If $\int^{x}_{0}2x(f(t))^2dt = \bigg(\int^{x}_{0}2f(x-t)dt\bigg)^2$ and $f(1) = 1,$ Then $f(x)$ is 
If  $\displaystyle \int^{x}_{0}2x(f(t))^2dt = \bigg(\int^{x}_{0}2f(x-t)dt\bigg)^2$ and $f(1) = 1,$ Then $f(x)$ is 

Try: Using $\displaystyle  \int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx$
We can write it as $$\displaystyle x\int^{x}_{0}f^2(t)dt = \bigg(\int^{x}_{0}2f(t)dt\bigg)^2 = 4 \bigg(\int^{x}_{0}f(t)dt\bigg)^2\;\;\;(*)$$
Using Leibnitz Rule of Differentiation
$$xf^2(x)+\int^{x}_{0}f^2(t)dt = 8\int^{x}_{0}f(t)dt\cdot f(x)$$
Again Differentiate w r to $x$
$$x\cdot 2f(x)\cdot f'(x)+f^2(x)+f^2(x)=8f^2(x)+8\int^{x}_{0}f(t)dt$$
Could some help me how to solve it, Thanks 
 A: @Angle made a mistake in their answer, but the method works. You have
$$ x\int_0^x f^2  = 2\left(\int_0^x f \right)^2 $$
and upon differentiating
$$ xf^2 + \int_0^x f^2 = 4f\int_0^x f $$
Multiplying through by $x$ and substituting the original relation
$$ x^2f^2 + 2\left(\int_0^x f\right)^2 = 4xf\int_0^x f $$
Let $g = \int_0^x f$, then
$$ x^2(g')^2 + 2g^2 = 4xgg' $$
Divide through by $g^2$
$$ \left(x\frac{g'}{g} \right)^2 - 4x\frac{g'}{g} + 2 = 0 \implies x\frac{g'}{g} = 2 \pm \sqrt{2} $$
Solving this, we find $g = cx^{2\pm\sqrt{2}}$. Differentiating and applying the given initial condition, the solution shows itself
$$ f(x) = x^{1\pm\sqrt{2}} $$
You can check that both solutions satisfy the original integral equation
A: After Leibniz, it should be:
$$xf^2(x)+\int^{x}_{0}f^2(t)dt = \color{red}{4}\int^{x}_{0}f(t)dt\cdot f(x)\;\;\;\;\;(**)$$
Let $$g(x) = \int^{x}_{0}f(t)dt$$
if we multiply $(**)$ with $x$ and use $(*)$ we get:
$$x^2f^2(x) + 4g^2(x) = 4xg(x)f(x)$$
implies $$xf(x)-2g(x)=0 \implies g(x)=xf(x)/2$$
Does this help?

So we have $$xf(x) = 2\int^{x}_{0}f(t)dt$$
and after differentiation we get $$f(x)+xf'(x) = 2f(x)\implies {f'(x)\over f(x)} ={1\over x}$$
So $$(\ln f(x))'  = (\ln x )' \implies \ln f(x) =\ln x+ c$$
Since $f(1)=1$ we get $c=0$ and so $\boxed{f(x)=x}$.
Plugging this in to staring formula we see that FE doesn't have a solution.
