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Math Help - Integration Involving Definite Integral

  1. #1
    Senior Member nycmath's Avatar
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    Integration Involving Definite Integral

    I know members on this site expect work or effort to be posted with question. I included my work here. I want to know where I went wrong. See picture.

    My Answer: (1/10)*ln|1/9|

    Book's Answer: (1/5)*ln 3
    Attached Thumbnails Attached Thumbnails Integration Involving Definite Integral-cam00250.jpg  
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  2. #2
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    Re: Integration Involving Definite Integral

    Have a look at this.
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    Re: Integration Involving Definite Integral

    I get $\dfrac{1}{25-x^2}=\dfrac{1}{10} \left(\dfrac{1}{x+5}-\dfrac{1}{x-5} \right)$

    which integrates to

    $\dfrac{1}{10}\ln\left(\left|\dfrac{x+5}{x-5}\right|\right)$ and evaluating

    $\dfrac{1}{10}\ln\left(\left|\dfrac{x+5}{x-5}\right|\right) \left|^4_0\right.=\\ \\

    \dfrac{1}{10}\ln\left(\left|\dfrac{4+5}{4-5}\right|\right)-\dfrac{1}{10}\ln\left(\left|\dfrac{5}{-5}\right|\right)=\\ \\

    \dfrac{1}{10}\ln(9)=\\ \\

    \dfrac{1}{10}\ln(3^2)=

    \dfrac{1}{5}\ln(3)$

    you should have been able to figure this out
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    Re: Integration Involving Definite Integral

    $\displaystyle \int \dfrac{dx}{a^2-x^2} = \ln\left|\dfrac{a+x}{a-x} \right| + C$

    You flipped the fraction on the RHS.

    Your answer should be:

    $\dfrac{1}{10}\ln\left(\dfrac{5+4}{5-4}\right) - \dfrac{1}{10}\ln\left(\dfrac{5+0}{5-0}\right) = \dfrac{1}{10}\ln 9$

    (Since all values are positive, I am ignoring the absolute value symbols).

    Next, using logarithm rules:

    $\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\cdot \dfrac{1}{2} \ln 9 = \dfrac{1}{5} \ln 9^{1/2} = \dfrac{1}{5}\ln 3$

    Romsek beat me to it.
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  5. #5
    Senior Member nycmath's Avatar
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    Re: Integration Involving Definite Integral

    Thanks everyone.

    However,where did 1/5 come from?

    How does (1/10)*ln 9 become (1/5)(1/2)ln9, which then becomes (1/5)*ln(sqrt{9}) leading to the answer (1/5)*ln 3?

    What led you to decide that 1/10 should be broken into two fractions (1/5)(1/2)?

    The 1/2 infront of ln becomes the exponent of 9, which is the same as the square root of 9.
    Last edited by nycmath; March 17th 2014 at 02:15 PM.
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  6. #6
    Senior Member nycmath's Avatar
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    Re: Integration Involving Definite Integral

    In addition, since (1/10)*ln|1/9)| =
    (1/5)*ln3, then my original answer is also correct, right? The 3 authors of the book decided to apply log rules to further simplify the answer, right?
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    Re: Integration Involving Definite Integral

    yes

    though your derivation was a mess. It's a simple application of partial fractions. No substitutions were needed.
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    Re: Integration Involving Definite Integral

    Quote Originally Posted by romsek View Post
    yes

    though your derivation was a mess. It's a simple application of partial fractions. No substitutions were needed.
    No, the OP's answer was wrong:

    $\dfrac{1}{10}\ln\left|\dfrac{1}{9}\right| = \dfrac{1}{10}\ln 9^{-1} = -\dfrac{1}{10}\ln 3^2 = -\dfrac{1}{5}\ln 3 \neq \dfrac{1}{5} \ln 3$

    nycmath: You used the wrong integral, as I pointed out in my post. That is why you got the wrong answer.

    As for why $\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\ln 3$, I broke up the fraction, but that is not needed. You can also do the following:

    $\dfrac{1}{10}\ln 9 = \dfrac{1}{10}\ln 3^2 = \dfrac{2}{10}\ln 3 = \dfrac{1}{5}\ln 3$
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  9. #9
    Senior Member nycmath's Avatar
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    Re: Integration Involving Definite Integral

    Quote Originally Posted by SlipEternal View Post
    No, the OP's answer was wrong:

    $\dfrac{1}{10}\ln\left|\dfrac{1}{9}\right| = \dfrac{1}{10}\ln 9^{-1} = -\dfrac{1}{10}\ln 3^2 = -\dfrac{1}{5}\ln 3 \neq \dfrac{1}{5} \ln 3$

    nycmath: You used the wrong integral, as I pointed out in my post. That is why you got the wrong answer.

    As for why $\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\ln 3$, I broke up the fraction, but that is not needed. You can also do the following:

    $\dfrac{1}{10}\ln 9 = \dfrac{1}{10}\ln 3^2 = \dfrac{2}{10}\ln 3 = \dfrac{1}{5}\ln 3$
    I thank you for your guidance through this questions and others here. I know my integration skills have improved since joining the site.
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  10. #10
    Senior Member nycmath's Avatar
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    Re: Integration Involving Definite Integral

    Thank you everyone. I have learned so much here. I could not be where I am in my self-study of calculus without your help and guidance.
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