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    \int r(r+2)(r-4)dr

    I am stuck with this. Initially, I thought I had to expand the brackets; to give me r^2-2r-8

    But, that prove to be wrong. I am unsure as to what I am to do with the "r" term before the brackets, I thought that I would need to use that to form a cubic function, to give me r^3-2r^2+8r; but I am not entirely sure.
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    (Original post by apronedsamurai)
    \int r(r+2)(r-4)dr

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    Multiply out the brackets then it is straightforward
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    (Original post by apronedsamurai)
    \int r(r+2)(r-4)dr

    I am stuck with this. Initially, I thought I had to expand the brackets; to give me r^2-2r-8

    But, that prove to be wrong. I am unsure as to what I am to do with the "r" term before the brackets, I thought that I would need to use that to form a cubic function, to give me r^3-2r^2+8r; but I am not entirely sure.
    It's -8r, and yes once you have the cubic function, integrate each term separately and don't forget the +C.
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    (Original post by apronedsamurai)
    \int r(r+2)(r-4)dr

    I am stuck with this. Initially, I thought I had to expand the brackets; to give me r^2-2r-8

    But, that prove to be wrong. I am unsure as to what I am to do with the "r" term before the brackets, I thought that I would need to use that to form a cubic function, to give me r^3-2r^2+8r; but I am not entirely sure.
    Have you got this now using the advice from the other posters and noting that your "+8r" should have been "-8r"?
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    r^3-2r^2-8r+c. Which integrates to;
    1/4r^4-2/3r^3-4r^2+c
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    \displaystyle\int^4_1 x^\frac{1}{2}\ dx
 
 
 
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