Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1).

Question

Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $$, $$0 \le x \le 1$$ and f(0) = 0, then $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $$ :

Accepted Answer

$$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } ,\,0 \le x \le 1$$

differentiating both the sides

$$\sqrt {1 - {{(f'(x))}^2}}  = f(x)$$

$$ \Rightarrow 1 - {(f'(x))^2} = {f^2}(x)$$

$${{f'(x)} \over {\sqrt {1 - {f^2}(x)} }} = 1$$

$${\sin ^{ - 1}}f(x) = x + C$$

$$\because$$ $$f(0) = 0 \Rightarrow C = 0 \Rightarrow f(x) = \sin x$$

Now, $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^x {\sin t\,dt} } \over {{x^2}}}\left( {{0 \over 0}} \right) = {1 \over 2}$$

Solutions · 1