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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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.)

*Last edited by ryos (2006-02-16 12:39:26)*

El que pega primero pega dos veces.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

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)]

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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.

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

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.

*Last edited by mikau (2006-02-16 13:03:49)*

A logarithm is just a misspelled algorithm.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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.

*Last edited by ryos (2006-02-16 16:42:24)*

El que pega primero pega dos veces.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

(3e³ - 2e²) - (e² - e²)

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.

*Last edited by mikau (2006-02-16 16:21:55)*

A logarithm is just a misspelled algorithm.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

Stupid typos. Again. It's fixed now.

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

El que pega primero pega dos veces.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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]

Why did the vector cross the road?

It wanted to be normal.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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

El que pega primero pega dos veces.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Theres a command I never heard of...

Thanks ryos.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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

El que pega primero pega dos veces.

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