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Thread: Substitution Indefinite Integral Problem- Just need work checked

  1. #1
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    Substitution Indefinite Integral Problem- Just need work checked

    Not sure if i've done this problem right.

    $\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

    $\displaystyle u=2x^3+7$

    $\displaystyle du=6x^2$

    $\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

    $\displaystyle \int (\frac{du}{u^{3/2}})$

    $\displaystyle \int (2u^{-3/2}+c$

    $\displaystyle 2(2x^3+7)^{-3/2}+c$

    $\displaystyle (4x^3+14)^{-3/2}+c$
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    Junior Member bebrave's Avatar
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    Yeah you are right i think. Because i solved and i found same
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  3. #3
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    Quote Originally Posted by Jim Marnell View Post
    Not sure if i've done this problem right.

    $\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

    $\displaystyle u=2x^3+7$

    $\displaystyle du=6x^2 \, {\color{red}dx}$ Mr F says: Note the red stuff.

    $\displaystyle \int (\frac{6x^2}{(2x^3+7)^{3/2}})dx$

    $\displaystyle \int (\frac{du}{u^{3/2}})$

    $\displaystyle \int (2u^{-3/2}+c$ Mr F says: Wrong from here onwards. See below.

    $\displaystyle 2(2x^3+7)^{-3/2}+c$

    $\displaystyle (4x^3+14)^{-3/2}+c$
    $\displaystyle \int\! \frac{du}{u^{3/2}} = \int u^{-3/2} \, du = -2 u^{-1/2} + C$ etc.

    Quote Originally Posted by bebrave View Post
    Yeah you are right i think. Because i solved and i found same
    Then you're wrong too.
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  4. #4
    Junior Member bebrave's Avatar
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    Thank you Mr.Fantastic...
    Yeah you are right i think. Because i solved and i found same
    Last edited by mr fantastic; Apr 11th 2009 at 12:07 AM.
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  5. #5
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    Exclamation

    so the answer instead would be:

    $\displaystyle (-4x^3-14)^{-1/2}+c$
    Last edited by Jim Marnell; Apr 11th 2009 at 06:27 AM.
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  6. #6
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    Quote Originally Posted by Jim Marnell View Post
    so the answer instead would be:

    $\displaystyle (-4x^3-14)^{-1/2}+c$
    No.

    When you substitute $\displaystyle u = 2x^3 + 7$ back into $\displaystyle -2 u^{-1/2} + C$ you do NOT get $\displaystyle (-4x^3-14)^{-1/2}+C$. The -2 does NOT and CANNOT get taken inside the brackets.

    By the way, even if the answer was $\displaystyle 2 u^{-1/2} + C$, this is NOT equal to $\displaystyle (4x^3+14)^{-1/2}+C$.
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