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anonymous

  • 5 years ago

anti-derivative of SQRT(1+(9x- (x^3/3))^2)

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  1. amistre64
    • 5 years ago
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    how you wanna do this lol the easy way or the hard way

  2. anonymous
    • 5 years ago
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    lets go with easy because this is the part of problems i can never do

  3. amistre64
    • 5 years ago
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    \[\sqrt{1+(9x-\frac{x^3}{3})^2}\] lets start by making sure we got the right problem is this it?

  4. anonymous
    • 5 years ago
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    yes

  5. amistre64
    • 5 years ago
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    well; its gonna be a trig sub that turns to int(sec(x)) dx inthe end lol

  6. anonymous
    • 5 years ago
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    omg i hate calculus lol

  7. amistre64
    • 5 years ago
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    lol...trig sub just cleans it up thats all

  8. anonymous
    • 5 years ago
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    yea but i integrated this on my calculator and i still got the wrong answer

  9. amistre64
    • 5 years ago
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    \[\int_{} \sqrt{1+tan^2(u)}du\] ;\[u= \frac{27x -x^3}{3}\]

  10. amistre64
    • 5 years ago
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    du = 27-3x^2/3 = 9-x^2 dx; maybe that aint a good idea lol

  11. amistre64
    • 5 years ago
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    but it shows me another idea lol

  12. anonymous
    • 5 years ago
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    yea im not so good with substitution either

  13. amistre64
    • 5 years ago
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    u = 9x-x^3 ; du = 9-3x^2 dx {S} sqrt(1+(u/3)^2) du -------------- 9-x^2 maybe?

  14. amistre64
    • 5 years ago
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    let me stare blindly at it for a few moments and let my thoughts congeal lol

  15. anonymous
    • 5 years ago
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    ok im just trying to follow the process here

  16. amistre64
    • 5 years ago
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    we might be better of expanding and combining everything under the radical into one shot

  17. anonymous
    • 5 years ago
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    sure i couldnt go anymore wrong with this problem lol

  18. amistre64
    • 5 years ago
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    \[\frac{\sqrt{x^6 + 729x^2 -54x^4 +9}}{3}\] is what i get with that ...maybe

  19. anonymous
    • 5 years ago
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    yea i got something completely off but u r def prolly right lol

  20. amistre64
    • 5 years ago
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    now we can factor under the radical to simply fy things

  21. amistre64
    • 5 years ago
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    sooo...how should we factor it :) all polynomials higher then degrees 2 can factor to linears and or irreducibla quads

  22. anonymous
    • 5 years ago
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    i honestly have no idea lol

  23. amistre64
    • 5 years ago
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    ima try some synthetic stuff and hope i get lucky

  24. anonymous
    • 5 years ago
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    lol ok im attempting to do other problems..im going to fail this quiz tomorrow lol

  25. amistre64
    • 5 years ago
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    well, the good news is; wolfram couldnt even figure it out lol

  26. anonymous
    • 5 years ago
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    whos that? some really smart kid im guessing :)

  27. amistre64
    • 5 years ago
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    www.wolframalpa.com type in: int(sqrt(1+(9x-(x^3)/3)^2))dx

  28. anonymous
    • 5 years ago
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    o gotcha lol..ok well maybe i went wrong somewhere..the question is find the length of the arch of the parabola y=9-x^2 that lies above the x-axis

  29. amistre64
    • 5 years ago
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    wolframalpha.com

  30. amistre64
    • 5 years ago
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    .... why you wanna keep secrets from me; i cant read your mind

  31. amistre64
    • 5 years ago
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    you had the wrong stuff inputed lol

  32. anonymous
    • 5 years ago
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    i did?? well the equation we use is SQRT(1+(dy/dx)^2

  33. amistre64
    • 5 years ago
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    \[\int\limits_{} \sqrt{1 + [f'x]^2} dx\]

  34. amistre64
    • 5 years ago
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    and is the derivative of 9-x^2 = to the integral of 9-x^2?

  35. anonymous
    • 5 years ago
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    yea and the derivative of 9-x^2 is 9x-(x^3/3) right?

  36. amistre64
    • 5 years ago
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    dy/dx = -2x

  37. anonymous
    • 5 years ago
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    ooo i did the anti derivative didnt i ..i do that all the time because we keep having to switch from derivativeto anti derivative lol

  38. amistre64
    • 5 years ago
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    anti derivative is a useless term; later chapters call them what they are inegrals lol

  39. amistre64
    • 5 years ago
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    integrals

  40. anonymous
    • 5 years ago
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    yea that what we do when we use integrals

  41. amistre64
    • 5 years ago
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    {S} sqrt(1+4x^2) dx ; [a,b] whatever your interval s its up there some where

  42. anonymous
    • 5 years ago
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    ok now i got it..wow i am retarded lol i do that a lot to

  43. anonymous
    • 5 years ago
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    so now how to integrate SQRT(1+4x^2)

  44. amistre64
    • 5 years ago
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    easier than you would that other monstrocity lol

  45. anonymous
    • 5 years ago
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    i got 2/3(1+4x^2)^(3/2) and then do you also multiply that by the derivative of the inside??

  46. amistre64
    • 5 years ago
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    sqrt(1+tan^2(u)) sec^2(u) du ------------------------- 4 sec(u). sec^2(u) du ---------------- 4 sec^3(u) du ---------- 4

  47. amistre64
    • 5 years ago
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    got that outta whack a little

  48. anonymous
    • 5 years ago
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    ok i g2g i got soccer..but thanks for the help :)

  49. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=int%28+sqrt%28+1%2B+%28-2x%29^2%29%29+dx+from+-3+to+3

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