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(D) Given the equation $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t$ for $0 \leq x \leq 1$.Differentiating both

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equals $\frac{1}{2}$

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(D) Given the equation $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t$ for $0 \leq x \leq 1$.Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus:$\sqrt{1-\left(f^{\prime}(x)\right)^{2}}=f(x)$Squaring both sides:$1-\left(f^{\prime}(x)\right)^{2}=f^{2}(x)$$\left(f^{\prime}(x)\right)^{2} = 1 - f^{2}(x)$$f^{\prime}(x) = \sqrt{1 - f^{2}(x)}$ (since $f$ is non-negative and $f(0)=0$ implies $f'(0)=1$)Separating variables:$\frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$Integrating both sides:$\sin^{-1}(f(x)) = x + C$Since $f(0)=0$,we have $\sin^{-1}(0) = 0 + C$,which gives $C=0$.Thus,$f(x) = \sin(x)$.Now,we evaluate the limit:$\lim_{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} \sin(t) \,dt = \lim_{x \rightarrow 0} \frac{[-\cos(t)]_{0}^{x}}{x^{2}} = \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^{2}}$Using the standard limit $\lim_{x \rightarrow 0} \frac{1 - \cos(x)}{x^{2}} = \frac{1}{2}$.

Let $f$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \,d t=\int_{0}^{x} f(t) \,d t$ for $0 \leq x \leq 1$ and $f(0)=0$,then $\lim_{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} f(t) \,d t$ is:

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