Finding the number of continuous functions \begin{gather*}
No.\ of\ continuous\ function( s) \ \ f:[ 0,1]\rightarrow \ \Re \ \ satisfying\\
\int ^{1}_{0} f( x) dx\ =\ \frac{1}{3} \ +\int ^{1}_{0} f^{2}\left( x^{2}\right) dx\ \ is/are
\end{gather*}
My approach:
I put $x^2$=$t$, giving $2xdx=dt$, but I am not able to find/ proceed further. Can anyone help please?
 A: You should do the same but inversely:
$$\int ^{1}_{0} f(x) dx = \int ^{1}_{0} f(t^2) 2tdt = \int ^{1}_{0} 2x f(x^2) dx$$
So we have:
$$\int ^{1}_{0} 2x f(x^2) dx = \frac{1}{3} +\int ^{1}_{0} f^{2}\left( x^{2}\right) dx \Longrightarrow \int ^{1}_{0} f^{2}\left( x^{2}\right) dx - \int ^{1}_{0} 2x f(x^2) dx + \frac{1}{3} = 0$$
$$\Longrightarrow 0 = \int ^{1}_{0} \left(f^{2}\left( x^{2}\right) - 2x f(x^2) \right)dx + \frac{1}{3} = \int ^{1}_{0} \left(\left(f\left( x^{2}\right) - x\right)^2 -x^2 \right)dx + \frac{1}{3} = $$
$$ = \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx - \int ^{1}_{0}x^2dx + \frac{1}{3} = \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx - \left[\frac{x^3}{3}\right]_0^1 + \frac{1}{3}$$
$$\Longrightarrow \int ^{1}_{0} \left(f\left( x^{2}\right) - x\right)^2 dx = 0$$
Since left side is a non negative function, it must be zero everywhere:
$$\left(f\left( x^{2}\right) - x\right)^2 = 0 \Longrightarrow f\left( x^{2}\right) - x = 0 \Longrightarrow f\left( x^{2}\right) = x \Longrightarrow f(x) = \sqrt x $$
