3psilon
  • 3psilon
How to integrate (6-2x)^2 . (indefinite integral). I'm trying U sub but Idk what I'm doing wrong
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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TuringTest
  • TuringTest
what did you use for u ?
3psilon
  • 3psilon
6-2x
TuringTest
  • TuringTest
can you show your sork? what did you get for du? what did you get for dx?

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3psilon
  • 3psilon
U = 6-2x du = -2dx \[(1/2) \int\limits_{}^{} -2U^2dx\] @TuringTest
TuringTest
  • TuringTest
I'm not seeing where you get the -2 from... u=6-2x du=-2dx -> dx=-du/2
3psilon
  • 3psilon
It's to get du right ? so I multiply by 1 pretty much 2/2
TuringTest
  • TuringTest
you have left your integral all mixed up with u integrated with respect to x; you can't do that. I think you are getting confused there. you need to sub in the values: u=6-2x du=-2dx -> dx=-du/2 subbing in for u and dx gives\[\int u^2(-\frac{du}2)=-\frac12\int u^2du\]
3psilon
  • 3psilon
Ohhh thank you turing!
TuringTest
  • TuringTest
Welcome!
3psilon
  • 3psilon
We weren't taught how to solve for dx and plug back in but I like that way!
TuringTest
  • TuringTest
I sort of never saw them do it in class that way either, I never understood why :p
3psilon
  • 3psilon
@TuringTest What answer are you getting ? I get \[\frac{-(6-2x)^{3}}{6} +c\] But my TI-89 is giving me\[\frac{4(x-3)^{3}}{3} +C\]
TuringTest
  • TuringTest
it is the same answer; they appear to have factored out a -2 in your TI-89
3psilon
  • 3psilon
Ohh okay! Thanks! :)

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