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Math Help - Integration!

  1. #1
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    Integration!

    The problem is:

    ∫(x^2)e^[-(x^3)/4]


    The attempt at a solution:

    Is it equals to (-4/3)e^[-(x^3)/4]???

    I am really confuse here. Any help would be appreciated
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by HeheZz View Post
    The problem is:

    ∫(x^2)e^[-(x^3)/4]


    The attempt at a solution:

    Is it equals to (-4/3)e^[-(x^3)/4]???

    I am really confuse here. Any help would be appreciated
    Yes, that is the correct answer (don't forget the "+C")! What is confusing you with this problem (or the substitution process in general)?
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  3. #3
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    ∫(x^2)e^[-(x^3)/4]

    Juz confusing that we didnt take (x^2) into account when integrating? Juz need to integrate e^[-(x^3)/4] so it becomes:

    [x^2] * [-4/3(x^2)]e^[-(x^3)/4]

    and so the (x^2) just cancel out itself to become (-4/3)e^[-(x^3)/4]
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by HeheZz View Post
    ∫(x^2)e^[-(x^3)/4]

    Juz confusing that we didnt take (x^2) into account when integrating? Juz need to integrate e^[-(x^3)/4] so it becomes:

    [x^2] * [-4/3(x^2)]e^[-(x^3)/4]

    and so the (x^2) just cancel out itself to become (-4/3)e^[-(x^3)/4]
    You actually do take the x^2 term into account!

    When you make the substitution u={\color{blue}-\tfrac{1}{4}x^3}, we see that \,du=-\tfrac{3}{4}{\color{red}x^2\,dx}

    At this stage, let's go back to the original integral:

    \int {\color{red}x^2}e^{-x^3/4}{\color{red}\,dx}.

    Observed that I highlighted the part that appeared in our \,du term! So we see that \,du=-\tfrac{3}{4}{\color{red}x^2\,dx}\implies -\tfrac{4}{3}\,du={\color{red}x^2\,dx}.

    Therefore, \int {\color{red}x^2}e^{{\color{blue}-x^3/4}}{\color{red}\,dx}=\int e^u(-\tfrac{4}{3}\,du)=-\tfrac{4}{3}\int e^u\,du.

    Then integrating results in -\tfrac{4}{3}e^u+C and back substitution gives us the desired result \int x^2e^{-x^3/4}\,dx=-\tfrac{4}{3}e^{-x^3/4}+C.

    Does this clarify what's going on here?
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  5. #5
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    Ahhh!! Thank You very much. I kind of forget these stuff ><. Thanks again for explaining it to me
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