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$$\int_0^1\frac{f(x)}{f(x) + f(1- x)}dx$$if we know that $$ f > 0, f - continuous function $$

ivanchai
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    Hint: Make a change of variable $y = 1-x$ – NN2 Apr 24 '23 at 19:39
  • @NN2's advice is a special case of the King property,$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx=\int_a^b[pf(x)+(1-p)f(a+b-x)]dx$$(of which the case $p=\frac12$ is the most frequently useful). – J.G. Apr 24 '23 at 19:55
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    ... and "King property" is a term that seems to be used only in India. But @NN2's advice is applicable the world over. – GEdgar Apr 24 '23 at 20:04

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using u=1-x you get the following: $$I=-\int_1^0 \frac{f(1-u)}{f(1-u)+f(u)}du=\int_0^1\frac{f(1-x)}{f(1-x)+f(x)}dx $$ so now:

$$2I = \int_0^1\frac{f(1-x)}{f(1-x)+f(x)}dx + \int_0^1\frac{f(x)}{f(x)+f(1-x)}dx=\int_0^1 1dx=1-0$$ so I=1/2

Michael Douhji
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