Let f0 on [0,1], twice differentiable on (0,1).

Question

Let $f\ge0$ on $[0,1]$, twice differentiable on $(0,1)$. If $\int_0^x\sqrt{1-(f'(t))^2}\,dt=\int_0^xf(t)\,dt$, $f(0)=0$, then $\lim_{x	o0}\dfrac{1}{x^2}\int_0^xf(t)\,dt$:

Accepted Answer

Differentiating: $\sqrt{1-(f')^2}=f \implies f'=\sqrt{1-f^2}$; solving gives $f(x)=\sin x$ (using $f(0)=0$).

$$\lim_{x\to0}\frac{\int_0^x\sin t\,dt}{x^2} = \lim_{x\to0}\frac{\sin x}{2x} = \frac12.$$

Solutions · 1