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Math Help - Integration by Substition

  1. #1
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    Integration by Substition

    I find myself sitting at my desk, pondering why I decided to take Calc 2. Anyway, I'm having trouble with simple integration by substitution, which does not bode well for the rest of my semester.

    Find the Following Integral: t^3(t^4-2)^5dt
    So far I have done
    u=t^4-2
    du=4t^3dt

    So i have ∫(u)^5(du/4)dt=1/4 ∫(t^4-2)^5*4t^3dt
    simplifying...
    1/4((t^4-2)^6)/6*t^4
    Giving me a final (wrong) answer of
    ((t^4-2)^6*t^4)/24
    What am I doing wrong here? Any help is extremely appreciated

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  2. #2
    Member roshanhero's Avatar
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    I hope that it helps. Sorry for the size being small.
    Attached Thumbnails Attached Thumbnails Integration by Substition-codecogseqn.gif  
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  3. #3
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    Quote Originally Posted by john84 View Post
    I find myself sitting at my desk, pondering why I decided to take Calc 2. Anyway, I'm having trouble with simple integration by substitution, which does not bode well for the rest of my semester.

    Find the Following Integral: t^3(t^4-2)^5dt
    So far I have done
    u=t^4-2
    du=4t^3dt

    So i have ∫(u)^5(du/4)dt=1/4 ∫(t^4-2)^5*4t^3dt
    simplifying...
    1/4((t^4-2)^6)/6*t^4
    Giving me a final (wrong) answer of
    ((t^4-2)^6*t^4)/24
    What am I doing wrong here? Any help is extremely appreciated

    *

    *
    \int{t^3(t^4 - 2)^5\,dt} = \frac{1}{4}\int{4t^3(t^4 - 2)^5\,dt}

    Let u = t^4 - 2 so that \frac{du}{dt} = 4t^3.

    The integral becomes

    \frac{1}{4}\int{u^5\,\frac{du}{dt}\,dt}

     = \frac{1}{4}\int{u^5\,du}

     = \frac{1}{4}\cdot\frac{1}{6}u^6 + C

     = \frac{1}{24}u^6 + C

     = \frac{1}{24}(t^4 - 2)^6 + C.
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