Definite integrals : how do we approach in solving a problem

Question

While practicing definite integrals I came across a question and now I am stuck

Question:

let f be a continous satisfying $f(x+y) = f(x) + f(y) + f(x)\cdot f(y)$ for all real $x$ and $y$ and $f'(0)= -1$.

Find the value of $\int_0^1 f(x) \, dx$.

I tried to solve by find the function but no success and i am confused.

Please tell me how to solve these kind of questions.

Answer 1

Rearrange

$$f(x+y)-f(x)=f(y)[1+f(x)]$$

Divide both sides by $y$ and take the limit $$\lim_{y\to0}\frac{f(x+y)-f(x)}{y}=\lim_{y\to0}=\frac{f(y)[1+f(x)]}{y}$$ $$f'(x)=[1+f(x)]\cdot\left[\lim_{y\to0}\frac{f(y)}{y}\right]$$

Can you proceed from here?

Notice $f(0)=0$ and given $f'(0)=-1$ $$\lim_{y\to0}\frac{f(y)}{y}=-1$$ So $$f'(x)=-[1+f(x)]$$ $$\int^1_0f'(x)dx=\int^1_0-1-f(x)dx$$ $$\int^1_0f(x)dx=[-x]^1_0-[f(x)]^1_0=-1-f(1)$$