insane Posted January 12, 2007 Share Posted January 12, 2007 I've spent 3 hours on this problem, and I'm guessing I'm missing something extremely simple. If any of you know how to integrate functions, I'd be obliged if you could give me a hand. ÃÆâÃâ¹Ã¢â¬Â Ãâë [x^3 * ÃÆâÃâ¹Ã¢â¬Â Ãâ¦Ã¡(x^2 + 1)] dx I'm totally stuck. And please don't respond unless you actually know how to integrate :P Link to comment Share on other sites More sharing options...
runescapeloser22 Posted January 12, 2007 Share Posted January 12, 2007 :shock: *huddles in corner* Link to comment Share on other sites More sharing options...
kido14 Posted January 12, 2007 Share Posted January 12, 2007 We just started learning integration this week =P. But isn"t there supposed to be a closed interval, like with 2 numbers on the top and bottom of the integral symbol? I guess that must only be with the ones we are doing. And please dont respond unless you actually know how to integrate :P I do know how to integrate, just not thise one :anxious: Sorry if you consider this spam Runescape, what was the point of that, cant you read? Last.fm: http://www.last.fm/user/Aaronm14/MY FAVORITE BAND:http://profile.myspace.com/index.cfm?fu ... d=64310717And the bible is the big book of lies, call me a racist if you must. Link to comment Share on other sites More sharing options...
insane Posted January 12, 2007 Author Share Posted January 12, 2007 But isn"t there supposed to be a closed interval, like with 2 numbers on the top and bottom of the integral symbol? I guess that must only be with the ones we are doing. That's a definite integral, which returns a value. My question is an indefinite integral, which returns a function. Link to comment Share on other sites More sharing options...
Gamemonster Posted January 12, 2007 Share Posted January 12, 2007 X^5 +1 that what i have so far.I'll get back to you when I have a different anwser Link to comment Share on other sites More sharing options...
insane Posted January 12, 2007 Author Share Posted January 12, 2007 X^5 +1 that what i have so far.I'll get back to you when I have a different anwser You are extremely far off, to the point that I don't think you know what integration is. Link to comment Share on other sites More sharing options...
azntemplar_00 Posted January 13, 2007 Share Posted January 13, 2007 X^5 +1 that what i have so far.I'll get back to you when I have a different anwser You are extremely far off, to the point that I don't think you know what integration is. Meh, get on MSN =P. I'm using U-substitution right now, but uh I'm stuck on a step. Might have to review me rules a bit more, but to be honest, so far I really only know how to do just simple anti-derivs and u-subs :( Link to comment Share on other sites More sharing options...
menot Posted January 13, 2007 Share Posted January 13, 2007 you'll have to use integration by parts, but i gave up maths and i cba to work it out for you. babelfish - level 180 60th placestrongguy - level 173 69th place Link to comment Share on other sites More sharing options...
Markkei Posted January 13, 2007 Share Posted January 13, 2007 I tried integration by parts with u = x^2 and dv = xÃÆâÃâ¹Ã¢â¬Â Ãâ¦Ã¡(x^2 + 1) to no avail. Though I'm incompetent with integration so perhaps you can do better :P Link to comment Share on other sites More sharing options...
wakka102 Posted January 13, 2007 Share Posted January 13, 2007 just curious, what class is this for? Link to comment Share on other sites More sharing options...
Viktorkrum77 Posted January 13, 2007 Share Posted January 13, 2007 just curious, what class is this for? Calc. :? Me doing staff. Link to comment Share on other sites More sharing options...
misterxman Posted January 13, 2007 Share Posted January 13, 2007 It's been a while since I done this, but here goes Pixel sigs by me.Pixel Art Link to comment Share on other sites More sharing options...
indy500fan Posted January 13, 2007 Share Posted January 13, 2007 just curious, what class is this for? It's dealing with integrals so it is some form of Calculus. Link to comment Share on other sites More sharing options...
punk4ever Posted January 13, 2007 Share Posted January 13, 2007 Well I think I have it figured out by I have no time to finish it really :P It's integration by parts, that I'm fairly sure of. u = (x^2+1)^(1/2) v' = x^3 u' = x(x^2+1)^(-1/2) v = (x^4)/4 Integral uv' dx = [ (x^2+1)^(1/2) * (x^4)/4 ] - [(1/4) Integral x(x^2+1)^(-1/2) * (x^4)/4 ] Then you have to do integration by parts again on [(1/4) Integral x(x^2+1)^(-1/2) * (x^4)/4 ] Does that help? The answer SHOULD be (1/15)*(x^2+1)^(3/2)*((3x^2)-2) but I didn't get it through going through the steps. There is a handy program which can do pretty much any integral for you without showing the steps (obviously) and it can be found at integrals.wolfram.com By The_Jeppoz :wink: Link to comment Share on other sites More sharing options...
insane Posted January 13, 2007 Author Share Posted January 13, 2007 The wolfram integrator puts it in extremely odd forms, I've found. Anyways, azn, mxm and punky, those were both extremely helpful, thanks a bunch! Link to comment Share on other sites More sharing options...
paragon_JOMO Posted January 13, 2007 Share Posted January 13, 2007 dont you have to use the product rule as your multiplying two functions... neway thts what we call it where im from... unfortunately im to lazy to work it out.... Link to comment Share on other sites More sharing options...
insane Posted January 13, 2007 Author Share Posted January 13, 2007 dont you have to use the product rule as your multiplying two functions... neway thts what we call it where im from... unfortunately im to lazy to work it out.... Product rule is differentiation. The integral method of product rule is called integration by parts. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now