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Math Help - quick integration question

  1. #1
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    quick integration question

    i am doing a problem by parts, and my dv= e^-2t dt

    i was just wondering how one knows that v, then, =-1/2 e^-2t
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  2. #2
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    Quote Originally Posted by buttonbear View Post
    i am doing a problem by parts, and my dv= e^-2t dt

    i was just wondering how one knows that v, then, =-1/2 e^-2t
    v=-1/2 e^-2t

    because the derivative of v is dv.

    v=-1/2 e^-2t
    then dv=e^{-2t} \cdot -2 \cdot \frac{-1}{2}=dv= e^{-2t}

    That's why.
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  3. #3
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    i know that the derivative of v is dv

    i'm sorry if this is a stupid question
    but if i'm starting with dv, i just don't see how you integrate dv to get v..
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    Quote Originally Posted by buttonbear View Post
    i am doing a problem by parts, and my dv= e^-2t dt

    i was just wondering how one knows that v, then, =-1/2 e^-2t
    b/c

    \int e^{-2t}dt

    =e^{-2t} \cdot \frac{1}{-2}+C

    and double check you have the right integral by differentiating it.

    there are many steps involving \int e^{f(x)}dx
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by buttonbear View Post
    i know that the derivative of v is dv

    i'm sorry if this is a stupid question
    but if i'm starting with dv, i just don't see how you integrate dv to get v..
    \int e^{kx}~dx = \frac 1k e^{kx} + C, for k \ne 0 a constant.

    we obtain this by doing a substitution of u = kx
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