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Math Help - Improper Integrals

  1. #1
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    Improper Integrals

    I'm stuck on this integral and it's due online tonight, thanks a lot guys!

    The Integral from 0 to infinity of: 1700te^(-0.5t)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by amaya View Post
    I'm stuck on this integral and it's due online tonight, thanks a lot guys!

    The Integral from 0 to infinity of: 1700te^(-0.5t)
    since this is to be handed in, we can't give you a full solution. instead, try to see if you can get by on hints.

    you want to find 1700 \int_0^\infty te^{-0.5t}~dt = 1700 \lim_{N \to \infty} \int_0^N te^{-0.5t}~dt

    now, the standard way to approach this integral is to use integration by parts. using u = t and dv = e^{-0.5t}

    or, as an alternative, you could find \frac d{dt} -te^{-0.5t} = \cdots

    integrate both sides and solve for the required integral
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    improper integrals..same problem!

    Quote Originally Posted by Jhevon View Post
    since this is to be handed in, we can't give you a full solution. instead, try to see if you can get by on hints.

    you want to find 1700 \int_0^\infty te^{-0.5t}~dt = 1700 \lim_{N \to \infty} \int_0^N te^{-0.5t}~dt

    now, the standard way to approach this integral is to use integration by parts. using u = t and dv = e^{-0.5t}

    or, as an alternative, you could find \frac d{dt} -te^{-0.5t} = \cdots

    integrate both sides and solve for the required integral


    Ok thanks. I feel like I should know this and I hope I'm not just missing something obvious.
    I'm doing integration by parts but I'm having trouble figuring out the integral of e^0.5t (this willl give me v). I know the integral of e^t will just be e^t but I'm not sure what to do when we have the -0.5 in front of the t.

    Thanks a lot.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by amaya View Post
    Ok thanks. I feel like I should know this and I hope I'm not just missing something obvious.
    I'm doing integration by parts but I'm having trouble figuring out the integral of e^0.5t (this willl give me v). I know the integral of e^t will just be e^t but I'm not sure what to do when we have the -0.5 in front of the t.

    Thanks a lot.
    in general: \int e^{kt}~dt = \frac 1{k}e^{kt} + C for k \in \mathbb{Z},~k \ne 0
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  5. #5
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    Quote Originally Posted by amaya View Post
    The Integral from 0 to infinity of: 1700te^(-0.5t)
    Let's just take what really matters: \int_0^\infty te^{-t/2}\,dt.

    Substitute u=\frac t2, the integral becomes

    4\int_0^\infty  {ue^{ - u} \,du}  = 4\cdot\Gamma (2) = 4.
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