View Full Version : Calculus
AbysmalGuilt
02-08-2008, 04:00 AM
Someone want to help meh out? I have two probs, I know some of you are math wizzes from reading the last math thread.
Anyways, problem:
Evaluate the integral. Solve by the simplest method--not all require integration by parts. (I don't know how to make the integral sign so you'll have to assume it's there.)
(x^4)(lnx)dx
and
1/x(lnx)^3 dx
First problem, toughest thing is, since I'm trying to make lnx my dv, I dunno how to integrate lnx. I looked this up and it said make lnx my u since it's easier to find the derivative for it (which it is) however, it never comes to a 0...
Suggestions please. You don't have to answer it for me, but any hints will work. Thanks much.
Jerov
02-08-2008, 04:10 AM
Someone want to help meh out? I have two probs, I know some of you are math wizzes from reading the last math thread.
Anyways, problem:
Evaluate the integral. Solve by the simplest method--not all require integration by parts. (I don't know how to make the integral sign so you'll have to assume it's there.)
(x^4)(lnx)dx
and
1/x(lnx)^3 dx
First problem, toughest thing is, since I'm trying to make lnx my dv, I dunno how to integrate lnx. I looked this up and it said make lnx my u since it's easier to find the derivative for it (which it is) however, it never comes to a 0...
Suggestions please. You don't have to answer it for me, but any hints will work. Thanks much.
Ok, so the second one is really [(1/x)(ln|x|)^3]dx, and you want to integrate. Well, first hand I'll give ya the derivative of ln|x|, which is 1/x. And the derivate you're given has 1/x in it, so it has to come from somewhere. Hope this gives you a big enough hint.
Edit: whoops! sorry if I'm confusin ya too much bud, been a long time. Here's the integration by parts rule:
Given an integral, we can treat it as the integral of f(x)g'(x)dx, which then we can then do f(x)g(x) - the integral of f'(x)g(x). Now, personally, here's what I'd do. You see that ln|x|? You can now do something with it and make this a whole lot nicer. Once again, sorry for the gibberish earlier, I needed to get my facts straight
AbysmalGuilt
02-08-2008, 04:22 AM
Ok, so the second one is really [(1/x)(ln|x|)^3]dx, and you want to integrate. Well, first hand I'll give ya the derivative of ln|x|, which is 1/x. And the derivate you're given has 1/x in it, so it has to come from somewhere. Hope this gives you a big enough hint.
For the first one, now that's a pain in the ass. Mmm....there's a section on integration by parts. I think how it goes is say you have an integral you can treat as dudv. What you want to do is udv - the integral of vdu. So what I would do to start is pick your u to be the easiest one to use, while leave in the integral that's giving you a pain which you can integrate alright by itself.
If you want a hint on a proper u, just ask
Edit: Oh, I know how to do this problem now. Alright, in order to integrate ln|x| properly, given vdu, with v being ln|x| and du being x^4, here's a big hint. What's the derivative of (ln|x|)^2. Then take a look at du. Something is counting down as you integrate by parts over and over again
That's my problem, I'm not able to integrate lnx for dv, I know to pick easiest as u.
Jerov
02-08-2008, 04:29 AM
Nonono, don't integrate ln|x|, that's not the point behind integration by parts. Treat ln|x| as your v, so the original integral right now is vdu. So now what this will turn into is vu - the integral udv. Try that bud. If it works, then the new integral you create in the difference should be MUCH easier to integrate
fuck math they are lying to you, you will never really use it.
Chaotix
02-08-2008, 04:30 AM
what. the. fuck.
Just when the teacher comes to check your homework be like, See i woulda done it but then it hit me I didnt want to do it and id hate to punish my self like that, she'll just call your house and you can handle that with the simple SHUT THE FUUront door
jcpiaya
02-08-2008, 04:51 AM
If you follow the simple formula for integration by parts, shown
http://en.wikipedia.org/wiki/Integration_by_parts
You should not use ln(x) as your v because the problem becomes increasingly difficult if you try to integrate ln(x)
By setting your problem with:
u=ln(x)
du=1/x
v=x^5/5
dv=x^4
The problem becomes much easier.
If you want to solve the problem on your own, don't look further.
ln(x)(x^5/5)- integral((x^5/5)(1/x)dx)
ln(x)*(x^5/5)-(x^5/25)
now to check your answer, we use derivative on this and see if you get the original answer.
Since the first portion follows the multiplication rule.
=(1/x)(x^5/5)+ln(x)(x^4)-(x^4/5)
=(x^4/5)-(x^4/5)+ln(x)(x^4)
=ln(x)(x^4)
Hint for the second problem:
Notice that the derivative of ln(x) is 1/x
Notice that the ln(x) is being "powered"
We can justify using substitution rule for this problem.
by setting
u=ln(x), we get
du=1/x
That should be more than enough information so solve the problem.
Syetactic
02-08-2008, 04:55 AM
I was going to take that in grade 12...
-goes and changes schedule-
AbysmalGuilt
02-08-2008, 04:58 AM
I was going to take that in grade 12...
-goes and changes schedule-
Yeah, this is actually the first year that I don't catch right on to everything we're doing. I'm usually like top of my class, now I'm like on the bottom. Next to my bad attendance. Darn AP Calculus -_- Good thing I dont NEED it, still gonna try to pass that AP Exam, one way or another.
Thanks for all of the suggestions.
Raiyne
02-08-2008, 05:33 AM
I'm gonna learn that sometime this year. :(
Jerov
02-08-2008, 05:34 AM
np, and if you're lucky (depending on your major) you should only have to go up to Calc 2, where you'll meet the wonderful world of matrices and how to manipulate them. Here's an example of a problem where solving with matrices helps
Solve the series of equations:
x + 2y + 3z = 3
2x + y + z = 2
3x + 2y + 2z = 5
Edit: upon retrospect, you could solve this problem with algebra, but when you do all the substitution, it looks hideous, hence why matrix work is better
ncyphe
02-08-2008, 06:24 AM
The sad part is that Integration is society is everywhere. I have to pretty much use it in 50% of my Physics equations.
Integrate a graphical equation utilizing the area of a circle and you can magically determine the volume of a graph rotated about an axis. Fun and complicated stuff.
Just wait till you get to mixture/flow rate problems, those are Hell.
Note: I barely passed Calc II with a C. I could have passed with a D, but believe me, you don't want a D on you transcript.
Mr. First Name Basis
02-08-2008, 11:26 AM
Just when the teacher comes to check your homework be like, See i woulda done it but then it hit me I didnt want to do it and id hate to punish my self like that, she'll just call your house and you can handle that with the simple SHUT THE FUUront door
Yeah. your cool....
There's this new trend going around that's called going somewhere in life. You should try it.
Sorry haven't taken Calculus yet but my sis is so I know a little bit about derivatives (sp?) but yeah that's pretty much it. I'm not really sure how to integrate so yea... I'm taking Calc next year.. taking Pre Calc now.
hobosexual
02-08-2008, 12:58 PM
I'm taking Pre Calculus right now. It sucks ass, even though I'm getting a B in there I absolutely hate it. I think i'm just gonna take fucking business math next year...
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