"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 $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}$ :"

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

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 $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}$ :

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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 $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}$ :

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