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Math Help - Integral Help!

  1. #1
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    Integral Help!

    Hey, can you show me how to do the integral for this question

    (t^2)(e^(-t^3)


    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mudd_101 View Post
    Hey, can you show me how to do the integral for this question

    (t^2)(e^(-t^3)


    Thanks
    Using the substitution u=-t^3, I leave it for you to show that \int t^2e^{-t^3}\,dt\xrightarrow{u=-t^3}{}-\tfrac{1}{3}\int e^u\,du

    Can you continue with the calculations?
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  3. #3
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    I almost got the answer but I'm negative off uhh I put my steps down their if you could Pleaassse look at them but uhh I'm new to integrals so I think they will be difficult to follow cuz I don't really know what I'm doing.

    okay so I have t^2 e^u

    u = -t^3
    -3t^2 dt = du
    t^2 dt= (-1/3)du

    so then I have: (-1/3)du *e^u
    then I did the integral of e^u (leaving behind the -1/3 from the du)
    (-1/3) ((e^(-t^3))/(-1))

    The answer in my text is (-1/3)((e^(-t^3))
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mudd_101 View Post
    I almost got the answer but I'm negative off uhh I put my steps down their if you could Pleaassse look at them but uhh I'm new to integrals so I think they will be difficult to follow cuz I don't really know what I'm doing.

    okay so I have t^2 e^u

    u = -t^3
    -3t^2 dt = du
    t^2 dt= (-1/3)du

    so then I have: (-1/3)du *e^u
    then I did the integral of e^u (leaving behind the -1/3 from the du)
    (-1/3) ((e^(-t^3))/(-1))

    The answer in my text is (-1/3)((e^(-t^3))
    I'm not sure how you got that -1...

    When you evaluate -\tfrac{1}{3}\int e^u\,du, you end up with -\tfrac{1}{3}e^u+C. Now back substitute to get \boxed{-\tfrac{1}{3}e^{-t^3}+C}
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  5. #5
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    ooo thanks
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