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Math Help - What am I doing wrong in this integration problem?

  1. #1
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    What am I doing wrong in this integration problem?

    Hi, I've been working on this integration problem, but when I upload it to our homework program the computer keeps saying that it's wrong. It's:

    \int \frac{z^5}{\sqrt[3]{9+z^6}} du

    So what I did was:

    u = 9 + z^6

    \frac{du}{6} = z^5 dz

    \frac{1}{6}\int u^{\frac{-1}{3}} du

    \frac{-1}{18 \sqrt[3]{(9+z^6)^4}}

    Do you think it's a syntax problem, or am I making an obvious mistake?
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  2. #2
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    Quote Originally Posted by Dana_Scully View Post
    Hi, I've been working on this integration problem, but when I upload it to our homework program the computer keeps saying that it's wrong. It's:

    \int \frac{z^5}{\sqrt[3]{9+z^6}} du

    So what I did was:

    u = 9 + z^6

    \frac{du}{6} = z^5 dz

    \frac{1}{6}\int u^{\frac{-1}{3}} du

    \frac{-1}{18 \sqrt[3]{(9+z^6)^4}}

    Do you think it's a syntax problem, or am I making an obvious mistake?
    The answer is (9+z^6)^(2/3) /4

    You made a mistake. Your substitution is good, but you went astray. After your substitution, simply treat the entity as a polynomial. Raise the exponent by one and divide by the new exponent. Then, resubstitute the original expression back in. Lastly, the sure-fire proof is to differentiate your answer and see that you get the original expression.
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  3. #3
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    Quote Originally Posted by Dana_Scully View Post
    Hi, I've been working on this integration problem, but when I upload it to our homework program the computer keeps saying that it's wrong. It's:

    \int \frac{z^5}{\sqrt[3]{9+z^6}} du

    So what I did was:

    u = 9 + z^6

    \frac{du}{6} = z^5 dz

    \frac{1}{6}\int u^{\frac{-1}{3}} du

    \frac{-1}{18 \sqrt[3]{(9+z^6)^4}}

    Do you think it's a syntax problem, or am I making an obvious mistake?
    I think you'll find that \int{u^{-\frac{1}{3}}\,du} = \frac{3}{2}u^{\frac{2}{3}}.
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  4. #4
    Super Member Aryth's Avatar
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    The mistake is somewhat obvious, we'll start at this part:

    \frac 16 \int u^{-\frac 13} ~du

    Now, the integral of u^n is:

    \frac{u^{n+1}}{n+1}

    So, that gives us:

    \frac 16 \left(\frac{u^{\frac{2}{3}}}{\frac{2}{3}}\right)

    Now we simplify:

    \frac 16 \left(\frac{3u^{\frac{2}{3}}}{2}\right)

    = \frac{u^{\frac{2}{3}}}{4}

    \frac{\sqrt[3]{(9 + z^6)^2}}{4}

    And there you go.
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  5. #5
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    Ah, thank you all so much. There goes me confusing derivation and integration again...not sure what's wrong with my head.
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