Give that $f$ is a decreasing continuous function and that $f(x+y) = f(x) + f(y) -f(x)f(y)$ and $f'(0)=-1;$ Then find $\int_{0}^{1}f(x)dx$ Give that $f$ is a decreasing continuous function and that 
$$f(x+y) = f(x) + f(y) -f(x)f(y)$$
and $f'(0)=-1;$ Then it is to be found what is $\int_{0}^{1}f(x)dx$
I am at a loss on how to approach this sum. I have tried substituting $y=0$ & $x=x$ to no avail. 
The given answers are - 
A)$1$
B)$1-e$
C)$2-e$
 A: Rewrite the equation as follows :
$$1-f(x+y)=(1-f(x))(1-f(y))$$
Now denote $1-f(x)=g(x)$ so the equation is now :
$$g(x+y)=g(x)g(y)$$
Now it follows that $g(x) \geq 0$ because :
$$g(x)=g\left(\frac{x}{2} \right )^2 \geq 0$$
If for some $a$ we have $g(a)=0$ then :
$$g(x)=g(a)g(x-a)=0$$ for every $x$ so $f(x)=1$ which contradicts the fact that $f'(0)=-1$
This means that $g(x)>0$ for every $x$ so let's denote :
$$g(x)=e^{h(x)}$$
Because $f$ is continuous , $h$ will be continuous as well and the equation is :
$$h(x+y)=h(x)+h(y)$$
But this is Cauchy's equation and because $h$ is continuous it follows that $$h(x)=cx$$ for some constant $c$ and so :
$$f(x)=1-e^{cx}$$
Now because $f'(0)=-1$ we get $c=1$ and then :
$$\int_{0}^{1} f(x) dx=\int_{0}^{1} \left(1-e^x \right ) dx=(1-e)-(0-1)=2-e$$
A: Hint: Consider what functional equation $g(x)=1-f(x)$ satisfies.
A: First, we can find $f(0)$ using properties of $f$. Given
\begin{equation}
f(x+y)=f(x)+f(y)-f(x)f(y),
\end{equation}
Let $x=y=0$. Then we get $f(0)=0$ or $f(0)=1$. Again, let $x=x$ and $y=0$, then we get $f(0)f(x)=f(0)$. If $f(0)=1$, then $f(x)$ is a constant function, contradicting that $f$ decreases. Thus $f(0)=0$.
Second, we can show that $f$ is differentiable for all $x\in\mathbb{R}$. Derivative of $f$ is
\begin{align}
f'(x)&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\\
&=\lim_{h\to 0}\frac{f(x)+f(h)-f(x)f(h)-f(x)}{h}\\
&=\lim_{h\to 0}\frac{f(h)(1-f(x))}{h}\\
&=\lim_{h\to 0}\frac{f(h)-f(0)}{h-0}\lim_{h\to 0}(1-f(x))\\
&=f(x)-1.
\end{align}
This is an first-order differential equation with initial value $f(0)=0$. Solve it, then we get
\begin{equation}
f(x)=1-e^x.
\end{equation}
Therefore,
\begin{equation}
\int_0^1 f(x)dx =\int_0^1 (1-e^x)dx =2-e.
\end{equation}
A: Let first pick $x=y=0$ which yields
$$
f(0)=f(0)+f(0)-f(0)f(0).
$$
Solving for $f(0)$ yields $f(0)=0$ or $f(0)=1$. Now setting only $y=0$, then
$$
f(x)=f(x)+f(0)-f(x)f(0).
$$
For $f(0)=0$ this yields $f(x)=f(x)$ and for $f(0)=1$ this yields $f(x)=1$. Since $f'(0)=-1$, then $f(x)=1$ would be impossible, thus $f(0)=0$.
Now setting $y=dx$ an infinitesimal small number and using a first order Taylor expansion yields
$$
f(x+dx)=f(x)+f(dx)-f(x)f(dx),
$$
$$
f(x)+f'(x)dx=f(x)+f'(0)dx-f(x)f'(0)dx,
$$
$$
f'(x)=f'(0) \left(1-f(x)\right).
$$
Substituting in $-1$ for $f'(0)$ yields
$$
f'(x)=f(x)-1.
$$
Solving this inhomogeneous first order differential equation, with the initial condition $f(0)=0$ yields
$$
f(x)=1-e^x.
$$
Solving the integral should then be trivial.
