I worked out a solution for a problem I am studying for an entrance exam, but I am not sure if it is correct. I would appreciate if some knowledgeable person could help me out.
Given
$ \int_0^x t^{2}f(t)dt + 2 \int_0^x tf(t)dt = x-4 \int_0^x f(t)dt $
- Obtain $f(x)$.
- Obtain the maximum Value of $f(x)$.
- Calculate the improper integral $\displaystyle \int_0^{+\infty} f(t)dt $.
I differentiated both sides (using the Fundamental Theorem of Calculus) and solved for $f(x)$. Came up with this: $ x^{2}f(x) + 2xf(x) = 1 - 4 f(x) $ $ f(x) = 1/(x^{2}+2x+4) $
I tried integrating it but it is complicated so I used the formula I got after differentiating
$ x^{2}f(x)+2xf(x) = 1 - 4f(x) $
Setting $x$ to $1$ solved the equation. (So I think it is right, but not sure)
For 2. my answer is $\frac 14$ because $f(0)$ is $\frac 14$. If $x$ increases it will tend to $0$.
For 3. I don't have a solution because I did not manage to integrate the function yet.
Any help or advice is appreciated. How can I integrate $f(x)$ to calculate the improper integral. Is $f(x)$ right at all?
Best regards