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Evaluate the integral

  1. Jan 26, 2007 #1

    tony873004

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    [tex]
    \int_1^9 {\frac{1}{{2x}}\,dx}
    [/tex]

    [tex]
    F(x) = \frac{{\ln \left| x \right|}}{2}
    [/tex]

    [tex]
    \frac{{\ln \left| 9 \right|}}{2} - \frac{{\ln \left| 1 \right|}}{2} = \frac{{\ln \left| 9 \right|}}{2} = \ln 3
    [/tex]
    The answer ln3 came from the back of the book. I realize from using my calculator that ln9 / 2 = ln3 , but I'm not sure why. I guess I forgot my rules of ln.

    Also, since the anti-derivate section gives 1/x as ln abs(x), was I correct in carrying the absolute value brackets to the ln abs(9) / 2 ? Why did the back of the book drop the absolute value brackets from the answer?

    Thanks!!
     
    Last edited: Jan 26, 2007
  2. jcsd
  3. Jan 26, 2007 #2

    Dick

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    Because the absolute value of 3 is 3.
     
  4. Jan 26, 2007 #3

    Dick

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    Also ln(9)=ln(3^2)=2*ln(3), since ln(a^b)=b*ln(a).
     
  5. Jan 26, 2007 #4

    radou

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    Recall that a*ln(x) = ln (a^x).

    Edit: ooops, late.
     
  6. Jan 26, 2007 #5

    tony873004

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    That makes sense. I will only need the abs brackets around a variable. If it is a fixed number, just get rid of the minus sign if any, and the brackets...

    Thanks... the toughest part about this calculus is remembering the all the pre-calc!
     
  7. Jan 27, 2007 #6
    Actually, I recall it as a*ln(x) = ln(x^a). :wink:
     
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