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I stumbled upon this gem whilst doing tasks, and I can't seem to quite grasp the key to solving this. The answer should be $\dfrac{1}{2}$. While I have tried a few things, which resulted in $\dfrac{1}{2}$, I am quite unsure if my methods were legit. I have thought of substitution, but I get stuck half way. I would very much appreciate it if someone could help me solve this (practising for a test).

$$\int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}\text{d}x$$

Edit: Let $f$ be a positive continuous function. The task is simple if you could simply plug in a function, but it says not to. I think it means to solve this generally.

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    $\begingroup$ What is your question? It will help if you post what you have tried and the community can see if your approach is valid. $\endgroup$ – Erik M Mar 7 '17 at 22:59
  • $\begingroup$ If we know nothing about $f(x)$, there is nothing to solve. First of all, we even do not know if this integration is meaningful - e.g. is f(x)/(f(x) + f(1-x)) measurable? $\endgroup$ – Yujie Zha Mar 7 '17 at 23:09
  • $\begingroup$ $f(1-x)$ is just $f(x)$ run backwards on the domain $[0,1]$, for whatever that may be worth. $\endgroup$ – Alfred Yerger Mar 7 '17 at 23:22
  • $\begingroup$ I really do not know where to start. I have posted the whole task as it is supposed to be solved. This is the last place I can ask, my friends are all sleeping..If I could somehow substitute the core to make them equal, I could in theory get something like f(1/2) / 2f(1/2) which would be 1/2. $\endgroup$ – MCrypa Mar 7 '17 at 23:22
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    $\begingroup$ Hint: $$I = \int_{0}^{1}\frac{f(x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\frac{f(x)+f(1-x)-f(1-x)}{f(x)+f(1-x)}dx=\int_{0}^{1}\left(1 - \frac{f(1-x)}{f(x)+f(1-x)}\right)dx=\;\;\cdots\;\;=1 - I$$ $\endgroup$ – dxiv Mar 7 '17 at 23:44
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Notice that if you change the variable in the integration by $x = 1 - x$, you get:

$$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx= \int_1^0 \frac{f(1 - x)}{f(1 - x) + f(x)}d(1 - x) = \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx$$

And $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx + \int_0^1 \frac{f(1 - x)}{f(1 - x) + f(x)}dx = \int_0^1 \frac{f(x) + f(1 - x)}{f(x) + f(1 - x)}dx = 1$$ So $$\int_0^1 \frac{f(x)}{f(x) + f(1 - x)}dx = \frac {1}{2}$$

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  • $\begingroup$ An oldie, but a goodie! (+1) $\endgroup$ – Mark Viola Mar 7 '17 at 23:55
  • $\begingroup$ Aha, of course! I should probably study the book more next time around. Thank you very much, you helped me a lot :)) Didn't think of adding the the two expressions hehe.. $\endgroup$ – MCrypa Mar 8 '17 at 0:06
  • $\begingroup$ No worries. Glad to help :) $\endgroup$ – Yujie Zha Mar 8 '17 at 2:24

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