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Math Help - simple intergration help

  1. #1
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    simple intergration help

    ok
    what is the integral of ( 1/2t)

    i did a u sub

    so u = 2t
    du = 2

    2 * ln2t

    is this correct?
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  2. #2
    o_O
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    \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \int \frac{1}{t} \ dt

    which is a standard integral.
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  3. #3
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    Quote Originally Posted by o_O View Post
    \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \int \frac{1}{t} \ dt

    which is a standard integral.
    ah so it is lnt/2

    ?
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  4. #4
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    yeah it is.. ty sir.
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  5. #5
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    No. You will get 1/2 ln(t).
    They are not the same.
    1/2 ln(t) <> ln(t/2). They maybe different.
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  6. #6
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    Quote Originally Posted by oswaldo View Post
    No. You will get 1/2 ln(t).
    They are not the same.
    1/2 ln(t) <> ln(t/2). They maybe different.
    ok but what am i doing wrong then


    int (1/2t)

    so if u just do a u sub instead of pulling the 1/2 out.

    u = 2t
    du = 2du

    (1/2) int [(2du)/(2t)

    =

    1/2 lnu

    1/2 * ln 2t


    why is it 1/2 * lnt and not 1/2 * ln2t

    what did i do wrong?
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  7. #7
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    You don't need any kind of u or whatever conversion. It is a simple basic integral. All you need to know is:
    int(1/t) = ln(t). THAT'S IT.

    1/2 is just a constant. So the answer of your questions is:
    int(1/2t)=(1/2) int(1/t)=(1/2) ln(t). AND THAT'S IT.
    Which may be different than ln(t/2).

    a*ln(t)=ln(a*t) is NOT always true.

    So (1/2)*ln(t) = ln((1/2)*t) is NOT always true.

    -O
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  8. #8
    o_O
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    If for some reason you decided to use a u-substitition:

    {\color{red}u = 2t} \ \Rightarrow \ {\color{blue}\tfrac{1}{2}du = dt}

    We then get: \int \frac{1}{{\color{red}2t}} \ {\color{blue}dt} \ = \ \frac{1}{2} \int \frac{{\color{blue}du}}{{\color{red}u}} \ = \ \frac{1}{2} \ln u + C \ = \ \frac{1}{2} \ln 2t + C

    Looks a bit different. But recall: \ln (ab) = \ln a + \ln b

    So: \frac{1}{2} \ln 2t + C = \frac{1}{2} \ln 2 + \frac{1}{2} \ln t + C

    But \tfrac{1}{2}\ln 2 is nothing more than a constant. So we'll combine it with C to make a new constant of integration which we'll call K.

    i.e. \int \frac{1}{2t} \ dt \ = \ \frac{1}{2} \ln 2t + C \ = \ \frac{1}{2} \ln t + K

    In other words, \tfrac{1}{2} \ln 2t and \tfrac{1}{2} \ln t are two legit antiderivatives of \tfrac{1}{2t}
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