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3psilon

  • 2 years ago

How to integrate (6-2x)^2 . (indefinite integral). I'm trying U sub but Idk what I'm doing wrong

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  1. TuringTest
    • 2 years ago
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    what did you use for u ?

  2. 3psilon
    • 2 years ago
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    6-2x

  3. TuringTest
    • 2 years ago
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    can you show your sork? what did you get for du? what did you get for dx?

  4. 3psilon
    • 2 years ago
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    U = 6-2x du = -2dx \[(1/2) \int\limits_{}^{} -2U^2dx\] @TuringTest

  5. TuringTest
    • 2 years ago
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    I'm not seeing where you get the -2 from... u=6-2x du=-2dx -> dx=-du/2

  6. 3psilon
    • 2 years ago
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    It's to get du right ? so I multiply by 1 pretty much 2/2

  7. TuringTest
    • 2 years ago
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    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\[\int u^2(-\frac{du}2)=-\frac12\int u^2du\]

  8. 3psilon
    • 2 years ago
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    Ohhh thank you turing!

  9. TuringTest
    • 2 years ago
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    Welcome!

  10. 3psilon
    • 2 years ago
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    We weren't taught how to solve for dx and plug back in but I like that way!

  11. TuringTest
    • 2 years ago
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    I sort of never saw them do it in class that way either, I never understood why :p

  12. 3psilon
    • 2 years ago
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    @TuringTest What answer are you getting ? I get \[\frac{-(6-2x)^{3}}{6} +c\] But my TI-89 is giving me\[\frac{4(x-3)^{3}}{3} +C\]

  13. TuringTest
    • 2 years ago
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    it is the same answer; they appear to have factored out a -2 in your TI-89

  14. 3psilon
    • 2 years ago
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    Ohh okay! Thanks! :)

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