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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

Samar Imam Zaidi
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1 Answers1

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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.

Michael Hoppe
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