Answer
$2$
Work Step by Step
$y = \sqrt{x-x^{2}} + \sin^{-1} \sqrt{x}$
then $dy/dx = \frac{1-2x}{2\sqrt{x-x^{2}}} + \frac{1}{2\sqrt{x} \sqrt{1-x}}$
$ = \frac{2-2x}{2\sqrt{x}\sqrt{1-x}} $
$= \sqrt{\frac{1-x}{x}}$
and $1+(dy/dx)^{2} = 1+ \frac{1-x}{x} = \frac{1}{x}$
The curve has endpoint (0, 0) and (1, $\frac{\pi}{2}$)
Thus,
$L = \int^{1}_{0} \sqrt{1/x} dx = \lim\limits_{t \to 0^{+}} \int ^{1}_{t} \sqrt{1/x} dx$
$= \lim\limits_{t \to 0^{+}} [2\sqrt{x}]^{1}_{t} $
$= \lim\limits_{t \to 0^{+}} [2\sqrt{1} - 2\sqrt{t}] $
$= 2$
