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The Mathematics Club ~ *ΜΑΘ* ~

Snow Phoenix

The transient snowman
982
Posts
15
Years
  • Well, I sure forgot about this place @-@ Nyeh... I remembered it because I have somewhat mehish news.

    I only have two more math classes left until I've finished all of the math I need in college :/ Calculus II and Stats. Everything after would be an empty credit. So I'm quite sad, yet happy.
     

    Ascaris

    boogey
    381
    Posts
    15
    Years
  • Username: i dont know
    Overall Education Level: 12th grade
    Mathematics Education Level (Or most recent/advanced math subject): im pretty much thorough on all forms of algebra, euclidean and cartesian geometry and basic multivariable calculus.
    Do you think you can be asked for help in your level or lower?: sure. as long as you dont ask me questions about statistics. i hate statistics.
     

    Corvus of the Black Night

    Wild Duck Pokémon
    3,416
    Posts
    15
    Years
  • Username: Corvidae
    Overall Education Level: Going into Freshman year college next year
    Mathematics Level (Or most recent/advanced math subject): I took stats and calculus, but I was horrible at stats and my calculus is rusty.
    Do you think you can be asked for help in your level or lower?: Pre-calc and I'm there. I'm taking calculus in college again for a reason.
     

    2Cool4Mewtwo

    Pwning in Ubers since 1996.
    1,182
    Posts
    13
    Years
  • This thread needs a deserved bump :)

    Username: 2cool4mewtwo
    Overall Education Level: Senior in high school.
    Mathematics Education Level (Or most recent/advanced math subject): AP Calculus BC, AP Statistics
    Do you think you can be asked for help in your level or lower?: Calc AB I'm somewhat comfortable with, though I'm not that reliable on what I'm learning right now :P
     

    Impo

    Playhouse Pokemon
    2,458
    Posts
    14
    Years
  • I remember posting here before, don't know why I stopped...

    Anyways, I've got my advance math classes these days, and they are just so unappealing.
    Stupid Differential Calculus or whatever it's called.

    I'm stuck on these questions, I'm afraid to admit, I need to work over them again x)
     

    Snow Phoenix

    The transient snowman
    982
    Posts
    15
    Years
  • I remember posting here before, don't know why I stopped...

    Anyways, I've got my advance math classes these days, and they are just so unappealing.
    Stupid Differential Calculus or whatever it's called.

    I'm stuck on these questions, I'm afraid to admit, I need to work over them again x)
    Ooo! I took that last semester at the community college I attend :3 I might be able to help :D I made an A in the class and I'm currently in Integral Calculus myself.
     

    Renii

    Se(Renii)ty
    83
    Posts
    12
    Years
  • I'm studying calculus right now. I might be able to help as well.

    In fact, I have an exam based on Differentiation and Integration in a few days.
     

    Renii

    Se(Renii)ty
    83
    Posts
    12
    Years
  • Could you guys give this question a try:

    BJicO.gif


    I was able to integrate it using the substitution t^20 = x and got a long answer. Is there an easier way to solve it?
     
    3,655
    Posts
    16
    Years
  • I've been doing some more thinking on your problem and have come up with a possible simplification.

    You could try the substitution t = x^k^-1, where k is any real number.

    What that does is inverse the power so you still end up with t^5 and t^4 on the denominator but it should simplify the numerator and make the integrand easier to tackle. I haven't run through the maths or anything but if you want, try it and see.
     

    Renii

    Se(Renii)ty
    83
    Posts
    12
    Years
  • You mean take t = x^(1/20)? (t = x^20^-1 ?)

    I tried that, gives a long answer. It requires applying the binomial theorem with a power of 15)

    Or something else like t = 1/x^k? How? :/
     
    3,655
    Posts
    16
    Years
  • I meant what I typed out exactly.

    t = x^k^-1

    So for the first term you have x^(1/5) so the transformation gives t^5 and the second term goes from x^(1/4) to t^4

    If that makes sense.

    Edit: The transformation inverses the power of x or in other words you flip the fractions over.
     
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