anonymous
  • anonymous
Tough Calculus Question: Let's see who can get it first!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
?
anonymous
  • anonymous
\[ \int_2^4 \frac{\sqrt{ln(9-x)} dx}{\sqrt{ln(9-x)}+\sqrt{ln(3+x)}}=? \]
anonymous
  • anonymous
In case you are curious this is a Putnam Problem.... Hint.... It isn't nearly as "tough" as it looks

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anonymous
  • anonymous
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 :)
ganeshie8
  • ganeshie8
\[I=\int_2^4 \frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(3+x)}} \\~\\ =\int_2^4 \frac{\sqrt{\ln(9-(2+4-x))} dx}{\sqrt{\ln(9-(2+4-x))}+\sqrt{\ln(3+(2+4-x))}}\\~\\ =\int_2^4 \frac{\sqrt{\ln(3+x)} dx}{\sqrt{\ln(3+x)}+\sqrt{\ln(9-x)}} \] Add first and last parts and get \[2I = \int_2^41\,dx = 2 \implies I=1\]
anonymous
  • anonymous
Hmmmm interesting.... yes correct :D Im glad I found a taker though you method is different from mine:
anonymous
  • anonymous
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: \[I= \int\limits^{-1}_1\frac{\sqrt{\ln(9-(3-u)}(-du)}{\sqrt{\ln(9-(3-u)}+\sqrt{\ln(3+(3-u)}} \\ \ \ =-\int\limits^{-1}_1\frac{\sqrt{\ln(6-u)}du}{\sqrt{\ln(6-u)}+\sqrt{\ln(6-u)}} \\ \ \ = \int\limits^{1}_{-1} \frac{du}{2}=\frac{2}{2}=1\]
anonymous
  • anonymous
Darnit I meant the subsitution 3-u :/ but im sure thats obvious :D
anonymous
  • anonymous
I guess I wound up getting lucky last night stopping prematurely because I see my error now
ganeshie8
  • ganeshie8
that's a clever substitution! :)

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