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Math Help - Problem #16: Another Integral

  1. #1
    Forum Admin topsquark's Avatar
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    Problem #16: Another Integral

    Let f(x) be a continuous, bounded function on  \[ \sqrt{5}, \sqrt{7} \] . Evaluate:
    \int_2^4 \frac{f \left ( \sqrt{9 - x}~ \right )}{f \left ( \sqrt{9 - x}~ \right ) + f \left ( \sqrt{x + 3} ~ \right )}~dx

    -Dan
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    Re: Problem #16: Another Integral

    Question: Are we assuming this is integrable? Because if f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right)=0 for any x \in [2,4], it is not integrable.
    Last edited by SlipEternal; July 29th 2014 at 04:45 PM.
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    Re: Problem #16: Another Integral

    Quote Originally Posted by SlipEternal View Post
    Question: Are we assuming this is integrable? Because if f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right)=0 for any x \in [2,4], it is not integrable.
    Hmmmm...I missed that. Yes, assume that the denominator doesn't go to 0 on that interval. The original problem, from which I generalized, was for f(x) = ln(x) which didn't have that issue. Thanks for the catch!

    -Dan
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    Re: Problem #16: Another Integral

    Spoiler:
    \begin{align*}\int_2^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx & = \int_2^3 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx + \int_3^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx \\ & = -\int_3^2 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx + \int_3^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx\end{align*}

    For the first integral, use the substitution u = 3-x, du = -dx. For the second, use the substitution u = x-3, du = dx.

    \begin{align*}-\int_3^2 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx + \int_3^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx & = \int_0^1 \dfrac{f\left(\sqrt{6+u}\right)}{ f\left(\sqrt{6+u}\right) + f\left( \sqrt{-u+6} \right) }du + \int_0^1 \dfrac{f\left(\sqrt{6-u}\right)}{ f\left(\sqrt{6-u}\right) + f\left( \sqrt{u+6} \right) }du \\ & = \int_0^1 \dfrac{f\left(\sqrt{6-u}\right) + f\left( \sqrt{6+u}\right)}{f\left(\sqrt{6-u}\right) + f\left( \sqrt{6+u}\right)}du = 1\end{align*}
    Last edited by SlipEternal; July 29th 2014 at 07:18 PM.
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    Forum Admin topsquark's Avatar
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    Re: Problem #16: Another Integral

    Quote Originally Posted by SlipEternal View Post
    Spoiler:
    \begin{align*}\int_2^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx & = \int_2^3 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx + \int_3^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx \\ & = -\int_3^2 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx + \int_3^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx\end{align*}

    For the first integral, use the substitution u = 3-x, du = -dx. For the second, use the substitution u = x-3, du = dx.

    \begin{align*}-\int_3^2 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx + \int_3^4 \dfrac{f\left(\sqrt{9-x}\right)}{ f\left(\sqrt{9-x}\right) + f\left( \sqrt{x+3} \right) }dx & = \int_0^1 \dfrac{f\left(\sqrt{6+u}\right)}{ f\left(\sqrt{6+u}\right) + f\left( \sqrt{-u+6} \right) }du + \int_0^1 \dfrac{f\left(\sqrt{6-u}\right)}{ f\left(\sqrt{6-u}\right) + f\left( \sqrt{u+6} \right) }du \\ & = \int_0^1 \dfrac{f\left(\sqrt{6-u}\right) + f\left( \sqrt{6+u}\right)}{f\left(\sqrt{6-u}\right) + f\left( \sqrt{6+u}\right)}du = 1\end{align*}
    Not quite what I had, but in the same "spirit."

    Nice job.

    -Dan
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    Re: Problem #16: Another Integral

    By the way, to make it even more generalized, you don't even need f(x) to be continuous. You just need it to be Riemann integrable on \left[\sqrt{5},\sqrt{7}\right] and f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right) \neq 0 for all x \in [2,4].
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    Re: Problem #16: Another Integral

    Quote Originally Posted by SlipEternal View Post
    By the way, to make it even more generalized, you don't even need f(x) to be continuous. You just need it to be Riemann integrable on \left[\sqrt{5},\sqrt{7}\right] and f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right) \neq 0 for all x \in [2,4].
    I hadn't considered Riemann integrable, thanks for that. I had thought of piece-wise continuous, but then I decided to keep it simple.

    -Dan
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    Re: Problem #16: Another Integral

    Thanks to Slip Eternal for a correct solution. Mine is quite similar, but doesn't involve splitting up the interval:

    Let
    (1) I = \int_2^4 \frac{f \left ( \sqrt{9 - x} \right )}{f \left ( \sqrt{9 - x} \right ) + f \left ( \sqrt{3 + x} \right ) }~dx

    Let u = 6 - x. Then
    I = -\int_4^2 \frac{f \left ( \sqrt{3 + u} \right )}{f \left ( \sqrt{3 + u} \right ) + f \left ( \sqrt{9 - u} \right ) }~du

    Neaten up the limits on the integration and switch the dummy variable from u back to x gives:
    (2) I = \int_2^4 \frac{f \left ( \sqrt{3 + x} \right )}{f \left ( \sqrt{3 + x} \right ) + f \left ( \sqrt{9 - x} \right ) }~dx

    Add lines (1) and (2):
    I + I = \int_2^4 \frac{f \left ( \sqrt{9 - x} \right ) + f \left ( \sqrt{3 + x} \right )}{f \left ( \sqrt{3 + x} \right ) + f \left ( \sqrt{9 - x} \right ) }~dx = \int_2^4 dx = 2

    And thus I = 1.

    -Dan
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