<s>[INST] Function satisfying the relation $f(x+y)=f(x)+f(y)-(e^{-x}-1)(e^{-y}-1)+1$ Let f be the differentiable function satisfying the relation  $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$; $\forall x,y \in R$ and $\mathop {\lim }\limits_{h \to 0} \frac{{f'\left( {1 + h} \right) + f\left( h \right) - {e^{ - 1}}}}{h}$ exist. The value of $\int\limits_0^1 {f\left( x \right)dx}  = \_\_\_\_\_\_\_$.
My approach is as follow
$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$\Rightarrow f\left( {x + y} \right) - f\left( x \right) = f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + y} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( y \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - y}} - 1} \right) + 1}}{h}$
y=h
$ \Rightarrow f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) - \left( {{e^{ - x}} - 1} \right)\left( {{e^{ - h}} - 1} \right) + 1}}{h}$
How will I proceed from here
 [/INST] Continuing from your method $$f'(x)=\lim_{h\to 0} \frac{f(h)+1 -(e^{-x}-1)(e^{-h}-1))}{h}$$ for derivative to exist $f(0)=-1$ hence by L-Hospitals rule $$f'(x)=\lim_{h\to 0}\frac{f(h)+1}{h}-\lim_{h\to 0}(e^{-x}-1)\frac{(e^{-h}-1)}{h}$$
$$f'(x)=f'(0)+e^{-x}-1=e^{-x}+C\Rightarrow f(x)=-e^{-x}+Cx$$
Now for the given limit to exist $f'(1)+f(0)=e^{-1}$...hence find $C$...
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