AmTran_Bus
  • AmTran_Bus
Integration help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
?
AmTran_Bus
  • AmTran_Bus
|dw:1434926190793:dw|
triciaal
  • triciaal
products

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AmTran_Bus
  • AmTran_Bus
|dw:1434926209378:dw|
AmTran_Bus
  • AmTran_Bus
\[uv-\int\limits vdu\]
idku
  • idku
more steps, but less mess and confusion. u=-3t then by parts
AmTran_Bus
  • AmTran_Bus
\[t \frac{ e ^{-3t} }{ 3 }-\int\limits \frac{ e ^{-3t} }{ 3 }dt\]
idku
  • idku
u can do a power series too, if you would like.
idku
  • idku
anyway, u got people here....
AmTran_Bus
  • AmTran_Bus
This is what I'm doing right now, this is the IBP section.
AmTran_Bus
  • AmTran_Bus
So do a U sub first? ok.
idku
  • idku
yes, u=-3t
idku
  • idku
du=(-3)dx -> du/(-3) = dx
idku
  • idku
u got this:)
idku
  • idku
(don't forget to sub back the x)
AmTran_Bus
  • AmTran_Bus
So do I need to say dt = -du/3?
idku
  • idku
yes that suffices
AmTran_Bus
  • AmTran_Bus
|dw:1434926595047:dw|
AmTran_Bus
  • AmTran_Bus
Supposed to be a neg there... Then IBP?
idku
  • idku
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) \[u=-3t~~~du=-3dt\] \[\Large \frac{1}{-9}\int\limits_{ }^{ }\color{blue}{-3}\color{red}{(-3t)}e^{\color{red}{-3t}}\color{blue}{dt}\]
idku
  • idku
the blue part is the part that gets replaced by du. the red - each thing is a u
AmTran_Bus
  • AmTran_Bus
Where is the 1/9 from? I would like to do it the way I have just started.
idku
  • idku
oh, 1/9 not 1/-9
AmTran_Bus
  • AmTran_Bus
|dw:1434926879794:dw|
idku
  • idku
i multiply times -3 times -3 and times 1/9
idku
  • idku
and write that all out, to have my u's
idku
  • idku
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.
idku
  • idku
\[\Large \frac{1}{9}\int\limits\limits_{ }^{ }-3(-3t)e^{-3t}dt\] (the only thing is like this) no -9 on bottom, it is positive
idku
  • idku
bye
anonymous
  • anonymous
Refer to the attachment from WolframAlpha.
1 Attachment
AmTran_Bus
  • AmTran_Bus
Thanks...but I really want to work it from scratch on my own.
AmTran_Bus
  • AmTran_Bus
|dw:1434927052945:dw|
AmTran_Bus
  • AmTran_Bus
If it were me, I'd do IBP first.|dw:1434927094434:dw|
AmTran_Bus
  • AmTran_Bus
|dw:1434927121584:dw|
AmTran_Bus
  • AmTran_Bus
At this point, what do I do?
AmTran_Bus
  • AmTran_Bus
Well, this is no help. Closed.
Zale101
  • Zale101
\[\Large \int\limits_{}^{}te^{-3t}dt\] 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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