Find the area bounded by $y=f(x)$ from $x=1$ to $x=3$ where $f(x)$ satisfies the equation $\int_0^1(x-f(x))f(x)dx=\frac1{12}$

Question

Find the area bounded by $y=f(x)$ from $x=1$ to $x=3$ where $f(x)$ satisfies the equation $\int_0^1(x-f(x))f(x)dx=\frac1{12}$

I tried solving $\int xf(x)dx$ by taking $f(x)$ as first function, I get $$f(x)\frac{x^2}2-\int f'(x)\frac{x^2}2dx$$

I also tried solving $\int (f(x))^2dx$ by taking $1$ as the second function. I get $$(f(x))^2x-\int 2f(x)f'(x)xdx$$

I don't think I am reaching anywhere.

Answer 1

So we have $$\int^{1}_{0}(x-f(x))f(x)dx=\frac{1}{12}.$$ Let's do the following manipulations: $$\int^{1}_{0}(4x-4f(x))f(x)dx=\int^{1}_{0}(4xf(x)-4f(x)^2)dx-\frac{1}{3}=0.$$ Notice that $$\int^{1}_{0}x^2dx=\frac{1}{3},$$ we can plug this above:

$$\int^{1}_{0}(4xf(x)-4f(x)^2-x^2)dx=-\int^{1}_{0}(x-2f(x))^2dx=0.$$

If $f(x)\geq 0$, then $\int^{b}_{a}f(x)dx=0$, implies $f(x)=0$. (I am assuming this)

Thus: $-\int^{1}_{0}(x-2f(x))^2dx=0$, then $x-2f(x)=0$, therefore: $$f(x)=\frac{x}{2}.$$

Finally we need:$$\int^{3}_{1}\frac{x}{2}dx=2.$$