My particular question is:
If $f$ be a decreasing continuous function satisfying $$f(x+y)=f(x)+f(y)-f(x)f(y) \quad \forall x,y \in \mathbb{R}$$ and $$f'(0)=-1 \text,$$ then $$\int_0^1f(x) \, \mathrm d x =?$$
Answer to this question is given to be $2-e$.
My attempt:
$f(0)=0$ by putting value of $x$ and $y$ as $0$ and noticing that $f(x)$ can not be constant.
Then I had some options in question. All of the answers involved $e$ so I tried some trivial values that $f(x)$ might assume but did not succeed in my attempt.
Since $f(0)=0$ and $f$ is a decreasing function hence integral will give a negative value. I managed to do it this far (which appears to be nothing).
Another attempt:
I tried to differentiate the property of $f$ $$f(x+y)= f(x)+ f(y) -f(x)f(y)$$ $$\implies f'(x+y) = f'(x) + f'(y) -f'(x)f(y) -f(x)f'(y) \text.$$ Putting $x , y = 0$ we get $f(0) = \frac 1 2$. So I think that this method is incorrect, but I don't know where?
Would anyone guide me please? Thanks.
Some hints.
a) Put $g(x)=1-f(x)$. Show that $g(x+y)=g(x)g(y)$ for all $x,y$.
b) Put $y=x$ and show that $g(y)\geq 0$ for all $y$.
c) Suppose that for a $u\in {\mathbb R}$ $g(u)=0$, and show then that $g(x)=0$ for all $x$, and that this imply a contradiction.
d) Now you can put $h(x)=\log g(x)$, and find the form of the (continuous ) fonction $h$.
e) I leave to you the end of the proof.