Hint for integral of a function Let $f:[0,1] \to \mathbb R$ be a continuous function and $f(x) > 0$ for all $x \in [0,1]$
Evaluate the integral below
 $$ \int_0^1 \frac{f(x)dx}{f(x)+f(1-x)} $$ 
I'm not sure about what should I do for this integral. I couldn't write Riemann sums because I don't know infimum and supremum. We only know that it's continuous and positive. How can I use continuity and positiveness for evaluating. I need some hints. Thanks a lot. 
 A: Hint: Positivity is really only there to make sure $f(x)+f(1-x)\neq 0$ for all $x\in[0,1]$, and similarly continuity can be replaced by weaker assumptions.  Substituting $u=1-x$ gives
$$
\int_0^1\frac{f(x)\,\mathrm{d}x}{f(x)+f(1-x)}=\int_0^1\frac{f(1-u)\,\mathrm{d}u}{f(u)+f(1-u)}
$$
A: This is just another way of making use of the fact that
$$ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx $$
Which you can proved  very easily using substitution $u = a +b -x$. 
For your problem $a = 0$ and $b=1$ and so 
$$ \int_0^1 \frac{f(x)}{f(x) + f(1-x)} dx = \int_0^1 \frac{f(0+1-x)}{f(0+1-x) + f(1-(0+1 -x))} = \int_0^1 \frac{f(1-x)}{f(1-x) + f(x)} $$
So if you set 
$$ I =  \int_0^1 \frac{f(x)}{f(x) + f(1-x)} dx $$ 
then 
$$ 2I = I + I = \int_0^1 \frac{f(x)}{f(x) + f(1-x)} dx + \int_0^1 \frac{f(1-x)}{f(1-x) + f(x)} dx \\
= \int_0^1 \frac{f(x)}{f(x) + f(1-x)} + \frac{f(1-x)}{f(1-x) + f(x)} dx\\
= \int_0^1 1 dx = 1 $$ 
Hence,
$$ I =  \int_0^1 \frac{f(x)}{f(x) + f(1-x)} dx  = \frac{1}{2}$$
This kind of problem make use of symmetry. Something that always make the problem much easier to solve in mathematics! 
For example: Suppose you want to solve: 
$$ \int_4^8 \frac{\log x}{\log x + \log(12-x)} dx$$
This integral looks very daunting, especially if you are thinking about using the Fundamental Theorem of Calculus (finding the antiderivative, then evaluate it at the boundary points). However, because of symmetry, and the fact that $ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx $. We can say that:
$$ I = \int_4^8 \frac{\log x}{\log x + \log(12-x)} dx =  \int_4^8 \frac{\log 12 - x}{\log 12 - x + \log(x)} dx$$
From here, we can see that 
$$ 2I = \int_4^8 \frac{\log x}{\log x + \log(12-x)} + \frac{\log 12 - x}{\log 12 - x + \log(x)} dx = \int_4^8 1 dx = 4 $$
and therefore, 
$$ I = \int_4^8 \frac{\log x}{\log x + \log(12-x)} dx = 2$$
There are so many definite integrals out there that can be solved very easily by this very simple but powerful technique, which is nothing more than U-Sub and making use of symmetry. It's kinda strange to me that so many calculus teachers never really mention this in class. Maybe it is very obvious for them. 
A: By cheating:
The question hints that the value of the integral does not depend on the particular $f$. So if we take $f(x)=x$,
$$\int_0^1 x\,dx=\frac12.$$
