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Math Help - definite integration of log

  1. #1
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    definite integration of log

    Hi i've got this question of definite integration (1/5x-6) dx where x=0 from x=1

    Ive got [1/5 log (5x+6)] x=1 x=0

    = [1/5 log (5+6)] - [1/5log(0+6)]
    = 0.208 - 0.1556
    =0.0526

    Am i supposed to have it as a decimal.. and how do i avoid using a decimal cos my friend got an answer of -log(6)/5


    And sorry not sure how to do the integration sign on the computer haha
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  2. #2
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    Is there some sort of formula to which we keep log in the equation? as the final answer consists of it.

    Or can i simply put it into the calculator leaving it as a decimal.
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  3. #3
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    Quote Originally Posted by fvaras89 View Post
    Hi i've got this question of definite integration (1/5x-6) dx where x=0 from x=1

    Ive got [1/5 log (5x+6)] x=1 x=0

    = [1/5 log (5+6)] - [1/5log(0+6)]
    = 0.208 - 0.1556
    =0.0526

    Am i supposed to have it as a decimal.. and how do i avoid using a decimal cos my friend got an answer of -log(6)/5


    And sorry not sure how to do the integration sign on the computer haha
    I assume you got F(x) = \frac{1}{5}log(5x+6) between 0 and 1.

     \frac{1}{5}log(5+6) -  \frac{1}{5}log(0x+6) = \frac{1}{5}(log(11)-log(6))

    1/5 is a factor so we can factor that out.
    Leave the logs in unless specifically asked to round to a decimal as the exact answer leaves the logs in. You can use the laws of logs to simplify

    log_c(a) - log_c(b) = log_c \left(\frac{a}{b}\right)

    I've no idea how your friend got their answer based on that simplification. What was the original question because it's also unusual to see logs that are not base e in integration
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    I assume you got F(x) = \frac{1}{5}log(5x+6) between 0 and 1.

     \frac{1}{5}log(5+6) - \frac{1}{5}log(0x+6) = \frac{1}{5}(log(11)-log(6))

    1/5 is a factor so we can factor that out.
    Leave the logs in unless specifically asked to round to a decimal as the exact answer leaves the logs in. You can use the laws of logs to simplify

    log_c(a) - log_c(b) = log_c \left(\frac{a}{b}\right)

    I've no idea how your friend got their answer based on that simplification. What was the original question because it's also unusual to see logs that are not base e in integration

    the original question is definite integral of (1/5x-6) dx where x=0 from x=1.

    So from your log laws. i got log(11/6) is that right?
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by fvaras89 View Post
    the original question is definite integral of (1/5x-6) dx where x=0 from x=1.

    So from your log laws. i got log(11/6) is that right?
    Assuming the equation in post 1 is correct then it should be.

    ------------------------------------------------------------

    I note you've put -6 in the original question and +6 in the OP, which is correct? I've took -6 but I can edit if it's +6

    \int^1_0\,\frac{1}{5x-6}\,dx

    I would use a u substitution to make it easier

    u = 5x-6 \: \rightarrow \: du = 5dx \: \rightarrow \: dx = \frac{1}{5}du

    Re-evaluate the limits in terms of u

    u = 5(1)-6 = -1
    u = 5(0) - 6 = -6

    So we get
    <br />
\frac{1}{5} \int^{-1}_{-6}\, \frac{1}{u}\, du

    = \frac{1}{5}[ln|u|] between -1 and -6

    \frac{1}{5}(ln|-1| - ln|-6|) = -\frac{ln6}{5}
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