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Don't know if this is correct (and final step)

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I really dont want to type all this out, but since I hate when people don't, I will. I hope someone takes the time to help me out :)

    [tex]\int_0^1 \frac{4ln(3x)}{\sqrt{x}}dx=\lim{t \to 0^{+}} \int_0^{t} \frac{4ln(3x)}{\sqrt{x}}dx[/tex]
    [tex]u=4ln3x, v=2\sqrt{x}\\du=\frac{4}{x}dx, dv=\frac{1}{\sqrt{x}}dx[/tex]
    [tex](4ln3x)(2\sqrt{x})-\int \frac{8\sqrt{x}}{x}dx=8\sqrt{x}(ln3x)-8\int x^{\frac{-1}{2}}dx=8\sqrt{x}(ln3x)-16\sqrt{x}[/tex]
    [tex]\lim_{t \to 0^{+}} (8\sqrt{x}(ln3x)-16\sqrt{x})|_{t}^{1}=\lim_{t \to 0^{+}} 8(ln3)-16-(8\sqrt{t}(ln3t)-16\sqrt{t})[/tex]

    Now I think that just equals 8ln3-16, but I'm not sure and wolfram gives me a different answer..

    Thanks.
     
  2. jcsd
  3. Mar 4, 2013 #2

    SammyS

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    The indefinite integral is correct. (I've checked your work that far.)

    What result does Wolfram give you?
     
  4. Mar 4, 2013 #3
    Wow I'm stupid I just did it again and I'm not sure why wolfram gives me the answer, but my calculator gives me a different answer?? Log=ln right?
     
  5. Mar 4, 2013 #4

    SammyS

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    That's true for Wolfram.

    For you calculator, it's probably log = log10 .
     
  6. Mar 4, 2013 #5
    That is so stupid -.-"
     
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