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Ephilo

  • one year ago

find the integral.

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  1. rosjua
    • one year ago
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    -5^-x/(log(5))+C

  2. Ephilo
    • one year ago
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    |dw:1361280612277:dw|

  3. mathsmind
    • one year ago
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    i have been trying to enter the site since yesterday did you all have the same problem?

  4. mathsmind
    • one year ago
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    anyways there are many ways to solve this problem, one of them would be Taylor series the other one would be substitution method

  5. Ephilo
    • one year ago
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    @mathsmind yea ive been having problems with it to so i did the problems i know how to do until it finally works

  6. Ephilo
    • one year ago
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    @mathsmind i figured that one out im stuck on another one now

  7. Ephilo
    • one year ago
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    evaluate the intergal |dw:1361284859202:dw|

  8. Ephilo
    • one year ago
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    @mathsmind that one ^^^

  9. mathsmind
    • one year ago
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    sorry i meant i was having problems entering the website since yesterday

  10. mathsmind
    • one year ago
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    i know how to solve both integral in my head

  11. Ephilo
    • one year ago
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    @mathsmind how long are you going to be one today ?

  12. mathsmind
    • one year ago
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    it depends on my work, if i am busy or not

  13. mathsmind
    • one year ago
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    can you answer my question please?

  14. mathsmind
    • one year ago
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    were you able to log on to the site

  15. Ephilo
    • one year ago
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    @mathsmind which one ? lol

  16. Ephilo
    • one year ago
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    oh no i wasnt i answered it above

  17. mathsmind
    • one year ago
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    this website? openstudy

  18. Ephilo
    • one year ago
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    it wasnt loading

  19. Ephilo
    • one year ago
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    i felt like crying lol, with out this website i don't know what id do

  20. mathsmind
    • one year ago
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    ok let me reply to the last integral

  21. mathsmind
    • one year ago
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    the answer for the last integral would be pi/6

  22. mathsmind
    • one year ago
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    now by looking at the last integral you can find out from the table that it is equal to arcsin(x/2)

  23. Ephilo
    • one year ago
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    for the 4-x?

  24. mathsmind
    • one year ago
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    \[\int\limits_{0}^{1}\frac{ dx }{ \sqrt{4-x^2} }\]

  25. mathsmind
    • one year ago
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    is that the integral or not?

  26. Ephilo
    • one year ago
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    yepp

  27. mathsmind
    • one year ago
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    the answer is pi/6

  28. Ephilo
    • one year ago
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    how did you get it?

  29. mathsmind
    • one year ago
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    ok there are two ways to do that

  30. mathsmind
    • one year ago
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    one way is to look at the arctrigs integrals

  31. mathsmind
    • one year ago
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    you know the table of integrals right!

  32. mathsmind
    • one year ago
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    the other way is to go through full substitution method

  33. Ephilo
    • one year ago
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    lol no but ill look them up

  34. mathsmind
    • one year ago
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    ok do you want detailed solution then

  35. Ephilo
    • one year ago
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    the u sub thing ?

  36. mathsmind
    • one year ago
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    yep

  37. mathsmind
    • one year ago
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    you let x=2u

  38. mathsmind
    • one year ago
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    x^2 = 4u^2

  39. Ephilo
    • one year ago
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    okay i get it so far

  40. mathsmind
    • one year ago
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    can you differentiate the last bit?

  41. mathsmind
    • one year ago
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    dx = 2du

  42. mathsmind
    • one year ago
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    \[\int\limits_{0}^{1}\frac{ 2du }{\sqrt{4-4u^2} }\]

  43. mathsmind
    • one year ago
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    \[\int\limits\limits_{0}^{1}\frac{ du }{\sqrt{1-u^2} }\]

  44. Ephilo
    • one year ago
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    your a math genuis lol

  45. mathsmind
    • one year ago
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    thanks that was kind of you, and you are a bright student

  46. mathsmind
    • one year ago
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    let me give you a third way how to do that

  47. Ephilo
    • one year ago
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    i understand the u-sub the most

  48. Ephilo
    • one year ago
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    i have three more integral questions

  49. mathsmind
    • one year ago
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    ok go ahed

  50. mathsmind
    • one year ago
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    you need to know different methods to solve one problem , that for future note...

  51. Ephilo
    • one year ago
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    altho the differentiate part is a bit hard

  52. mathsmind
    • one year ago
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    i have a phone call but still submit your work

  53. Ephilo
    • one year ago
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    but ill look back it later because i have to get this course done to graduate

  54. mathsmind
    • one year ago
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    no i meant you make y =2sinx sub

  55. mathsmind
    • one year ago
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    different substitution

  56. mathsmind
    • one year ago
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    if you look at the unit triangle you can figure out that you need a sin sub

  57. Ephilo
    • one year ago
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    okay

  58. Ephilo
    • one year ago
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    find the intergal |dw:1361286908472:dw|

  59. mathsmind
    • one year ago
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    again you are going to get an arctan but now we need to look at the factors

  60. mathsmind
    • one year ago
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    just by looking at this integral we will get 4/3arctan(3x) +c

  61. mathsmind
    • one year ago
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    the question is what was x=? in this u sub

  62. mathsmind
    • one year ago
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    you need to assume that 3x=u, 9x^2=u^2, 3dx=du

  63. mathsmind
    • one year ago
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    \[\int\limits \frac{ 4du }{ 3(1+u^2) }=\frac{ 4 }{ 3}\arctan(u)+c\]

  64. mathsmind
    • one year ago
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    \[u=3x, \therefore \frac{ 4 }{ 3 }\arctan(3x)+c\]

  65. mathsmind
    • one year ago
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    that's your final answer

  66. Ephilo
    • one year ago
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    okay i just did the other one so i have one more integral question

  67. Ephilo
    • one year ago
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    |dw:1361287534741:dw|

  68. mathsmind
    • one year ago
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    ok

  69. mathsmind
    • one year ago
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    again you can see we will get the arctan + something

  70. mathsmind
    • one year ago
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    your answer should be\[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2}(x-1))+c\]

  71. mathsmind
    • one year ago
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    \[\int\limits\limits\limits \frac{ dx }{4+(1-x)^2}\]

  72. mathsmind
    • one year ago
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    now 2u=1-x, 4u^2=(1-x)^2,

  73. mathsmind
    • one year ago
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    now one thing you need to know is how to let the 4 be a one in the denominator

  74. Ephilo
    • one year ago
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    how do i know that?

  75. mathsmind
    • one year ago
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    ok let me show, it is clear from the u-sub

  76. mathsmind
    • one year ago
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    remember that 2u =1-x, 2du=-dx

  77. Ephilo
    • one year ago
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    okay so far so good lol

  78. mathsmind
    • one year ago
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    \[\int\limits\frac{ 2du }{4(1+(u)^2)}\]

  79. mathsmind
    • one year ago
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    can you see it better this way?

  80. mathsmind
    • one year ago
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    \[\frac{ 1 }{ 2 }\arctan(u)+c\]

  81. mathsmind
    • one year ago
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    \[2u=1-x, \space u=\frac{ 1 }{ 2 }(1-x)\]

  82. mathsmind
    • one year ago
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    \[\frac{ 1 }{ 2 }\arctan(\frac{ 1 }{ 2 }(1-x))+c\]

  83. mathsmind
    • one year ago
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    that's your final answer

  84. Ephilo
    • one year ago
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    okay thanks ! im working on some ill let you know if i need more help

  85. mathsmind
    • one year ago
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    you don't sound like you got the last one?

  86. Ephilo
    • one year ago
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    lol i havent fully looked at it yet im finishing a problem

  87. mathsmind
    • one year ago
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    ok

  88. mathsmind
    • one year ago
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    so i read your mind well hehehe

  89. Ephilo
    • one year ago
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    lol yeahh i was wondering how you knew that,but im looking at it now

  90. mathsmind
    • one year ago
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    i still don't understand how the engagement scheme works on open study

  91. Ephilo
    • one year ago
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    i get that one

  92. mathsmind
    • one year ago
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    good

  93. Ephilo
    • one year ago
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    i think its by the questions you answer, and medals you receive

  94. mathsmind
    • one year ago
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    no i am not asking about those two, i am asking about the engagement part

  95. mathsmind
    • one year ago
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    the one that has learner next to it

  96. Ephilo
    • one year ago
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    lol i thinks its from asking questions cause thats how you learn right

  97. Ephilo
    • one year ago
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    okay i found two i need help on

  98. mathsmind
    • one year ago
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    go ahead

  99. Ephilo
    • one year ago
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    determine which of the integrals can be found using the basic integration formulas you have studied so far in the text.

  100. mathsmind
    • one year ago
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    i think you need to open a new thread because this one is getting heavy with BW

  101. Ephilo
    • one year ago
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    lol okay i will

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