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> 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 
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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