Let $f$ be a non-negative function defined on the interval $[0, 1]$. If $\int_{0}^{x} \sqrt{1-(f'(t))^{2}}\,dt = \int_{0}^{x} f(t)\,dt, \quad 0 \le x \le 1$, and $f(0) = 0$, then

$f \geq 0 \text{ on } [0,1],\qquad f(0) = 0$ $\int_{0}^{x} \sqrt{1-(f'(t))^{2}}\,dt = \int_{0}^{x} f(t)\,dt$ $\text{To find: compare } f\!\left(\tfrac{1}{2}\right) \text{ with } \tfrac{1}{2},\ \text{ and } f\!\left(\tfrac{1}{3}\right) \text{ with } \tfrac{1}{3}$ $\text{Plan: differentiate both sides (FTC)}$ $\sqrt{1-(f')^{2}} = f \Rightarrow \text{solve for } f$ $\text{then compare } f(\tfrac12),\ f(\tfrac13)$ $\dfrac{d}{dx}\!\left[\int_{0}^{x}\sqrt{1-(f'(t))^{2}}\,dt\right] = \sqrt{1-(f'(x))^{2}}$ $\sqrt{1-(f'(x))^{2}} = f(x)$ $1 - (f')^{2} = f^{2} \Rightarrow (f')^{2} + f^{2} = 1$ $f' = \sqrt{1-f^{2}} \Rightarrow \dfrac{df}{\sqrt{1-f^{2}}} = dx$ $\sin^{-1} f = x + C$ $f(0) = 0 \Rightarrow C = 0 \Rightarrow f(x) = \sin x$ $\text{check: } (f')^{2}+f^{2} = \cos^{2}x + \sin^{2}x = 1$ $\sin x < x \ \text{for } x > 0$ $f\!\left(\tfrac12\right) = \sin\tfrac12 < \tfrac12,\qquad f\!\left(\tfrac13\right) = \sin\tfrac13 < \tfrac13$ $\boxed{\text{option (3)}}$