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Math Help - New more challenging and interesting calculus puzzle

  1. #1
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    New more challenging and interesting calculus puzzle

    You're not going to believe the answer to this integral:

    \int u^2\sqrt{(a + bu)} du where a and b are constants
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  2. #2
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    You can just do v = a + bu.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    You can just do v = a + bu.
    Can you go on to solve it? (I'm sure you can, see what answer pops up)
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  4. #4
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    Um, sure. Here goes. I get:

    Spoiler:

    \displaystyle\int u^{2}\sqrt{a+bu}\,du=\frac{1}{b^{3}}(a+bu)^{3/2}\left(\frac{2(a+bu)^{2}}{7}-\frac{4a(a+bu)}{5}+\frac{2a^{2}}{3}\right)+C.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    Um, sure. Here goes. I get:

    Spoiler:

    \displaystyle\int u^{2}\sqrt{a+bu}\,du=\frac{1}{b^{3}}(a+bu)^{3/2}\left(\frac{2(a+bu)^{2}}{7}-\frac{4a(a+bu)}{5}+\frac{2a^{2}}{3}\right)+C.
    I'll let you know tomorrow as Howard Anton's calculus book seems to give a different answer.
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  6. #6
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    This answer is entirely equivalent to the following:

    Spoiler:

    \displaystyle\sqrt{a+bu}\left(\frac{16a^{3}}{105 b^{3}}-\frac{8a^{2}u}{105 b^{2}}+\frac{2au^{2}}{35b}+\frac{2u^{3}}{7}\right)  +C.


    Just multiply my answer out to get the other. Is this what your book has?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    This answer is entirely equivalent to the following:

    Spoiler:

    \displaystyle\sqrt{a+bu}\left(\frac{16a^{3}}{105 b^{3}}-\frac{8a^{2}u}{105 b^{2}}+\frac{2au^{2}}{35b}+\frac{2u^{3}}{7}\right)  +C.


    Just multiply my answer out to get the other. Is this what your book has?
    The book has \dfrac{2}{105b^3}(15b^2u^2 - 12abu +8a^2)(a + bu)^{3/2} + C which looks similar to your answer and is probably the same (after algebra verification).

    It's interesting to note that such a simple looking integral leads to a monstrous result.
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