If a continuous function $f$ satisfies the relation ;

$\overset{t}{\underset{0}{\int}} \left(f(x) - \sqrt{f'(x)}\right)dx = 0$ and $f(0) = \dfrac{-1}{2}$

Then $f(x)$ is equal to

Given,

$\overset{t}{\underset{0}{\int}} \left(f(x) - \sqrt{f'(x)}\right)dx = 0$

Now using Leinbnitz's rule of differentiation under integral sign we get,

$f(t)-\sqrt{f'(t)}=0$

or, $\dfrac{f'(t)}{[f(t)]^2}=1$

or, $\dfrac{d[f(t)]}{[f(t)]^2}=dt$

Now integrating both sides we get,

$-\dfrac{1}{f(t)}=t+c$ [ Where $c$ being integrating constant]......(1)

Given,

$f(0)=-\dfrac{1}{2}$.

Using this condition in (1) we get,

$c=2$.

So the solution takes the form,

$-\dfrac{1}{f(t)}=t+2$

or, $f(t)=-\dfrac{1}{t+2}$.

So,

$f(x)=-\dfrac{1}{x+2}$.