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The sign of the integral

  1. May 23, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    Find
    [tex]\int_2^4{log_{10}xdx[/tex]

    2. Relevant equations
    Use the change of base formula
    [tex]log_ba=\frac{log_ca}{log_cb}[/tex]

    3. The attempt at a solution
    Using the formula:
    [tex]\frac{1}{ln10}\int_2^4lnxdx[/tex]

    [tex]=\frac{1}{ln10}[\frac{1}{x}]_2^4[/tex]

    [tex]=\frac{1}{ln10}(\frac{1}{4}-\frac{1}{2})[/tex]

    [tex]=\frac{-1}{4ln10} \approx -0.1086[/tex]

    What my problem is, is that the integral is giving a negative value, but the function of [itex]y=log_{10}x[/itex] is positive for [itex]2\leq x\leq 4[/itex].

    So is it that I've made a mistake in calculating this integral, or is the negative value I'm getting legitimate? At this point I would need to reconsider when the integral is ever negative.
     
  2. jcsd
  3. May 23, 2009 #2
    [tex] \int ln(x)dx = xln(x) - x [/tex]
     
  4. May 23, 2009 #3

    cristo

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    What've you done here? You seem to have differentiated instead of integrated. As Random Variable mentions above, the 'trick' to integrating the logarithm function is to write it as 1.ln(x) and use parts.
     
  5. May 24, 2009 #4

    Mentallic

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    Oh hehe, oops xD

    Uhh... I still don't understand how you obtained that result Random Variable. Can you guys please explain this
    a little more?
     
  6. May 24, 2009 #5

    cristo

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    You can write ln(x) as 1.ln(x), to which you can then apply integration by parts. Let u=ln(x), dv=1dx, which will give du=1/x dx, v=x. Then, recall that

    [tex]\int u dv=uv-\int v du\Rightarrow\int \ln x dx=x\ln x-\int x\frac{1}{x} dx = x\ln x - x (+C)\,.[/tex]
     
  7. May 24, 2009 #6

    Mentallic

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    This has never been done in class before. Maybe we were meant to experiment or just know this result by natural instinct? :smile: Thanks.
     
  8. May 24, 2009 #7

    Mark44

    Staff: Mentor

    Do you mean that this integral hasn't been done in class, or the technique of integration by parts hasn't been done in class?
     
  9. May 25, 2009 #8

    Mentallic

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    Integration by parts. And this integral hasn't been done either or else it wouldn't make sense for the teacher to throw it in an assignment soon after showing us how to solve it (which would mean she would probably have taught us integration by parts at that point :tongue:).
     
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