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Christos

  • 2 years ago

Please fellow open study members, solve with me . Finals countdown 3 hours :P https://www.dropbox.com/s/vwu945w28tgowlp/Screenshot%202014-01-15%2015.09.53.jpg (ANXIETY LEVEL OVER THE TOP)

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  1. hartnn
    • 2 years ago
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    so where are you stuck ?

  2. Christos
    • 2 years ago
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    @hartnn

  3. Christos
    • 2 years ago
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    They are the only problems to which I don't have solutions , and I also have infinite series to cover , if you give me what I lack, it would be of a great help at the current situation ! :(

  4. ganeshie8
    • 2 years ago
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    link is not opening for meh

  5. Christos
    • 2 years ago
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    http://screencast.com/t/WittgfBhztpV

  6. Christos
    • 2 years ago
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    just tell me what do I do

  7. Christos
    • 2 years ago
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    for the third one

  8. Christos
    • 2 years ago
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    after the long division , then what

  9. ganeshie8
    • 2 years ago
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    before we dive in, two things to keep in mind :- 1) we can integrate ANY/EVERY rational function 2) to use partial fractions, degreem of numerator must always be LESS than degree of denominator

  10. ganeshie8
    • 2 years ago
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    *degree

  11. Christos
    • 2 years ago
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    *reading carefully*

  12. hartnn
    • 2 years ago
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    yeah, thats the reason she did long division for 3rd

  13. hartnn
    • 2 years ago
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    so, you must have something like x/x^2+1 ,right ??

  14. hartnn
    • 2 years ago
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    hint : d/dx (x^2+1) = 2x

  15. hartnn
    • 2 years ago
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    so take u = x^2+1 , du =...

  16. ganeshie8
    • 2 years ago
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    for first example, since degree of numerator < degree of denominator, next do below steps :- step1 : factor the denominator ive seen hartnn explaining partial fractions excellently before... so i give u hartnn :)

  17. ganeshie8
    • 2 years ago
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    @hartnn ... plz continue if u have time :)

  18. hartnn
    • 2 years ago
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    i have like 3 minutes :P

  19. ganeshie8
    • 2 years ago
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    and im sure, u can make Christos understand half of calculus in less than 3 minutes ;)

  20. hartnn
    • 2 years ago
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    probably, if she is sitting right in front of me ;)

  21. Christos
    • 2 years ago
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    she is intenting a new study method called parallel processing !!

  22. Christos
    • 2 years ago
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    I am just kidding

  23. hartnn
    • 2 years ago
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    times up :P i need to rush b4 my bus leaves without me :P be back online when i reach home..

  24. ganeshie8
    • 2 years ago
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    oh hartnn you're in office ha ... runn.. il see if i can do some justice in explaining partial fractions :)

  25. ganeshie8
    • 2 years ago
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    \(\large \frac{1}{x^2+3x-4}\) factor the denominator \(\large \frac{1}{(x-1)(x+4)}\)

  26. ganeshie8
    • 2 years ago
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    step2 : break left fraction into sum of fractions :- \(\large \frac{1}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}\) find A and B

  27. Christos
    • 2 years ago
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    answer: 1/5ln|x-2| - 1/5 ln|x+4| + C

  28. Christos
    • 2 years ago
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    is it ?

  29. ganeshie8
    • 2 years ago
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    thats right !

  30. Christos
    • 2 years ago
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    lets go 2nd

  31. Christos
    • 2 years ago
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    do I need to factor the denominator here ?

  32. ganeshie8
    • 2 years ago
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    yes, factoring IS the first step

  33. ganeshie8
    • 2 years ago
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    and ofc, the 0th step is long-division :)

  34. Christos
    • 2 years ago
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    A/x b/x+1 C/x+1

  35. Christos
    • 2 years ago
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    B/x-1 * * *

  36. ganeshie8
    • 2 years ago
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    step1 : factor denominator \(\large \frac{1}{x(x^2-1)} \) \(\large \frac{1}{x(x-1)(x+1)} \)

  37. Christos
    • 2 years ago
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    A B C: http://screencast.com/t/a5HayvjoOi

  38. ganeshie8
    • 2 years ago
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    step2 : wrote the fraction as sum of simple fractions \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) solve A, B, C

  39. ganeshie8
    • 2 years ago
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    yes looks good, solve A, B, C

  40. Christos
    • 2 years ago
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    I hope you are not leading me into a teaching trap at the current moment huh :D

  41. Christos
    • 2 years ago
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    ok ok I solve it now

  42. ganeshie8
    • 2 years ago
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    whats teaching trap lol im no teacher... guess im more toward a professor who screws his students hmm lol

  43. Christos
    • 2 years ago
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    ok I can't solve this, whats the next step :D

  44. Christos
    • 2 years ago
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    if x = 1 denominator is undefined A.k.A me missing something

  45. ganeshie8
    • 2 years ago
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    there is a small trick to solve this, it wont take any time to learn the trick. let me show u how to find A, B, C super fast

  46. ganeshie8
    • 2 years ago
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    \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to sovle A : Multuply the whole equation wid \(x\), and plug \(x = 0\) \(\large \frac{1}{(x-1)(x+1)} = A + \frac{Bx}{x-1} + \frac{Cx}{x+1}\)

  47. ganeshie8
    • 2 years ago
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    when u plug \(x = 0\), everything on right side cancels away, except \(A\)

  48. Christos
    • 2 years ago
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    -1

  49. ganeshie8
    • 2 years ago
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    \(\large \frac{1}{(0-1)(0+1)} = A + 0 + 0\) solve A

  50. ganeshie8
    • 2 years ago
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    yes, A = -1

  51. Christos
    • 2 years ago
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    yesyes

  52. ganeshie8
    • 2 years ago
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    similarly find B and C

  53. Christos
    • 2 years ago
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    x = ?

  54. ganeshie8
    • 2 years ago
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    \(\large \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\) to find B :- multiply the whole equation wid \((x-1)\) and plugin \(x = 1\) \(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\)

  55. Christos
    • 2 years ago
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    and then the main fraction is undefined..

  56. ganeshie8
    • 2 years ago
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    how ?

  57. ganeshie8
    • 2 years ago
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    \(\large \frac{1}{x(x+1)} = \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x+1}\) plugin x = 1 \(\large \frac{1}{1(1+1)} = 0 + B + 0\)

  58. ganeshie8
    • 2 years ago
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    il let u solve C :)

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