Calculus integration problem. please help. - Printable Version -Shoutbox ( https://shoutbox.menthix.net)+-- Forum: MsgHelp Archive (/forumdisplay.php?fid=58)+--- Forum: General (/forumdisplay.php?fid=11)+---- Forum: General Chit Chat (/forumdisplay.php?fid=14)+----- Thread: Calculus integration problem. please help. ( /showthread.php?tid=67263)
Calculus integration problem. please help. by Rubber Stamp on 10-13-2006 at 11:11 AM
can someone please solve this integration problem for me? i have been trying it since a week, gone mad trying to solve it. please help. RE: Calculus integration problem. please help. by Rubber Stamp on 10-14-2006 at 06:11 PM
could some one do it? RE: Calculus integration problem. please help. by andrewdodd13 on 10-14-2006 at 07:58 PM
Well... the square root of x^2 1/x^2 is just x + 1/x isn't it? RE: RE: Calculus integration problem. please help. by J-Thread on 10-14-2006 at 08:07 PM
quote: No it isn't... The square root of x * x = x... So the square root of x * x + a, where a > 0, is greater then x... Right? When x > 0, 1 / (x * x) > 0, so the square root will be greater then x. You see? The integral of that formula is hard to find... and it's been a while since I have been integrating formulas... RE: Calculus integration problem. please help. by markee on 10-15-2006 at 02:01 AM
Ok well I'm quite sure you have to use integration by parts. I will update this post when I get the answer for you. RE: Calculus integration problem. please help. by John Anderton on 10-15-2006 at 06:44 AM
quote:wtf? If that were true most of the maths problems in the world would be solved in a few steps RE: Calculus integration problem. please help. by Rubber Stamp on 10-15-2006 at 07:27 AM
i dont think this can be done by parts. i tried using it, but the expression we get inside the second integration sign is even more complicated than the question itself. RE: Calculus integration problem. please help. by Chrono on 10-15-2006 at 08:43 AM
have you tried changing variables like sen( u) = x ? Or maybe the good old Tan(x/2) ? i dunno it just looks like it could be possible with one of those, im too lazy to try now though maybe tomorrow RE: Calculus integration problem. please help. by John Anderton on 10-15-2006 at 01:46 PM
quote:sin wont do it. Bleh it could but i cba to work with (sin(u))^4 quote:Isnt that for places where you have trignometrics in the numerator and denominator? I was thinking of something like x^2 = (sin(u))^2 or something like that. cba atm. mabbe after my exams RE: Calculus integration problem. please help. by Rubber Stamp on 10-15-2006 at 04:22 PM
nups, i already tried various trigo properties, none work. RE: Calculus integration problem. please help. by J-Thread on 10-15-2006 at 08:08 PM
How do you know it is solvable? RE: Calculus integration problem. please help. by Chrono on 10-15-2006 at 08:51 PM
quote:lol thats the same than x=sen RE: RE: Calculus integration problem. please help. by andrewdodd13 on 10-15-2006 at 09:43 PM
quote: Yeah, wasn't really thinking straight when I wrote that... I used an on-line integrator (its the first hit for "Online integrator" on google ) and it gave (3x^4 - 1) ^-1 . I don't think it's right - doesn't differentiate to the correct result as far as I can tell: - (12x^3) / ((3x^4)^2) I'll have another bash it via substitution, but I think integration by parts is the way to go... but I can't do that yet. ^_^. RE: Calculus integration problem. please help. by markee on 10-15-2006 at 11:29 PM
quote: I think he was thinking of doing something with De'Mouive's but I don't think it will work in this case. What I did to get as far as I did was I had to first expand the equations and then use a reversal of the product rule and you should then try to simplify what you get. It is the only way I can think of doing this problem. EDIT: Just expanding out the equation I got it to x^3+1/x+x/(x^4+1) you should be able to integrate it in parts now. for some reason maple is giving me 1/4*x^4+ln(x)+1/2*arctan(x^2) but I didn't think the arctan was right, I thought it was meant to be a ln and for some reason it is different to my other expression I guess it might be that I messed up somewhere last time, I could get rid of the squareroot which i couldn't before. I don't know, lets hope I'm not as dodgy in my maths tutorial this afternoon, I'll ask my teacher to help if I remember RE: Calculus integration problem. please help. by Chrono on 10-16-2006 at 12:29 AM
ah maple, im glad i dont need to use that thing anymore . i really disliked that program for some reason, even though it was pretty useful for Diferential equations RE: Calculus integration problem. please help. by Rubber Stamp on 10-17-2006 at 05:23 PM
quote: its from a set of question from a pretty respectable book. RE: Calculus integration problem. please help. by kotjze on 10-17-2006 at 05:41 PM
quote: x^3+1/x+x/(x^4+1) does not equal (x+1/x)*sqrt(x^2+1/x^2) which is why maple gave you the wrong answer. RE: RE: RE: Calculus integration problem. please help. by Jesus on 10-17-2006 at 06:13 PM
quote:lol I did that too, got this: don't know if it's correct, looks far too complicated for me edit: image is gone, will try to get it back edit2: got it RE: Calculus integration problem. please help. by s7a5 on 10-17-2006 at 09:18 PM
never did this type of problem b4, and substitution technique doesnt seem to work for it so maybe that's how it goes: RE: Calculus integration problem. please help. by kotjze on 10-18-2006 at 01:50 AM
quote: Really? So... (5^2 + 1/6^2)^1/2 = (5 + 1/6)? No. If that were true, then (x + 1/x)^2 = (x^2 + 1/x^2), but if you work that out, it's (x^2 + 2 + 1/x^2) RE: Calculus integration problem. please help. by Vilkku on 10-18-2006 at 04:54 AM
I don't have enough time to actually to solve it atm, but I think I know how I would attempt it. RE: Calculus integration problem. please help. by Chrono on 10-18-2006 at 05:01 AM
quote:you cant do that, lol counterexample: int x^2 =(X^3)/3 int x * int x = (x^4)/4 i tried to solve it earlier but didnt manage to get to an answer i used to be good at those, but it's been a long time since i had to integrate a difficult expression quote:ew that's so wrong you cant do that either... RE: RE: Calculus integration problem. please help. by Vilkku on 10-18-2006 at 05:02 AM
quote:Oh yeah, it has to be + or - between them... was too long scince I did this. RE: RE: Calculus integration problem. please help. by J-Thread on 10-18-2006 at 11:55 AM
quote:I've got a pretty respectable book that asks if P=NP, but I didn't manage to solve that either However, I think I have enough mathematical knowledge to solve this problem, but as I stated before, it a pretty tough one. A lot of people have failed... But when I have some spare time I will try to solve it RE: RE: Calculus integration problem. please help. by andrewdodd13 on 10-18-2006 at 10:50 PM
quote: That's what I done.. (x^2 + 1/x^2) ^ 1/2 != (x+1/x) . Via parts, I get the following... I don't think it's right, I cba differentiating it atm, because I'd then have to do something fancy to get it into its original form . However, I also don't think by parts will work because of the sqrt... RE: Calculus integration problem. please help. by markee on 10-18-2006 at 11:45 PM
You sure the question wasn't (x + 1/x)(x^2 + 2 + 1/(x^2))^(1/2) ? RE: Calculus integration problem. please help. by J-Thread on 10-21-2006 at 05:52 AM
I tried and I thought I had the good idea, but I came this far (see attachement). Now I'm stuck again...it is really a difficult one! RE: Calculus integration problem. please help. by Chrono on 10-21-2006 at 05:59 AM
haha i did the exact same thing, that was actually my first try . RE: Calculus integration problem. please help. by markee on 10-21-2006 at 02:27 PM
I did the same thing in about as many steps as chrono. Will post my steps from there. code: see why I wanted to simplify it a bit..... RE: RE: Calculus integration problem. please help. by J-Thread on 10-22-2006 at 12:05 PM
quote:That's just what you like or not. I am used to writing my steps down as clear as possible. Especially when I am proving something, I like to do easy steps that everybody can check. If no step is wrong, I have proved the fact I wanted to prove. When I do "bigger" steps, the prove will be a bit hard to follow. Of course I see the bigger steps myself, but I know that not everybody does. However, it seems like this trick also didn't work. I honestly begin to believe that there isn't a solution to this problem, but I can't prove it yet. And I guess it would be easier to solve a problem (if a solution exists) then proving that no solution exists... RE: Calculus integration problem. please help. by Buttah N Bred on 02-15-2007 at 12:19 PM
Wow! It appears that everyone is getting something different. Here is what I did. I used substitution and realized that the 1/x^2 can be rewritten as 1x^-2. RE: Calculus integration problem. please help. by Rolando on 02-15-2007 at 01:04 PM
quote: Nice math but.. did you see the date of the other posts? Months ago.
RE: Calculus integration problem. please help. by Buttah N Bred on 02-15-2007 at 08:17 PM
no, I wasn't focused on the date that it was posted. this was for anybody else who may have had this problem. |