# Thread: Problem #16: Another Integral

1. ## Problem #16: Another Integral

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

-Dan

2. ## Re: Problem #16: Another Integral

Question: Are we assuming this is integrable? Because if $\displaystyle f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right)=0$ for any $\displaystyle x \in [2,4]$, it is not integrable.

3. ## Re: Problem #16: Another Integral

Originally Posted by SlipEternal
Question: Are we assuming this is integrable? Because if $\displaystyle f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right)=0$ for any $\displaystyle x \in [2,4]$, 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

4. ## Re: Problem #16: Another Integral

Spoiler:
\displaystyle \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 $\displaystyle u = 3-x, du = -dx$. For the second, use the substitution $\displaystyle u = x-3, du = dx$.

\displaystyle \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*}

5. ## Re: Problem #16: Another Integral

Originally Posted by SlipEternal
Spoiler:
\displaystyle \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 $\displaystyle u = 3-x, du = -dx$. For the second, use the substitution $\displaystyle u = x-3, du = dx$.

\displaystyle \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

6. ## 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 $\displaystyle \left[\sqrt{5},\sqrt{7}\right]$ and $\displaystyle f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right) \neq 0$ for all $\displaystyle x \in [2,4]$.

7. ## Re: Problem #16: Another Integral

Originally Posted by SlipEternal
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 $\displaystyle \left[\sqrt{5},\sqrt{7}\right]$ and $\displaystyle f\left(\sqrt{9-x}\right)+f\left(\sqrt{x+3}\right) \neq 0$ for all $\displaystyle x \in [2,4]$.
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

8. ## 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) $\displaystyle I = \int_2^4 \frac{f \left ( \sqrt{9 - x} \right )}{f \left ( \sqrt{9 - x} \right ) + f \left ( \sqrt{3 + x} \right ) }~dx$

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

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

$\displaystyle I + I = \int_2^4 \frac{f \left ( \sqrt{9 - x} \right ) + f \left ( \sqrt{3 + x} \right )}{f \left ( \sqrt{3 + x} \right ) + f \left ( \sqrt{9 - x} \right ) }~dx = \int_2^4 dx = 2$