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Math Help - Integrating the root of a power

  1. #1
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    Integrating the root of a power

    I'm working from a book and trying to integrate \sqrt[4]{2x^{3}}, unfortunately I always arrive at the wrong answer. Can anyone point out where I'm going wrong with this working?

    \sqrt[4]{2x^{3}}=(2x)^\frac{3}{4}

    Therefore

    \frac{dy}{dx}=\frac{3}{4}(2x)^\frac{-1}{4}=\frac{3}{4}(\frac{1}{\sqrt[4]{2x}})
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  2. #2
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    Quote Originally Posted by djnorris2000 View Post
    I'm working from a book and trying to integrate \sqrt[4]{2x^{3}},
    \displaystyle\int {x^{\frac{n}<br />
{m}} dx}  = \frac{m}<br />
{{m + n}}x^{\frac{{m + n}}<br />
{m}}  + c
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    Hmm, I've not come across that rule before, so would it give \frac{4}{7}x^\frac{7}{4}+2 or have I completely misinterpreted it?

    I was trying to use x^n=nx^{n-1}.
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    Quote Originally Posted by djnorris2000 View Post
    I was trying to use x^n=nx^{n-1}.
    Are you doing a derivative or an anti-derivative?
    In the title of the OP it says integrate which is an anti-derivative.
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  5. #5
    Super Member Quacky's Avatar
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    Quote Originally Posted by djnorris2000 View Post
    I'm working from a book and trying to integrate \sqrt[4]{2x^{3}}, unfortunately I always arrive at the wrong answer. Can anyone point out where I'm going wrong with this working?

    \sqrt[4]{2x^{3}}=(2x)^\frac{3}{4}

    Therefore

    \frac{dy}{dx}=\frac{3}{4}(2x)^\frac{-1}{4}=\frac{3}{4}(\frac{1}{\sqrt[4]{2x}})
    Quote Originally Posted by djnorris2000 View Post
    Hmm, I've not come across that rule before, so would it give \frac{4}{7}x^\frac{7}{4}+2 or have I completely misinterpreted it?

    I was trying to use x^n=nx^{n-1}.
    Quote Originally Posted by Plato View Post
    Are you doing a derivative or an anti-derivative?
    In the title of the OP it says integrate which is an anti-derivative.
    Regardless of the confusion, there is a gaping error in the OP, where you say that:

    \sqrt[4]{2x^{3}}=(2x)^\frac{3}{4}

    and I think you mean:

    \sqrt[4]{2x^{3}}=\sqrt[4]{2}\times(x)^\frac{3}{4} - the 2 isn't cubed. This should help with whatever it is you are trying to solve. I don't know whether you're trying to differentiate or integrate still, so please specify - copy the exact wording of the question.
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  6. #6
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    Thanks Quacky, I meant differentiate all along. That's what I get for browsing ahead in the book and posting when I'm brain dead from work.

    I figured I'd be making a simple mistake in there and now that you've pointed that out I'm arriving at the right answer.
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