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Thread: Integrate (1 + 3x)(x - 2)

  1. November 14th 2010, 02:28 AM #1
    jasonlewiz
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    Exclamation Integrate (1 + 3x)(x - 2)

    Evaluate this integral

    \int\((1+3x)(x-2) dx

    I use the general power formula for this one.

    Let u = (1+3x)(x-2)
    du = 6x-5

    I stop from here, I don't know what's next and I don't know if I do it right.
    Last edited by mr fantastic; November 14th 2010 at 10:42 AM. Reason: Useless title.
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  2. November 14th 2010, 02:39 AM #2
    Unknown008
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    What did you do exacly?

    Expand what you have in brackets and integrate as usual.
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  3. November 14th 2010, 03:18 AM #3
    TheChaz
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    Quote Originally Posted by jasonlewiz View Post
    Evaluate this integral

    \int\((1+3x)(x-2) dx

    I use the general power formula for this one.

    Let u = (1+3x)(x-2)
    du = 6x-5

    I stop from here, I don't know what's next and I don't know if I do it right.

    Sorry for the title/subject, I forgot to edit
    \int\((1+3x)(x-2) dx = \int\((3x^2 - 5x - 2)dx...

    It's strange that you would "know" u-substitution, but not how to FOIL and integrate a polynomial term by term...
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  4. November 14th 2010, 05:24 AM #4
    jasonlewiz
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    Quote Originally Posted by TheChaz View Post
    \int\((1+3x)(x-2) dx = \int\((3x^2 - 5x - 2)dx...

    It's strange that you would "know" u-substitution, but not how to FOIL and integrate a polynomial term by term...

    ahh, so I can change (1+3x)(x-2) to (3x+1)(x-2) ^^now I know..thanks
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  5. November 14th 2010, 05:33 AM #5
    Unknown008
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    That's not the point!

    You expand the brackets like this:

    (1+3x)(x-2) = 3x^2 - 5x - 2

    Then you integrate:

    \displaystyle \int 3x^2 - 5x - 2\ dx
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