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anonymous

  • 5 years ago

Find the antiderivative of f(x)= (x^3+cube root of x+3)/(3)

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  1. anonymous
    • 5 years ago
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    f'(x)= (x^3+cube root of x+3)/(3) --- forgot the prime

  2. anonymous
    • 5 years ago
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    Distribute the 3 into both components of the numerator and integrate the two parts separately like so: \[F(x)=\int\limits_{}^{}(x^3/3)dx+\int\limits_{}^{}(\sqrt{x+3}/3)dx\] The first integral is fairly straightforward. To do the second one, you have to choose a substitution u = x + 3, find that du = dx and make the substitution to find the antiderivative of sqrt(u)/3. In the end, you should get: \[F(x)=x^4/12+2/9(x+3)^{3/2}+C\]

  3. anonymous
    • 5 years ago
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    it says math processing error so i cant see your work

  4. anonymous
    • 5 years ago
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    Okay, I tried doing it with a square root anyway, so it is incorrect. I'll work on it.

  5. anonymous
    • 5 years ago
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    thank you. Im workin on it too

  6. anonymous
    • 5 years ago
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    If f'(x) = (x^3 + cuberoot(x + 3))/3, then... f(x) = integral(x^3 / 3)dx + integral(cuberoot(x + 3) / 3)dx Find the antiderivative of each of these. The second one you must use a u-substitution of u = x + 3. f(x) = x^4 / 12 + (x + 3)^(4/3) / 4 + C However, the problem wants the antiderivative of f, so we have to take another antiderivative. I'll put the second part of the integral back into the u-substitution to get: F(x) = integral(x^4 / 12)dx + integral(u^(4/3) / 4)du + C F(x) = x^5 / 60 + 3u^(7/3) / 28 + Cx + D I hope I did that right... :/

  7. anonymous
    • 5 years ago
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    Ha, silly me. You don't need to do the part of finding the antiderivative of the antiderivative, I don't think.

  8. anonymous
    • 5 years ago
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    oh yea thats right, its just one antiderivative

  9. amistre64
    • 5 years ago
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    f(x) = [x^3 + cbrt(x+3)] --------------- 3 is this the equation?

  10. anonymous
    • 5 years ago
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    yes

  11. anonymous
    • 5 years ago
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    thats it

  12. anonymous
    • 5 years ago
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    please explain an easier way of looking at antiderivatives. My teacher just wants us to do trial and error. I know all the derivative rules by the way

  13. amistre64
    • 5 years ago
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    (1/3)x^3 + (1/3)(x+3)^(1/3) sound good?

  14. anonymous
    • 5 years ago
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    thats F(x)?

  15. amistre64
    • 5 years ago
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    no thats a modified version of f(x); trying to get it to look more "do-able"

  16. anonymous
    • 5 years ago
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    oh okay

  17. anonymous
    • 5 years ago
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    so now you take it backwards?

  18. amistre64
    • 5 years ago
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    as much as you can :) the first term is easy... (1/3)(1/4) x^4

  19. amistre64
    • 5 years ago
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    the second one we want to make sure we get all the parts right: (S) (1/3)(x+3)^(1/3) dx (1/3) (S) (x+3)^(1/3) dx lets try to put this into a (S) u du type format.. u = x+3 du = 1 dx which is good, it means that we aint got to add nothing fancy to it :)

  20. amistre64
    • 5 years ago
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    (1/3) (S) u^(1/3) du (1/3) (4/3)(x+3)^(4/3) does that make sense?

  21. anonymous
    • 5 years ago
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    what is S

  22. amistre64
    • 5 years ago
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    lol .... the integral sign is a deformed "S" so I just use that to indicate the integral sign :) (S) = integral sign... thats all

  23. anonymous
    • 5 years ago
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    oh ok

  24. amistre64
    • 5 years ago
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    and it should be (3/4)(x+3)^(4/3) thats better

  25. amistre64
    • 5 years ago
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    check to make sure that derives down to our intended form...

  26. anonymous
    • 5 years ago
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    thats the 2nd term right?

  27. anonymous
    • 5 years ago
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    so the first term is 1/12 (x^4)

  28. amistre64
    • 5 years ago
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    2nd term: (1/3)(3/4) (x+3)^(4/3) right? clean it up some....

  29. amistre64
    • 5 years ago
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    youre keeping good track of this stuff :)

  30. anonymous
    • 5 years ago
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    i can do the chain rule to clean that up?

  31. amistre64
    • 5 years ago
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    no chain rule needed, just squish it all together to get something resembling one term :)

  32. anonymous
    • 5 years ago
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    what do you get?

  33. amistre64
    • 5 years ago
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    (x+1) cbrt(x+1) ------------- is what I get for the 2nd term 4

  34. anonymous
    • 5 years ago
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    we cant get rid of the cbrt?

  35. amistre64
    • 5 years ago
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    nope, its stuck there :) but we were aboe to pull out a chunk of that (x+3)^4 :) namely that (x+3)^3 its fine, unless we missed something. better double check our work to make sure :)

  36. anonymous
    • 5 years ago
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    yea let me do the derivative of it

  37. amistre64
    • 5 years ago
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    im gonna go get a coffee while you whittle away :)

  38. anonymous
    • 5 years ago
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    kk

  39. anonymous
    • 5 years ago
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    this is what i got: F'(x)= x^3+1/4[(x+1)[1/3(x+1)^-2/3+(x+1)^1/3]

  40. anonymous
    • 5 years ago
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    that doesnt seem like the original

  41. amistre64
    • 5 years ago
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    let me do this on paper to check your work...and mine :)

  42. anonymous
    • 5 years ago
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    k sounds good

  43. amistre64
    • 5 years ago
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    x^4 (x+3)^(4/3) F(x) = ---- + ---------- 12 4 This is what I want to "derive" to get back to our original equation....

  44. amistre64
    • 5 years ago
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    I get: 4x^3 4 (x+3)^(1/3) f(x) = ----- + ------------- (4)(3) (4)(3) Do you see that it works? or did I miss something?

  45. anonymous
    • 5 years ago
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    take the derivative

  46. anonymous
    • 5 years ago
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    its not working for me

  47. amistre64
    • 5 years ago
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    check my work, and se if its right :)

  48. anonymous
    • 5 years ago
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    k the first term def works

  49. anonymous
    • 5 years ago
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    the second term would be 1/12(4) 1/3(x+3)^-2/3

  50. anonymous
    • 5 years ago
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    so that is 1/3* (1/3(x+3)^-2/3

  51. amistre64
    • 5 years ago
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    remember to take the (1/4) to the side and derive the rest of it, dont try to confuse this with the quotient rule

  52. anonymous
    • 5 years ago
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    im doing chain

  53. anonymous
    • 5 years ago
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    let me try again

  54. anonymous
    • 5 years ago
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    1 sec

  55. amistre64
    • 5 years ago
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    the only thing to chain would be the derivative of (x+3) which equals 1 :)

  56. anonymous
    • 5 years ago
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    1/4*4/3(x+3)

  57. anonymous
    • 5 years ago
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    1/4*4/3(x+3)^1/3*1

  58. anonymous
    • 5 years ago
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    thats the chain

  59. amistre64
    • 5 years ago
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    there you go :)

  60. anonymous
    • 5 years ago
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    i dont see how that makes the original

  61. amistre64
    • 5 years ago
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    cross your 4s, and whats left?

  62. amistre64
    • 5 years ago
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    (1/4)(4/3) = (1/3)

  63. anonymous
    • 5 years ago
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    yea but its x^3 on the bottom

  64. amistre64
    • 5 years ago
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    (x+3)^(1/3) = cbrt(x+3) right?

  65. anonymous
    • 5 years ago
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    the 3 isnt in the cbrt. Its not suppose to be

  66. amistre64
    • 5 years ago
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    theres no x term on the bottom.... clean it up so you can see what your looking at :)

  67. anonymous
    • 5 years ago
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    but f(x) is (x^3+cbrt(x)+3)/x^3

  68. amistre64
    • 5 years ago
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    now now, I asked you if I had the equation right to begin with.... and now you wanna change it on me? .... :)

  69. anonymous
    • 5 years ago
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    lol sorry

  70. amistre64
    • 5 years ago
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    lol .... well it does make it easier

  71. anonymous
    • 5 years ago
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    i wrote it wrong its my fault

  72. amistre64
    • 5 years ago
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    integratings integrating, doesnt matter if its the right one or not, well get it done :)

  73. amistre64
    • 5 years ago
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    f(x) = (x^3+cbrt(x)+3)/x^3 we good with this one then?

  74. anonymous
    • 5 years ago
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    yes

  75. amistre64
    • 5 years ago
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    just remember, you start at the beginning, and when you get to the end....stop :)

  76. anonymous
    • 5 years ago
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    k

  77. amistre64
    • 5 years ago
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    thats from alice in wonderland, through the looking glass :)

  78. anonymous
    • 5 years ago
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    can we set it up like this x^3*(x)^1/3+3*x^-3

  79. amistre64
    • 5 years ago
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    x^3 +x^(1/3) +3 f(x) = --- --------- --- x^3 x^3 x^3

  80. amistre64
    • 5 years ago
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    f(x) = 1 + x^(-8/3) + 3x^(-3) right?

  81. anonymous
    • 5 years ago
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    which form are you using

  82. anonymous
    • 5 years ago
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    thats right

  83. anonymous
    • 5 years ago
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    1/3/3=-8/3

  84. anonymous
    • 5 years ago
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    1/3-3 i mean

  85. amistre64
    • 5 years ago
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    the rest is cake walk :) f(x) =1 + x^(-8/3) + 3x^(-3) F(x) = [x] + [3x^(-5/3)]/5 - [3x^(-2)]/2 + C good?

  86. anonymous
    • 5 years ago
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    let me take the derivative

  87. amistre64
    • 5 years ago
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    second term should be (-) not (+)

  88. anonymous
    • 5 years ago
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    why negatives, everything needs to be positive that wont make sense

  89. amistre64
    • 5 years ago
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    for instance, take the last term: (S) 3x^(-3) dx 3 x^-2 ------ do you see why its negative now? -2 <----- makes a difference

  90. anonymous
    • 5 years ago
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    oh ok

  91. amistre64
    • 5 years ago
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    2nd term: (S) x^(-8/3) dx x^[(-8/3 + 3/3)] x^(-5/3) 3x^(-5/3) -------------- = ------- = --------- (-8/3 + 3/3) -5/3 -5

  92. anonymous
    • 5 years ago
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    sounds good. lol it was so confusing im sorry

  93. amistre64
    • 5 years ago
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    usually when I miss something on a test, its cause I forgot to keep track of me signs :)

  94. anonymous
    • 5 years ago
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    so it the F(x)=x-1/5*(3x)^-5/3-[3x)^-2]+c

  95. anonymous
    • 5 years ago
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    ?

  96. amistre64
    • 5 years ago
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    it looks good except for this: ....-[3x)^-2]___+c somethings missing :) can you tell me what you left out? and its prolly just a typo error...

  97. anonymous
    • 5 years ago
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    -1/2*(3x)^-2?

  98. amistre64
    • 5 years ago
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    thats what I was looking for :) good job, now get the derivative and see if it matches....which it will :)

  99. anonymous
    • 5 years ago
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    lol my brain isnt working forwards now

  100. amistre64
    • 5 years ago
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    F(x) = x - [3x^(-5/3)]/5 - [3x^-2]/2 + C f(x) = 1 - [(3/5)(-5/3) x^(-8/3)] - [(3/2)(-2) x^-3 ] f(x) = 1 + x^(-8/3) +3x^(-3) which we can modify back to your original stuff....

  101. anonymous
    • 5 years ago
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    oh ok

  102. amistre64
    • 5 years ago
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    does it make sense to you? or are you just agreeing with me :)

  103. anonymous
    • 5 years ago
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    how does x^(-8/3) become cbrt of x?

  104. anonymous
    • 5 years ago
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    never mind, you divide everything by x^3

  105. anonymous
    • 5 years ago
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    correct?

  106. amistre64
    • 5 years ago
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    put all your negative exponents underneath again.... but make sure you take out the x^(-3) from x^(-8/3) remember when bases multiply together, their exponents "add" up. to split them simply undo the process like this: x^(-8/3) = x^(1/3) x^(-9/3)

  107. amistre64
    • 5 years ago
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    yes, in essense, yes :)

  108. anonymous
    • 5 years ago
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    how does the 1 become x^3

  109. anonymous
    • 5 years ago
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    1/x^3=x^3 with the common denomonator

  110. anonymous
    • 5 years ago
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    omg i am acting dumb, sorry

  111. amistre64
    • 5 years ago
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    (x^3)/(x^3) = 1 right? thats where we got it from

  112. anonymous
    • 5 years ago
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    yea thats right 1/x^3= x^3+1/x^3

  113. amistre64
    • 5 years ago
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    had to go hunt down which version we were using :)

  114. anonymous
    • 5 years ago
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    yea it works out, complicated but works out

  115. amistre64
    • 5 years ago
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    its easier with a fresh pair of eyes :)

  116. anonymous
    • 5 years ago
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    yea thank you sooooooo much

  117. amistre64
    • 5 years ago
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    youre welcome :)

  118. anonymous
    • 5 years ago
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    Have a good night, ill ttyl

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