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Math Help - [SOLVED] Area under a curve, confused

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Area under a curve, confused

    Find the area under this curve : y=\frac {1}{x^2-4} in the region x\geq 3 and y\geq 0.
    I've already checked that y\geq 0 is no restriction, so they simply ask me to find \int_3^{+\infty} \frac{dx}{x^2-4}. Looking at a formula, I found the integral's worth \frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -.... I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
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    How did you get \frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -...?

    It should be \frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{b-2}{b+2}\right) \right] -... according to mathcad.
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  3. #3
    MHF Contributor arbolis's Avatar
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    It should be according to mathcad.
    You're right, sorry about it. I made a mistake looking at my book. Anyway, I'm still stuck .
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by arbolis View Post
    Find the area under this curve : y=\frac {1}{x^2-4} in the region x\geq 3 and y\geq 0.
    I've already checked that y\geq 0 is no restriction, so they simply ask me to find \int_3^{+\infty} \frac{dx}{x^2-4}. Looking at a formula, I found the integral's worth \frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -.... I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
    I get the value of the integral to be \frac{1}{4}\cdot\bigg[\lim_{b\to{\infty}}\ln\bigg[\frac{b-2}{b+2}\bigg]+\ln(5)\bigg]

    The limit converges to zero, so the integral has a value of \frac{1}{4}\ln(5)

    Does this make sense?

    --Chris
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by arbolis View Post
    Find the area under this curve : y=\frac {1}{x^2-4} in the region x\geq 3 and y\geq 0.
    I've already checked that y\geq 0 is no restriction, so they simply ask me to find \int_3^{+\infty} \frac{dx}{x^2-4}. Looking at a formula, I found the integral's worth \frac{1}{4}\cdot \lim_{b\to +\infty} \left[ \ln \left( \frac{2-b}{2+b}\right) \right] -.... I gave up here, I'm not able to solve this limit... mostly due to the restriction of the logarithm domain. I'd be very grateful if you could help me!
    \int_3^{\infty}\frac{dx}{x^2-4}=\int_0^{\infty}\frac{-dx}{4-x^2}=\frac{-1}{4}\int\frac{dx}{1-\left(\frac{x}{2}\right)^2}

    Now let \varphi=\frac{x}{2}\Rightarrow{2\varphi=x}

    So dx=2d\varphi

    So we get that

    \int\frac{dx}{x^2-4}=\frac{-1}{2}{\rm{arctanh}}\left(\frac{x}{2}\right)\bigg|_  {3}^{\infty}=\frac{\ln(5)}{4}
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  6. #6
    MHF Contributor arbolis's Avatar
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    I get the value of the integral to be

    The limit converges to zero, so the integral has a value of

    Does this make sense?
    Yes this does make sens! I'm too tired, how could have missed it from what Plato said...
    Thank you Chris.
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  7. #7
    Eater of Worlds
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    If we solve the integral:

    \int_{3}^{L}\frac{1}{x^{2}-4}dx

    =\frac{ln(5(L-2))}{4}-\frac{ln(L+2)}{4}

    Let t=L-2

    \lim_{L\to \infty}\left[\frac{ln(5t)}{4}-\frac{ln(t+4)}{4}\right]

    =\frac{1}{4}\lim_{t\to {\infty}}ln(\frac{5t}{t+4})

    =\lim_{t\to {\infty}}\frac{ln(5-\frac{20}{t+4})}{4}

    As t\to {\infty}, we can see we get

    \frac{ln(5)}{4}

    Kind of roundabout, but I thought it would help you see.
    Last edited by galactus; July 14th 2008 at 02:20 PM. Reason: chris and mathstud beat me. Their methods will certainly help. Just another way.
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  8. #8
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    Hello, arbolis!

    You did fine ... up to where you quit.


    Find the area of this region: . y\:=\:\frac {1}{x^2-4},\;x\geq 3,\;y\geq 0
    You integral is correct: . \int^{\infty}_3\frac{dx}{x^2-4}

    The formula is usually written: . \int\frac{dx}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C

    . . but the important feature is the absolute value.


    We have: . \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^{\infty}_3 \quad\Rightarrow\quad \lim_{b\to\infty}\,\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^b_3

    . =\;<br />
\lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg]  \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(  5\cdot\frac{b-2}{b+2}\right)


    Divide top and bottom by b

    . . \lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(5\cdot\fr  ac{1 - \frac{2}{b}}{1 + \frac{2}{b }}\right) \;\;=\; \;\frac{1}{4}\ln\left(5\cdot\frac{1-0}{1+0}\right) \;\;=\;\;\boxed{\frac{1}{4}\ln(5)}

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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, arbolis!

    You did fine ... up to where you quit.

    You integral is correct: . \int^{\infty}_3\frac{dx}{x^2-4}

    The formula is usually written: . \int\frac{dx}{u^2-a^2} \;=\;\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C

    . . but the important feature is the absolute value.


    We have: . \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^{\infty}_3 \quad\Rightarrow\quad \lim_{b\to\infty}\,\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right|\,\bigg]^b_3

    . =\; \lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)" alt="
    \lim_{b\to\infty}\:\frac{1}{4}\,\bigg[\ln\left(\frac{b-2}{b+2}\right) - \ln\left(\frac{1}{5}\right)\bigg] \;\;=\;\;\lim_{b\to\infty}\:\frac{1}{4}\,\ln\left( 5\cdot\frac{b-2}{b+2}\right)" />


    Divide top and bottom by b

    . . \lim_{b\to\infty}\:\frac{1}{4}\,\ln\left(5\cdot\fr  ac{1 - \frac{2}{b}}{1 + \frac{2}{b }}\right) \;\;=\; \;\frac{1}{4}\ln\left(5\cdot\frac{1-0}{1+0}\right) \;\;=\;\;\boxed{\frac{1}{4}\ln(5)}
    Honestly I don't see what the big to do is over this.

    \frac{b-2}{b+2}\sim{1}\quad\text{As }b\to\infty

    \lim_{b\to\infty}\ln\left(\frac{b-2}{b+2}\right)\sim\ln(1)=0

    Giving us

    \frac{1}{4}\bigg[0-\ln\left(\frac{1}{5}\right)\bigg]=\frac{\ln(5)}{4}
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