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Evaluate the integral

  • Thread starter tony873004
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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:

Answers and Replies

Dick
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Because the absolute value of 3 is 3.
 
Dick
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Also ln(9)=ln(3^2)=2*ln(3), since ln(a^b)=b*ln(a).
 
radou
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Recall that a*ln(x) = ln (a^x).

Edit: ooops, late.
 
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!
 
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