## Calculus 8th Edition

$2$
$y = \sqrt{x-x^{2}} + \sin^{-1} \sqrt{x}$ then $dy/dx = \frac{1-2x}{2\sqrt{x-x^{2}}} + \frac{1}{2\sqrt{x} \sqrt{1-x^{2}}} = \frac{2-2x}{2\sqrt{x}\sqrt{1-x}} = \sqrt{\frac{1-x}{x}}$ and $1+(dy/dx)^{2} = 1+ \frac{1-x}{x} = \frac{1}{x}$ The curve has endpoint (0, 0) and (1, $\frac{\pi}{2}$) So $L = \int^{1}_{0} \sqrt{1/x} dx = \lim_{t->0+} \int ^{1}_{t} \sqrt{1/x} dx= \lim_{t->0+} [2\sqrt{x}]^{1}_{t} = lim_{t->0+} [2\sqrt{1} - 2\sqrt{t}] = 2$