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anonymous

  • 5 years ago

Can someone explain me the steps of solving integrals using partial fractions?

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  1. anonymous
    • 5 years ago
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    Should i give you am example?

  2. anonymous
    • 5 years ago
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    the integration is usually the easy part. finding the decomposition is a pain often

  3. anonymous
    • 5 years ago
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    ya i don't get the steps. my book didnt tell me

  4. anonymous
    • 5 years ago
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    |dw:1326817894193:dw|

  5. anonymous
    • 5 years ago
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    \[\frac{1}{(x-2)(x-3)}=\frac{a}{x-2}+\frac{b}{x-3}\] and you need a and b. we know that \[a(x-3)+b(x-2)=1\] so this is an easy one. since this equality has to be true for all values of x, it must be true if x = 2, so you get \[a(2-3)+b(2-2)=1\] \[-a=1\] \[a=-1\]

  6. anonymous
    • 5 years ago
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    lol they got a diff answer is that ok?

  7. anonymous
    • 5 years ago
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    How about using heavy side cover up for this one?

  8. anonymous
    • 5 years ago
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    repeat the process making x = 3, and you will find \[b=1\] so we have \[\frac{1}{(x-2)(x-3)}=\frac{-1}{x-2}+\frac{1}{x-3}\] and now integrate term by term. now it is not ok

  9. anonymous
    • 5 years ago
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    *no

  10. anonymous
    • 5 years ago
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    They got 1/3 and -1/3

  11. anonymous
    • 5 years ago
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    then maybe i made a mistake, let me write it out

  12. TuringTest
    • 5 years ago
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    is that a 5 on the second set of parentheses?

  13. TuringTest
    • 5 years ago
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    sat used a 3

  14. anonymous
    • 5 years ago
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    doh, i used 3!!

  15. anonymous
    • 5 years ago
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    LOL ya sorry abt that

  16. anonymous
    • 5 years ago
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    ohhh didnt even notce that sat

  17. TuringTest
    • 5 years ago
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    sat has done it correctly for sure, just with the wrong number

  18. anonymous
    • 5 years ago
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    no that is my fault, but idea is clear right?

  19. anonymous
    • 5 years ago
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    \[a(x-5)+b(x-2)=1\] let \[x=2\] get \[-3a=1,x=-\frac{1}{3}\]

  20. anonymous
    • 5 years ago
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    ummmm so y did u chose the value 2?

  21. anonymous
    • 5 years ago
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    i am going to wager you can figure out why i picked 2

  22. anonymous
    • 5 years ago
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    and what i will pick next also

  23. anonymous
    • 5 years ago
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    need a hint?

  24. anonymous
    • 5 years ago
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    well cuz it is in the denomanator? lol

  25. anonymous
    • 5 years ago
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    i have this equality \[a(x-5)+b(x-2)=1\] and i am looking for a and b. so how can i find a easily?

  26. anonymous
    • 5 years ago
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    by eliminating one variable by making one part =0

  27. anonymous
    • 5 years ago
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    K i got that LOL

  28. anonymous
    • 5 years ago
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    zactly. so first we let x =2, and then we let x = 5

  29. anonymous
    • 5 years ago
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    oh ok got that :D Thanks :D

  30. anonymous
    • 5 years ago
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    yw

  31. anonymous
    • 5 years ago
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    I thought u guys started teaching today

  32. anonymous
    • 5 years ago
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    This one a is a simple one if I have a more difficult one I will be back :D

  33. anonymous
    • 5 years ago
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    with some practice you can do this: write \[\frac{1}{(x-2)(x-5)}=\frac{a}{x-2}+\frac{b}{x-5}\] now to find a, put your finger over the x - 2 in the first expression, put x = 2 and get \[a=\frac{1}{2-3}=-\frac{1}{3}\]

  34. myininaya
    • 5 years ago
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    its called office hours

  35. anonymous
    • 5 years ago
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    Thanks sat :D

  36. anonymous
    • 5 years ago
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    i really need to get back to work, but i have 30 emails to send out and got bored bored bored. later

  37. anonymous
    • 5 years ago
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    LOL bye

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