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anonymous

  • one year ago

Tough Calculus Question: Let's see who can get it first!

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  1. anonymous
    • one year ago
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    ?

  2. anonymous
    • one year ago
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    \[ \int_2^4 \frac{\sqrt{ln(9-x)} dx}{\sqrt{ln(9-x)}+\sqrt{ln(3+x)}}=? \]

  3. anonymous
    • one year ago
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    In case you are curious this is a Putnam Problem.... Hint.... It isn't nearly as "tough" as it looks

  4. anonymous
    • one year ago
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    If no one gets it in a day or two I'll post the answer. I just figured my fellow maths aficionados would find this one as fun as I did when I came across it :)

  5. ganeshie8
    • one year ago
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    \[I=\int_2^4 \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}} \\~\\ =\int_2^4 \frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln(3+(2+4-x))}}\\~\\ =\int_2^4 \frac{\sqrt{\ln(3+x)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}} \] Add first and last parts and get \[2I = \int_2^41\,dx = 2 \implies I=1\]

  6. anonymous
    • one year ago
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    Hmmmm interesting.... yes correct :D Im glad I found a taker though you method is different from mine:

  7. anonymous
    • one year ago
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    I noted the arguments of the log went from 5 to 7 and 7 to 5 depending on either the 9-x or x+3 so this suggested the substitution x=u-6... this gives: \[I= \int\limits^{-1}_1\frac{\sqrt{\ln(9-(3-u)}(-du)}{\sqrt{\ln(9-(3-u)}+\sqrt{\ln(3+(3-u)}} \\ \ \ =-\int\limits^{-1}_1\frac{\sqrt{\ln(6-u)}du}{\sqrt{\ln(6-u)}+\sqrt{\ln(6-u)}} \\ \ \ = \int\limits^{1}_{-1} \frac{du}{2}=\frac{2}{2}=1\]

  8. anonymous
    • one year ago
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    Darnit I meant the subsitution 3-u :/ but im sure thats obvious :D

  9. anonymous
    • one year ago
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    I guess I wound up getting lucky last night stopping prematurely because I see my error now

  10. ganeshie8
    • one year ago
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    that's a clever substitution! :)

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