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anonymous

  • one year ago

Integrate this pls

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  1. freckles
    • one year ago
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    integration!!!!!

  2. freckles
    • one year ago
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    looks like we going to need some completing the square and some trig sub action

  3. welshfella
    • one year ago
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    yes

  4. freckles
    • one year ago
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    \[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c-\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]

  5. freckles
    • one year ago
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    that is generally how I complete the square

  6. freckles
    • one year ago
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    if you want you can use that as a formula too

  7. freckles
    • one year ago
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    you know of course assuming a isn't 0

  8. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{12s+0.025^2}~ds}\) it is like this?!

  9. SolomonZelman
    • one year ago
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    or what is exactly the integral?

  10. freckles
    • one year ago
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    there is a square root over the bottom part

  11. anonymous
    • one year ago
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    Yes @SolomonZelman

  12. freckles
    • one year ago
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    oh no square root?

  13. anonymous
    • one year ago
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    Trying to find the value of s

  14. freckles
    • one year ago
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    wait that 5 is a s?

  15. freckles
    • one year ago
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    I'm no confused

  16. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.025}^2}~ds}\) ??? I am also confused

  17. anonymous
    • one year ago
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    that's s

  18. SolomonZelman
    • one year ago
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    the last thing I wrote

  19. SolomonZelman
    • one year ago
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    ?

  20. SolomonZelman
    • one year ago
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    so square root is raised to the second power?

  21. freckles
    • one year ago
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    I thought it was: \[\int\limits_{0}^{2000}\frac{ds}{\sqrt{12s+0.02s^2}}\]

  22. SolomonZelman
    • one year ago
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    is that 5 or s at the end of the root?

  23. SolomonZelman
    • one year ago
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    can you draw it pliz? sliver x?

  24. anonymous
    • one year ago
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    @freckles was right. It isnt 5..

  25. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\int\limits_{0}^{2000}\frac{1}{\sqrt{12s+0.02s^2}}~ds}\)

  26. anonymous
    • one year ago
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    yes yes

  27. SolomonZelman
    • one year ago
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    I ended up with partial fractions which are solvable, but it took too many steps, I think it is not correct to post it here.

  28. freckles
    • one year ago
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    you show eventually wind up with this (if you do the completing the square and you aim for a trig sub) \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^2-1} }\]

  29. freckles
    • one year ago
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    should (not show)

  30. freckles
    • one year ago
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    \[\sec^2(\theta)-1=\tan^2(\theta) \\ \text{ so } \sec(\theta)=\frac{s+300}{300} \\ \]

  31. anonymous
    • one year ago
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    so its sec^2 theta

  32. SolomonZelman
    • one year ago
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    this is what I ended up after I have written all possible latex :o-:()

  33. SolomonZelman
    • one year ago
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    I completed the square in the root, and factored out of 1/ √(0.02) then I did u=s+300

  34. SolomonZelman
    • one year ago
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    then partial fractions decomposition/expansion

  35. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\frac{\sqrt{2}}{120} \int\limits_{30}^{2300}\frac{\sqrt{u^2-90,000}}{u-300}-\frac{\sqrt{u^2-90,000}}{u+300}~du}\)

  36. SolomonZelman
    • one year ago
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    (also factored out of 600)

  37. SolomonZelman
    • one year ago
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    won't disturb tho'

  38. anonymous
    • one year ago
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    And I know its a long long process..

  39. SolomonZelman
    • one year ago
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    I hate bad integrals like yours....

  40. SolomonZelman
    • one year ago
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    I would use geom. shapes if it was my choice. To many equations, sick of them:o-:(

  41. anonymous
    • one year ago
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    Well actually this was derived in a dynamics problem. :(

  42. SolomonZelman
    • one year ago
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    oh

  43. SolomonZelman
    • one year ago
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    if you sure u derived it correctly.... (don't think you would make a mistake on that... i have seen you solve DE)

  44. freckles
    • one year ago
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    I'm not really sure if you need anymore help me but let me know

  45. freckles
    • one year ago
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    also @SolomonZelman your way is equivalent but I think it still needs trig sub

  46. anonymous
    • one year ago
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    @freckles after noted that it was sectheta.. whats next?

  47. freckles
    • one year ago
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    differentiate

  48. SolomonZelman
    • one year ago
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    perhaps, I was thinking of separating them into two integrals and in one go w=u-300 and in other p=u+300

  49. freckles
    • one year ago
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    \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{ds}{\sqrt{(\frac{s+300}{300})^2-1} } \] \[\sec(\theta)=\frac{s+300}{300} \\ \sec(\theta) \tan(\theta) d \theta=\frac{1}{300} ds \\ 300 \sec(\theta) \tan(\theta) d \theta =ds\]

  50. freckles
    • one year ago
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    \[\frac{10}{\sqrt{2}} \int\limits_0^{arcsec(\frac{23}{3})} \sec(\theta) d \theta \]

  51. anonymous
    • one year ago
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    okay @freckles im following =)

  52. SolomonZelman
    • one year ago
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    ((( When I did those in my class I had a nice math professor who would let use simpson's trapezoidal and midpoint rules, but on a condition that we find the max possible error, and use a bigger n. ))) not a fan of these:) bye

  53. freckles
    • one year ago
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    should finally get: \[\frac{10}{\sqrt{2}} \ln|\frac{23}{3}+\frac{1}{3} \sqrt{23^2-3^2}|-\frac{10}{\sqrt{2}}\ln|1+0|\]

  54. anonymous
    • one year ago
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    lol @SolomonZelman thank you so much =)

  55. freckles
    • one year ago
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    the second term is just 0

  56. freckles
    • one year ago
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    \[\frac{10}{\sqrt{2}} \ln|\frac{23}{3}+\frac{2 \sqrt{130}}{3}|\]

  57. freckles
    • one year ago
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    I will let you play with that and dress it however you want

  58. anonymous
    • one year ago
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    Got 17.78

  59. SolomonZelman
    • one year ago
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    didn't do much.. yw

  60. freckles
    • one year ago
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    by the way you can drop the | |

  61. SolomonZelman
    • one year ago
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    close

  62. SolomonZelman
    • one year ago
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    wolfram gives 19.274

  63. anonymous
    • one year ago
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    Ops. Its 22.004

  64. freckles
    • one year ago
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    yeah you are a little bit off by the way since this was an ugly beast and you probably will come up again a lot more ugly foe I suggest wolfram for checking your answers

  65. freckles
    • one year ago
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    oops and solomon already said something about wolfram

  66. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=10%2Fsqrt%282%29*ln%7C23%2F3%2B2%28sqrt%28130%29%2F3%29%7C

  67. freckles
    • one year ago
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    which you could also put the initial integral in there and compare what I got with that should be same thing

  68. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=integrate%281%2Fsqrt%280.02s%5E2%2B12s%29%2Cs%3D0..2000%29 they actually round theirs to the nearest thousandth and don't off any more digits

  69. anonymous
    • one year ago
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    Aaah yes yes..

  70. anonymous
    • one year ago
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    Its 19..

  71. anonymous
    • one year ago
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    Thank you so much @freckles

  72. anonymous
    • one year ago
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    I still need to digest this..

  73. freckles
    • one year ago
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    if you want to be exact the answer is: \[\frac{10}{\sqrt{2}} \ln(\frac{23+2 \sqrt{130}}{3}) \\ \text{ or if you just want to give a better answer than wolfram then you can say something like } \\ 19.273952\]

  74. anonymous
    • one year ago
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    Thank you so much =)

  75. freckles
    • one year ago
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    np

  76. anonymous
    • one year ago
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    @freckles I wanna ask something.. Why did you use the completing the square?

  77. freckles
    • one year ago
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    I did the completing the square to do the trig sub that usually works for things in this form: \[\int\limits_{}^{}\frac{1}{\sqrt{(ax^2+bx+c)^n}}dx\]

  78. anonymous
    • one year ago
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    Oh yes thank you!

  79. anonymous
    • one year ago
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    @freckles how did you get the 300?

  80. anonymous
    • one year ago
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    and also the 1/30 sqrt of 2?

  81. freckles
    • one year ago
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    I followed what I said about completing the square: \[as^2+bs+c \\ a(s^2+\frac{b}{a}s)+c \\ a(s^2+\frac{b}{a}s+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ a(s+\frac{b}{2a})^2+c-\frac{ab^2}{4a^2} \\ a(s+\frac{b}{2a})^2+c-\frac{b^2}{4a} \\ a(s+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\] You can use this as a formula but I will go through the steps with the numbers instead just for another example of how to complete the square: \[0.02s^2+12s+0 \\ 0.02s^2+12s\\0.02(s^2+\frac{12}{0.02}s) \\ 0.02(s^2+\frac{12}{0.02}s+(\frac{12}{2(0.02)})^2)-0.02(\frac{12}{2(0.02)})^2 \\ 0.02(s+\frac{12}{2(0.02)})^2-0.02 \frac{12^2}{2^2(0.02^2)} \\ 0.02(s+\frac{12}{2} \frac{1}{0.02})^2-0.02 \frac{(2 \cdot 6)^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{1}{0.02})^2-0.02 \frac{2^2 6^2}{2^2(0.02)^2} \\ 0.02(s+6 \frac{100}{2})^2-0.02 \frac{ \cancel{2^2}6^2}{\cancel{2^2}(0.02)^2} \\ 0.02(s+3(100))^2-\cancel{0.02} \frac{6^2}{0.02^{\cancel{2}}} \\ 0.02(s+300)^2-\frac{36}{0.02} \\ 0.02(s+300)^2-\frac{3600}{2} \\ 0.02(s+300)^2-1800 \\ \frac{2}{100}(s+300)^2-1800 \\ 1800( \frac{2}{100(1800)}(s+300)^2-1) \\ 1800(\frac{2}{(100)18(100)}(s+300)^2-1) \\ 1800(\frac{2}{18(100)^2}(s+300)^2-1) \\ 1800(\frac{1}{9(100)^2}(s+300)^2-1) \\ \\ 1800(\frac{1}{(300)^2}(s+300)^2-1) \\ 1800((\frac{s+300}{300})^2-1)\]

  82. freckles
    • one year ago
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    \[\sqrt{1800}=\sqrt{900 \cdot 2}=\sqrt{900} \sqrt{2} =30 \sqrt{2}\]

  83. freckles
    • one year ago
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    is how I came to: \[\frac{1}{30 \sqrt{2}}\int\limits_{0}^{2000} \frac{1}{\sqrt{(\frac{s+300}{300})^2-1}} ds\]

  84. anonymous
    • one year ago
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    Thank you thank you soo much

  85. anonymous
    • one year ago
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    @freckles how did you get the upper limit (arc sec 23/3)? =)

  86. freckles
    • one year ago
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    s=2000 and \[\sec(\theta)=\frac{s+300}{300}\] plug in old limit into sub to find new limit

  87. anonymous
    • one year ago
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    Ahh thank you so much @freckles =)

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