires integration by parts?

As far as I know, you simple use u and V as substitutions for the variable of integration, when your done you replace it with its original value.

For instance ∫ x e^x dx from x = 2 to x = 3

This is probably the easiest integration by parts problem. Let u = x, du = dx.   Let dV = e^x dx   V = e^x

so we use the formula   ∫ u Dv dx = uv -   ∫ V du

insert the values back in and we get:

x e^x -  ∫ e^x dx

= x e^x - e^x

= e^x( x - 1)

Now we evaluate this from x = 2 to x = 3:

2e^3 - e^2

piece of cake.

ryos
2006-02-17 11:40:50

Yeah, it's straight from my book. And maybe it's because I learned from that book, but I find it clearer than Ricky's version.

There was an error in it though, which I've now corrected.

Ricky
2006-02-17 08:30:15

If what you copied is straight from your book, that's the weirdest defintion for integration by parts I've ever seen.

Here's one that should be more clear:



From this, it should be clear that you must first divide up your function into two parts, u and dv, then solve for v and du, and finally just plug it all back in.

irspow
2006-02-17 08:06:50

The uv outside of the integral sign in the formula is just a part of the integral "already solved".  So your first interpretation is correct.

[h(b) - h(a)] - [g(b) - g(a)]

ryos
2006-02-17 07:55:04

My book says that the definite version of the integration by parts formula is:



When actually evaluating the integral, I'm not sure how to interpret this. I think it could be either:



...or, it could be:



Is it one of those, or something else entirely?

(Good gads, LaTeX competes with regular expressions for ugliest mishmash of punctuation ever.)

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