[seanohue]

When you are integrating something like (u-2)/(sqrt(u)) over the interval 1 to 4, when you break up the terms do you add, subtract or multiply? The anti-derivative for the top is u^2/2 - 2u and for the bottom is 2u^(1/2). Do I add, subtract or multiply those terms? Sorry for the dumb question

 

[nibiyabi]

You might need the quotient rule of differentiation:

EDIT: Oh, looks like lipidicman had the same idea.

[size=xx-small]I have no idea what's going on.

[/size]

 

[lipidicman]

You might need the quotient rule of differentiation

I haven't solved it for you, but it looks like that form
http://en.wikipedia.org/wiki/Derivat...the_derivative

 

[Trippytiger]

Well, (u-2)/sqrt(u) is equal to [u/sqrt(u)]-[2/sqrt(u)]. That's just basic math, and easy to see, right?

The properties of integrals lets you integrate each term seperately - that is, the integral of (u-2)/sqrt(u) is equal to the integral of u/sqrt(u) minus the integral of 2/sqrt(u). You don't actually need to separate them like this, but it might help make things clearer for you.

From there, knowing that u/sqrt(u) = u*u^-0.5 = sqrt(u) makes solving the first integral a piece of cake. and the second one is easily solved as it is.

Make sense?

Unless I misread your post, I think the approach that you wanted to take shows that you really need to familiarize yourself with the properties of integrals, and maybe fractions for that matter. It's very basic math, but knowing little tricks like that can make integration a lot easier.

EDIT: And something very important that you might need to watch out for with that equation is the fact that the graph of your function crosses the x-axis between the limits of integration. If you have to get the total area between the graph and the x-axis, you'll need to remember to break up your boundaries.

 

[nibiyabi]

oops, delete

 

[seanohue]

Ah, I didn't think about splitting them up like that. Thanks Trippy

 

[Trippytiger]

No problem! Glad I could help.

 

[mrarroyo]

And they say there are no nerds here!

 

[uraflit]

omg calculus... i got A's and B's in thos classes, but i hated them with a fiery red passion

 

[Lazarus Short]

How dare you say "simple" and "calculus" in the same sentence? Were this my site, I would ban you forthwith.

Laz

 

[Trippytiger]

Oh, come on, basic differentiation and integration are as easy as arithmetic! Compared to what I'm doing in my classes now, that question is Mickey Mouse stuff.

I think someone is an artsy...

 

[Lazarus Short]

Quote:

Originally Posted by Trippytiger /img/forum/go_quote.gif Oh, come on, basic differentiation and integration are as easy as arithmetic! Compared to what I'm doing in my classes now, that question is Mickey Mouse stuff. I think someone is an artsy...

Artsy? Yes, I've had poems published, have written two short storys, but bombed calc two times running. Our brains may not be wired the same.

Laz

 

[Chef Medeski]

Quote:

Originally Posted by Trippytiger /img/forum/go_quote.gif Oh, come on, basic differentiation and integration are as easy as arithmetic! Compared to what I'm doing in my classes now, that question is Mickey Mouse stuff. I think someone is an artsy...

Yeah come on. Product Rule: Drop the numder down. Subtract 1. How much easier can it get?

 

[Trippytiger]

Quote:

Originally Posted by Lazarus Short /img/forum/go_quote.gif Artsy? Yes, I've had poems published, have written two short storys, but bombed calc two times running. Our brains may not be wired the same. Laz

True. I've only bombed calc once.

Quote:

Originally Posted by Chef Medeski /img/forum/go_quote.gif Yeah come on. Product Rule: Drop the numder down. Subtract 1. How much easier can it get?

That's actually not the product rule... The product rule deals with two differentiable functions multiplied together. What you've described is just the standard derivative of any given variable.

 

[gotchaforce]

Quote:

Originally Posted by seanohue /img/forum/go_quote.gif When you are integrating something like (u-2)/(sqrt(u)) over the interval 1 to 4, when you break up the terms do you add, subtract or multiply? The anti-derivative for the top is u^2/2 - 2u and for the bottom is 2u^(1/2). Do I add, subtract or multiply those terms? Sorry for the dumb question

The answer to this is simple, just plug the stupid thing into your calculator! (using fnint)