If $f(x)=x+ \int_{0}^{1} t(x+t)f(t)dt$, find $\int_{0}^{1} f(x).dx$

Question

If $$f(x)=x+ \int_{0}^{1} t(x+t)f(t)dt$$, then find the value of $\int_{0}^{1} f(x).dx$

After rearranging I get $f(x)=x(1+\int_{0}^{1} t \cdot f(t).dt)+\int_{0}^{1} t^2 \cdot f(t).dt$

which means $f(x)$ will be a linear function of type $ax+b$ but how to find $a=1+\int_{0}^{1} t \cdot f(t).dt$ and $b=\int_{0}^{1} t^2 \cdot f(t).dt$ to finally get the desired result?

Answer 1

You have figured out that it is a linear function. Now you just need to note that by putting $f(t)=at+b$ back into the original equation, we get $$a= 1+\int_{0}^{1} (at^2+bt) \mathrm{d}t=1+\frac{a}{3}+\frac{b}{2}$$$$b=\int_{0}^{1} (at^3+bt^2) \mathrm{d}t=\frac{a}{4}+\frac{b}{3}$$ And you can solve the system of equations. Can you continue from here?