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Math Help - Integration problem

  1. #1
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    Joined
    Jun 2008
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    Integration problem

    Hey everyone. This has been really bugging me for 4 hours now, and I cannot figure out where I am going wrong. If anyone can help me that would be greatly appreciated.

    The current flowing through a capacitor is:

    I = Ioe^-(kt)

    Io = 2 amps
    k = 0.2s
    T = time

    Find the charge flowed after the first 5 seconds


    Q = I dt



    2e^-(0.2t) dt


    = 1/-0.2e^-(0.2t)

    = -10e^-(0.2t)

    5
    | -10e^-(0.2t)
    0


    [ -10e^-(0.2t)] = {-10e^-(0.2x5)} - {-10e^-(0.2x0)}


    = -10e^-1 - 0

    Then I end up with a negative number for Q . If anyone could point me in the right direction i'd be thankful .
    Last edited by thecakeisalie; June 1st 2008 at 05:28 PM.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yuma, AZ, USA
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    5
    | -10e^-(0.2t)
    0


    [ -10e^-(0.2t)] = {-10e^-(0.2x5)} - {-10e^-(0.2x0)}


    = -10e^-1 - 0

    Then I end up with a negative number for Q . If anyone could point me in the right direction i'd be thankful .
    Here is your problem

    e^0=1 \ne 0

    -10e^{-0.2t} \bigg|_{0}^{5}=-10e^{-1}-[-10e^{0}]=-10e^{-1}+10

    =10-10e^{-1}=10\left( 1-\frac{1}{e}\right)=10\left( \frac{e-1}{e}\right)
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  3. #3
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    Jun 2008
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    Doh, silly me lol.

    Thank you for your help! Really appreciate it!
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