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anonymous

  • one year ago

I don't know how to approach this one... Help would be very much appreciated!!

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  1. anonymous
    • one year ago
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  2. Zale101
    • one year ago
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    18x-x^2 is a standard quadratic equation. If you converted (18x-x^2) into the vertex form, you can easily apply u-sub.

  3. Zale101
    • one year ago
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    Have you tried that?

  4. anonymous
    • one year ago
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    Ohhh just a sec! Im gonna give that a try :)

  5. anonymous
    • one year ago
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    So is it 81 - (x - 9)^2 ?

  6. Zale101
    • one year ago
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    |dw:1438583101045:dw|

  7. Zale101
    • one year ago
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    Correct

  8. anonymous
    • one year ago
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    Ah yes! Great! :)

  9. Zale101
    • one year ago
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    |dw:1438583282353:dw|

  10. anonymous
    • one year ago
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    Ohh I see! So then it turns into \[\int\limits \sqrt{-u^2 +81} du = \int\limits (-u^2+81)^{1/2}du = \int\limits (-u+9)du = -\frac{ 1 }{ 2 }u^2 + 9u\] Is that right?

  11. anonymous
    • one year ago
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    I think my 3rd step is wrong, isn't it?

  12. anonymous
    • one year ago
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    yea theres something wrong with the power

  13. anonymous
    • one year ago
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    I knew it xD lol How would I simplify it?

  14. anonymous
    • one year ago
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    |dw:1438584976279:dw|

  15. anonymous
    • one year ago
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    Right!

  16. anonymous
    • one year ago
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    I mean is my step from (-u^2 + 81)^(1/2) to (-u+9) correct?

  17. anonymous
    • one year ago
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    i remember you have to divide by 18+1

  18. anonymous
    • one year ago
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    \[\int\limits_{}^{}(ax+b)dx=\frac{ 1 }{ a }\times \frac{ (ax+b)^{a+1} }{ a+1 }\]\[a \neq-1\]

  19. anonymous
    • one year ago
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    \[\int\limits_{}^{}(ax+b)^ndx\]oh god the lag with the equation is horrible

  20. anonymous
    • one year ago
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    Lol xD I'm kinda lost

  21. anonymous
    • one year ago
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    the x^2 is the messy thing

  22. anonymous
    • one year ago
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    I don't exactly know what to do with it xD but it totally looks wrong

  23. anonymous
    • one year ago
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    where did ya power go?

  24. anonymous
    • one year ago
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    you need to add 1 to 1/2 and divide it by 1+1/2

  25. anonymous
    • one year ago
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    I considered it as (-u^2)^(1/2) and 81^(1/2) which is completely wrong, haha

  26. anonymous
    • one year ago
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    @saseal but then what about inside the brackets?

  27. anonymous
    • one year ago
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    I can? But isn't that just like breaking up a root like this... \[\sqrt{a+b} = \sqrt{a} + \sqrt {b} \] And I though that was incorrect.. Is it not?

  28. anonymous
    • one year ago
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    thought*

  29. Astrophysics
    • one year ago
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    \[\int\limits \sqrt{-u^2+81}du \implies \int\limits \sqrt{81-u^2}du \] do another substitution, requires trig.

  30. Jhannybean
    • one year ago
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    Oh, maybe a double substition could work?

  31. Jhannybean
    • one year ago
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    Dang it.

  32. anonymous
    • one year ago
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    yea i feel i should have gone \[\int\limits_{}^{}\sqrt{18x-x^2}dx = \int\limits_{}^{}(18x-x^2)^{\frac{ 1 }{ 2 }}dx\]\[u=x^2\]\[du=2x\]

  33. Astrophysics
    • one year ago
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    This is not a pleasant integral lol Your next substitution is \[u = 9\sin(t)\]

  34. anonymous
    • one year ago
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    Haha! Right!! Something to do with these... ? \[\sin^2x+\cos^2x =1\] and \[\tan^2x +1 = \sec^2x \] Or am I on the wrong track again xD lol

  35. anonymous
    • one year ago
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    @Astrophysics i tried that with a calculator and the answer looks a little crazy

  36. anonymous
    • one year ago
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    @Astrophysics how'd you get 9sin(t) ? :O

  37. anonymous
    • one year ago
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    @saseal which one did you try?

  38. Jhannybean
    • one year ago
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    \[\int \sqrt{-u^2+81}du \rightarrow \int (-u^2+81)^{1/2}du \rightarrow \\ v=-u^2+81 ~,~ dv=-\frac{2}{3}u^3du\rightarrow -\frac{3dv}{2} \\ \sf \text{ it gets messy with algebraic subs :(...}\]

  39. anonymous
    • one year ago
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    astro's trig integral

  40. anonymous
    • one year ago
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    Hmmm... @Jhannybean I'm trying to figure out what you did there lol

  41. Jhannybean
    • one year ago
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    Its called a double substitution.

  42. Jhannybean
    • one year ago
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    But don't use it, it's not going to be pretty.

  43. Jhannybean
    • one year ago
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    i got it.

  44. Jhannybean
    • one year ago
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    \[\sqrt{81-u^2}du \longrightarrow \sqrt{a-x^2}dx ~\therefore~ x=a\sin(t)\]

  45. anonymous
    • one year ago
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    @Astrophysics I tried putting in wolfram's answer before, but it said it was incorrect >.<

  46. Jhannybean
    • one year ago
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    It's a trig substitution rule that you learn in calculus.

  47. anonymous
    • one year ago
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    you wan symbolab's answer?

  48. Astrophysics
    • one year ago
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    Oooh nice

  49. anonymous
    • one year ago
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    always good to have more options

  50. Astrophysics
    • one year ago
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    Yeah try that answer, it uses the exact approach I was talking about

  51. anonymous
    • one year ago
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    Alrighty... thats a lot to type in xD haha hold up xD

  52. Astrophysics
    • one year ago
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    I mean maybe there is a simpler way, I don't know.

  53. anonymous
    • one year ago
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    i don't know how i forgot much of my integration stuff right after the test

  54. anonymous
    • one year ago
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  55. anonymous
    • one year ago
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    I don't know am I typing it in incorrectly or what? XD

  56. Astrophysics
    • one year ago
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    You put it in wrong, it's all over 2

  57. anonymous
    • one year ago
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    Ohhh shoooot! >.<

  58. anonymous
    • one year ago
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    Still wrong...

  59. Astrophysics
    • one year ago
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    Your doing the wrong integral

  60. anonymous
    • one year ago
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    it looks like you have enter your answer on a different question

  61. Astrophysics
    • one year ago
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    bahaha!

  62. anonymous
    • one year ago
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    XDDDDD

  63. anonymous
    • one year ago
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    Oh my goodness ahahaha

  64. anonymous
    • one year ago
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    dont blame me its almost 1 am

  65. anonymous
    • one year ago
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    pretty cool you can enter the answer over and over again

  66. anonymous
    • one year ago
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    hehehehe My bad :P

  67. Astrophysics
    • one year ago
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    Lol I always hated answering questions on the computer...that's why I like assignments where you can hand it in, and yay!

  68. anonymous
    • one year ago
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    yay!

  69. anonymous
    • one year ago
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    much wow, such correct

  70. Astrophysics
    • one year ago
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    xD

  71. anonymous
    • one year ago
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    I know! Lol but i still don't understand how they get that >.<

  72. Astrophysics
    • one year ago
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    Go over the symbolab result I guess

  73. Astrophysics
    • one year ago
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    it shows all the steps

  74. anonymous
    • one year ago
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    thats why i chose em over wolfram

  75. Astrophysics
    • one year ago
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    here is the trick though... http://www.sosmath.com/calculus/integration/trigsub/trigsub.html

  76. anonymous
    • one year ago
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    Yeah, i guess i should! It sucks that wolfram doesn't show all the steps unless you go Pro XD

  77. Astrophysics
    • one year ago
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    Yeah :(

  78. Astrophysics
    • one year ago
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    Nice work everyone! Collective work is always good, especially for bad integrals!

  79. anonymous
    • one year ago
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    Yeah! Thanks all!! @Jhannybean @saseal @Astrophysics :D

  80. Astrophysics
    • one year ago
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    and @Zale101 :P

  81. anonymous
    • one year ago
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    I was going all nuts over this one lol

  82. anonymous
    • one year ago
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    Oh yes!! And thanks to @Zale101 as well! :D Thanks for reminding me! Haha

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