<s>[INST] Solving the equation $f(x) = \int_0 ^x \sqrt{4-2f(t)} dt$ A continuous function $f(x)$ satisfies the following.
For all reals $x \le b$, $f(x)=a(x-b)^2+c$.
For all reals, $f(x) = \int_0 ^x \sqrt{4-2f(t)} dt$.
If $\int_0^6 f(x) dx = \frac{q}{p}$, where $p,q$ are relatively prime positive integers, find $p+q$
My approach is as follow $f\left( x \right) = \int\limits_0^x {\sqrt {4 - 2f\left( t \right)} dt} $
$f'\left( x \right) = \sqrt {4 - 2f\left( x \right)} $
$\frac{d}{{dx}}f\left( x \right) = \sqrt {4 - 2f\left( x \right)} $
$\frac{{d\left( {f\left( x \right)} \right)}}{{\sqrt {4 - 2f\left( x \right)} }} = dx$
Not able to proceed from here
 [/INST] We have
$$\frac{f'(x)}{\sqrt{4-2f(x)}}=1.$$
From here
$$-\sqrt{4-2f(x)}=x+c$$
hence
$$f(x) = 2-\frac12(x+c)^2.$$
As $f(0)=0$ we have $c=\pm2$.  (As the vertex is $(\pm2,2)$ the condition $4-2f(x)\ge0$ is satisfied for all $x$.)  Now integrate.
</s>