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Math Help - integration using cute root

  1. #1
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    integration using cute root

    hey guys, im interested in another integration question using cute roots

    integrate:

    <br /> <br />
\int \frac{1}{\sqrt[3]{4-5x}}<br /> <br />

    any ideas? should i make that cute root in the form (4-5x)^{\frac{1}{3}} first?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,

    Quote Originally Posted by Title
    integration using cute root
    Cute root or cube root ?
    Quote Originally Posted by jvignacio View Post
    should i make that cute root in the form (4-5x)^{\frac{1}{3}} first?
    Yes, then you can try using \int u' u^\alpha \,\mathrm{d}x = \frac{u^{\alpha+1}}{\alpha +1}+C for \alpha \neq -1.
    Last edited by flyingsquirrel; October 24th 2008 at 02:19 AM. Reason: du ---> dx
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    Hi,


    Cute root or cube root ?

    Yes, then you can try using \int u' u^\alpha \,\mathrm{d}u = \frac{u^{\alpha+1}}{\alpha +1}+C for \alpha \neq -1.
    hhaha yes sorry cube root!!!
    im not quite sure what you mean in that formular?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jvignacio View Post
    im not quite sure what you mean in that formular?
    One has  \frac{1}{\sqrt[3]{4-5x}}=\frac{1}{(4-5x)^{1/3}}=(4-5x)^{-1/3} so \int \frac{\mathrm{d}x}{\sqrt[3]{4-5x}}=\int (4-5x)^{-1/3}\,\mathrm{d}x. It looks like \int u'u^\alpha \,\mathrm{d}x with u=4-5x and \alpha=-1/3. The only term which is missing is u'... but we can make it appear. u' is given by

    u'=\frac{\mathrm{d}(4-5x)}{\mathrm{d}x}=-5

    so if we write

    \int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x=-5\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x

    we're done since we know that \int u'u^\alpha \,\mathrm{d}x=\frac{u^{\alpha+1}}{\alpha+1}+C for \alpha\neq -1.
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  5. #5
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    Quote Originally Posted by flyingsquirrel View Post
    One has  \frac{1}{\sqrt[3]{4-5x}}=\frac{1}{(4-5x)^{1/3}}=(4-5x)^{-1/3} so \int \frac{\mathrm{d}x}{\sqrt[3]{4-5x}}=\int (4-5x)^{-1/3}\,\mathrm{d}x. It looks like \int u'u^\alpha \,\mathrm{d}x with u=4-5x and \alpha=-1/3. The only term which is missing is u'... but we can make it appear. u' is given by

    u'=\frac{\mathrm{d}(4-5x)}{\mathrm{d}x}=-5

    so if we write

    \int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x=-5\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x

    we're done since we know that \int u'u^\alpha \,\mathrm{d}x=\frac{u^{\alpha+1}}{\alpha+1}+C for \alpha\neq -1.
    ahh yes now i understand.

    does the answer end up being

    <br />
\frac {3(4-5x)^{\frac{2}{3}}}{2} + C<br />

    ?
    Last edited by jvignacio; October 24th 2008 at 09:50 PM.
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    \int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x=-5\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x
    I meant

    \int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x={\color{red}-\frac15}\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x

    ...

    Quote Originally Posted by jvignacio View Post
    does the answer end up being

    <br />
\frac {3(4-5x)^{\frac{2}{3}}}{2} + C<br />

    ?
    Check it by yourself . The derivative of <br />
x\mapsto\frac {3(4-5x)^{\frac{2}{3}}}{2} + C<br />
is <br />
x \mapsto -5(4-5x)^{-\frac{1}{3}}<br />
and it should be x\mapsto(4-5x)^{-\frac{1}{3}}. It turns out that you forgot the red thing :

    \int (4-5x)^{-1/3}\,\mathrm{d}x=\ldots={\color{red}-\frac15}\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x
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