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    Ok, I have to evaluate this:
    ∫(x - 1/x )dx within the boundaries of 6 and 3 (I dont know how to make them float by the integral sign!)
    Im not really sure what form I should be leaving my answer in...do I do it in decimals, or leave ln there, or what? Please could someone solve it for me so I can see what to do.
    Thank you very much
    franks xxx
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    (Original post by franks)
    Ok, I have to evaluate this:
    ∫(x - 1/x )dx within the boundaries of 6 and 3 (I dont know how to make them float by the integral sign!)
    Im not really sure what form I should be leaving my answer in...do I do it in decimals, or leave ln there, or what? Please could someone solve it for me so I can see what to do.
    Thank you very much
    franks xxx
    \int_{3}^{6} x- \frac{1}{x} dx \\

= [\frac{1}{2}x^{2}-\ln |x| ]_{3}^{6} \\

=18-\ln 6-\frac{9}{2}+\ln 3 \\

=\frac{27}{2}+\ln \frac{1}{2} \\

=\frac{27}{2}-\ln 2
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    ok, looks like i havent learnt about some of the rules about e and logs.
    Hmm, third line, how do you get -ln6 + ln3 = ln 0.5
    how does ln0.5 = -ln 2
    Sorry, the answers to these are probably really obvious, but ive only had half a lesson on the whole of the 'e' stuff and I havent bought myself a text book yet!
    Thank you!
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    lnA - lnB = ln(A/B)
    lnA+lnB = ln(AB)

    klnA = ln(A^K)
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    ok ive worked out the first part - about the -ln6 + ln3 = 0.5 , just had to remember some of those rules, but still stuck on the ln0.5 = -ln2
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    lnA - lnB = ln(A/B)
    lnA+lnB = ln(AB)

    klnA = ln(A^K)
    oh thanks - just worked that out !
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    (Original post by franks)
    oh thanks - just worked that out !
    If you remember the lna + lnb one, you can work out the klna one.

    e.g. 3lna = lna + lna + lna = ln(aaa) = lna^3
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    thanks everyone, but can anyone please explain to me how ln0.5 = -ln 2 . Im probably missing a really simple point, but I just need someone to explain quickly!
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    (Original post by franks)
    thanks everyone, but can anyone please explain to me how ln0.5 = -ln 2 . Im probably missing a really simple point, but I just need someone to explain quickly!
    (-1)*ln2 = ln (2-1) = ln(½)

    [log law: lnxn = nlnx]

    Aitch
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    (Original post by Aitch)
    (-1)*ln2 = ln (2-1) = ln(½)

    [log law: lnxn = nlnx]

    Aitch
    you can also use another rule.
    ln (0.5) = ln(1/2) = ln1 - ln2 = 0 - ln2 = -ln2

    (btw, ln1 is always 0, as e^0=1)

    [rule used = lnA - lnB = ln(A/B)]
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    thanks
 
 
 
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