does anyone know calculus here

Discussion in 'Community' started by Steven1621, Jun 3, 2004.

  1. Steven1621 macrumors 6502a

    Steven1621

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    #1
    hey i have a rather difficult integration problem that i just cannot solve and i figure that macrumors people are rather smart so i figured i would see if anyone here knew how to do this.

    the integration is (3x^(1/2))/(1+x^(3/4)) dx

    i really just need the first few steps and i should be able to figure it out from there.

    thanks!
     
  2. Vlade macrumors 6502a

    Vlade

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    #2
    Hehe, I just accidently differentiated it, but I can't figure out how to integrate it. Anyone have a TI89 calculator?
     
  3. yoman macrumors 6502a

    yoman

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    #3
    try splitting it up like this:

    3x^(1/2) * (1+x^(3/4))^-1

    then integrate.
     
  4. MoparShaha macrumors 68000

    MoparShaha

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    #4
    As a matter of fact, I do :D.

    Answer is:

    -4(ln(x^(3/4)+1)-x^(3/4))​

    I haven't actually tried it on paper, but it looks like an integration by parts problem.
     
  5. reaper macrumors 6502

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    #5
    Try thinking about the integration in terms of the 3x^(1/2) piece being a component of an initial (1+x^x)^y term. This way, that x^x component will have to in some way (using multipliers) differentiate out into the resultant 3x^(1/2). Also, some cancellations will probably occur to make it all look pretty.

    Backwards thinking is always the hardest, and yet, for me, was always the most rewarding. Hope this helps without giving you too much of a hint (i.e., doing the work for you). See where it takes you and don't be afraid to do trial-and-error... sometimes that's the only way with these sorts of problems.

    - reaper
     
  6. King Cobra macrumors 603

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    Mar 2, 2002
    #6
    I think it's substitution.

    TIP: The integral sign on a Mac keyboard is: option-b

    So, you have:
    ∫(3x^(1/2))/(1+x^(3/4))dx

    --> Let u = x^(3/4)

    That means x^(1/2) = u^(2/3)
    and x^(1/4) = u^(1/3)

    So, for du:

    du = 3/(4x^(1/4)) = 3/(4u^(1/3)) du
    EDIT: (error spotted by Veldek) it should read: du = 3/(4x^(1/4)) = 3/(4u^(1/3)) dx


    When you bring all that back into the integral and simplify, you should have:

    ∫u/(1+u) du

    That is an integration by parts problem.
     
  7. reaper macrumors 6502

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    #7
    Ack... you gave him the answer. Where't the fun in that? :p

    - reaper
     
  8. legion macrumors 6502a

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    #8
    Wow, don't feel like doing your homework the right way, huh? :rolleyes: I just realized I looked at 2 posts from you to do calculus homework. When you get to your first year in college and plan to do any more math classes, it won't get easier and it's time to learn all the tricks (substitution, by parts, etc) now. (and memorize the tables of integrals)

    I take it from your sig that you're currently a HS senior.
     
  9. Steven1621 thread starter macrumors 6502a

    Steven1621

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    #9
    i certainly know my calculus, but there are integrations that simply are rather challenging and i would say this is one of them. and with this being one of the last assignments of my senior year in high school, i would say that i would just like to be done with it.
     
  10. Neserk macrumors 6502a

    Neserk

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    #10
    Sadly, I skipped Calculus because I was too worried about my GPA :rolleyes: If I had it to do over I say: screw the GPA take Calculus. The even sadder thing is I'm actually very good in math.
     
  11. themadchemist macrumors 68030

    themadchemist

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    #11
    Actually, there's a careless mistake here of some importance. 3x^(1/2)=3(u-1)^(2/3). I thik that King Cobra forgot about the pesky integer within the power function. Actually, that integer complicates things QUITE a bit. However, bite your lip with repeated integration by parts (if I'm not mistaken, it'll take more than one) and you should be fine.

    Keep in mind: Make dv as complicated as you can while still being able to integrate it reasonably. u will generally be the more complicated of the two parts and with good reason: It's easier to differentiate than to integrate.

    Good luck!
     
  12. Veldek macrumors 68000

    Veldek

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    #12
    It’s substitution, but you don‘t substitute u=x^(3/4), but u=x^(3/2). Then you get

    du = 3/2 * x^(1/2) dx.

    Now, when you transfer the integral to

    2/(1+x^(3/4)) * 3/2 * x^(1/2) dx

    you can do the substitution and get

    2/(1+u^(1/2)) du

    The rest is up to you.
     
  13. King Cobra macrumors 603

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    Mar 2, 2002
    #13
    I don't see that error. I took the substitution: u = x^(3/4) -- for now, focus on that substitution, as I'll get to the recommended substitution in the above post shortly.

    Then I rose the power of both sides to the (2/3) power.
    u = x^(3/4) becomes
    u^(2/3) = x^(2/4), or simply x^(1/2)

    Likewise, to get to u^(1/3), I rose the power of both sides of -- u = x^(3/4) -- to the (1/3) power.
    u = x^(3/4) becomes
    u^(1/3) = x^(1/4) <-- that step is needed to substitite in a "u" of some form into x^(1/4) when finding du.


    When I performed that substitution (twice to check for errors), the resulting integral became:

    ∫ 2 /(1+u^(1/2)) du

    Usually, if the original integral has a radical, and you can't resolve that using integration by parts, the chain rule, or du/u, then let "u" = the radical.
     
  14. Veldek macrumors 68000

    Veldek

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    #14
    Oops, you’re right. I forgot that du = 3/2 * x^(1/2).

    But that doesn’t change much. As you said the resulting integral is

    2/(1+u^(1/2)) du
     
  15. Veldek macrumors 68000

    Veldek

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    #15
    So far, it has been correct but then you say that

    du = 3/(4u^(1/3)) du

    This is obviously not true, so it seems you made a mistake there.
     
  16. Abstract macrumors Penryn

    Abstract

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    #16
    Its okay, I think. He only asked on how he should start. The fact that we're giving him more is our business, and icing on the cake for him. :)

    That's the way I'd start. Then use substitution by parts twice.
     
  17. agreenster macrumors 68000

    agreenster

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    #17
    God. It's only been 6 years since high school calculus, and already this looks like greek to me.

    I wasnt GREAT in HS, but I scored a 3 on the AP test.

    :(
     
  18. 1macker1 macrumors 65816

    1macker1

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    #18
    You can to download MAPLESOFT's Maple program to "check" all your calculus problems. It's not to hard to learn maple. You get a free 30 day trial i think. It's totally worth the time.
     
  19. wPod macrumors 68000

    wPod

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    #19
    only do that if you are actually going to check your work!!! in middle school i wrote a calculator program to do quadratics for me (like x^2 +4x -2 =0 what is x?) it was really cool and did just about everything including irrational numbers. and to this day in my calc III course in college i still cannot solve such problems! (well now solving such little algebra is only a small small step, but i still get it wrong! or spend way too much time on it :-/ )

    as for the original question, the integration by parts looks good!
     
  20. Wes macrumors 68020

    Wes

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    #20
  21. themadchemist macrumors 68030

    themadchemist

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    #21
    Bah! Stop using calculator symbolic integration! Learn to do it by hand!
     
  22. Steven1621 thread starter macrumors 6502a

    Steven1621

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    #22
    hey everyone thanks for the help. i had an idea of where i was heading, but had trouble getting the first step, which usually is all i need. this was the tenth of ten very difficult integrations i had to sum up our studies in calc II. again, thanks a lot.
     
  23. virividox macrumors 601

    virividox

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  24. legion macrumors 6502a

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    #24
    This is Calc II? What did they teach you in Calc I... just derivatives? This is part of basic integration and should have been covered in your first 3 months of calculus.

    It's only been 10years since I left high school, but wow are they going down hill...
     
  25. legion macrumors 6502a

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    #25
    No, you aren't doing calculus. Your computer is. You're nothing but an input device... any 2nd grader could be taught to input things into a program to get an answer.

    Learn to do this by hand since it isn't complicated. Using programs and calculators to do calculus should be reserved for harder questions (especially those that don't have exact answers) or when you need to graph or interpolate data points into equations.
     

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