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AmTran_Bus

  • one year ago

Integration help

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  1. anonymous
    • one year ago
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    ?

  2. AmTran_Bus
    • one year ago
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    |dw:1434926190793:dw|

  3. triciaal
    • one year ago
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    products

  4. AmTran_Bus
    • one year ago
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    |dw:1434926209378:dw|

  5. AmTran_Bus
    • one year ago
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    \[uv-\int\limits vdu\]

  6. idku
    • one year ago
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    more steps, but less mess and confusion. u=-3t then by parts

  7. AmTran_Bus
    • one year ago
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    \[t \frac{ e ^{-3t} }{ 3 }-\int\limits \frac{ e ^{-3t} }{ 3 }dt\]

  8. idku
    • one year ago
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    u can do a power series too, if you would like.

  9. idku
    • one year ago
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    anyway, u got people here....

  10. AmTran_Bus
    • one year ago
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    This is what I'm doing right now, this is the IBP section.

  11. AmTran_Bus
    • one year ago
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    So do a U sub first? ok.

  12. idku
    • one year ago
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    yes, u=-3t

  13. idku
    • one year ago
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    du=(-3)dx -> du/(-3) = dx

  14. idku
    • one year ago
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    u got this:)

  15. idku
    • one year ago
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    (don't forget to sub back the x)

  16. AmTran_Bus
    • one year ago
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    So do I need to say dt = -du/3?

  17. idku
    • one year ago
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    yes that suffices

  18. AmTran_Bus
    • one year ago
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    |dw:1434926595047:dw|

  19. AmTran_Bus
    • one year ago
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    Supposed to be a neg there... Then IBP?

  20. idku
    • one year ago
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    you can also say, \[\Large \int\limits_{ }^{ }te^{-3t}dt\] \[\Large \frac{1}{-9}\int\limits_{ }^{ }-3(-3t)e^{-3t}dt\] (i used a magic 1) \[u=-3t~~~du=-3dt\] \[\Large \frac{1}{-9}\int\limits_{ }^{ }\color{blue}{-3}\color{red}{(-3t)}e^{\color{red}{-3t}}\color{blue}{dt}\]

  21. idku
    • one year ago
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    the blue part is the part that gets replaced by du. the red - each thing is a u

  22. AmTran_Bus
    • one year ago
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    Where is the 1/9 from? I would like to do it the way I have just started.

  23. idku
    • one year ago
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    oh, 1/9 not 1/-9

  24. AmTran_Bus
    • one year ago
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    |dw:1434926879794:dw|

  25. idku
    • one year ago
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    i multiply times -3 times -3 and times 1/9

  26. idku
    • one year ago
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    and write that all out, to have my u's

  27. idku
    • one year ago
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    u don't have to do it like this. I am preparing my integral for the sub. U also can do the sub and rearrange then.

  28. idku
    • one year ago
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    \[\Large \frac{1}{9}\int\limits\limits_{ }^{ }-3(-3t)e^{-3t}dt\] (the only thing is like this) no -9 on bottom, it is positive

  29. idku
    • one year ago
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    bye

  30. anonymous
    • one year ago
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    Refer to the attachment from WolframAlpha.

    1 Attachment
  31. AmTran_Bus
    • one year ago
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    Thanks...but I really want to work it from scratch on my own.

  32. AmTran_Bus
    • one year ago
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    |dw:1434927052945:dw|

  33. AmTran_Bus
    • one year ago
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    If it were me, I'd do IBP first.|dw:1434927094434:dw|

  34. AmTran_Bus
    • one year ago
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    |dw:1434927121584:dw|

  35. AmTran_Bus
    • one year ago
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    At this point, what do I do?

  36. AmTran_Bus
    • one year ago
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    Well, this is no help. Closed.

  37. Zale101
    • one year ago
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    \[\Large \int\limits_{}^{}te^{-3t}dt\] u=t and du=dt dv=e^(-3t) and v=(-1/3)e^(-3t) Now, use IBP \(uv-\int\limits_{}^{}vdu\) \(\large =-\frac{1}{3}te^{-3t}-\int\limits_{}^{}(-\frac{1}{3})e^{-3t}dt\) \(\large =-\frac{1}{3}te^{-3t}+\frac{1}{3}\int\limits_{}^{}e^{-3t}dt=-\frac{1}{3}te^{-3t}-\frac{1}{9}e^{-3t}+c\)

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