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(D) Let the curve be $y = f(x)$. The area $A$ bounded by the curve,the $x$-axis,and the ordinate at $x$ is given by $A = \int_0^x y \, dx$. The arc le

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$(B)$ and $(C)$ both

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(D) Let the curve be $y = f(x)$. The area $A$ bounded by the curve,the $x$-axis,and the ordinate at $x$ is given by $A = \int_0^x y \, dx$. The arc length $s$ of the curve from $x=0$ to $x$ is given by $s = \int_0^x \sqrt{1 + (y')^2} \, dx$. According to the problem,$A = s$,so $\int_0^x y \, dx = \int_0^x \sqrt{1 + (y')^2} \, dx$. Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus,we get $y = \sqrt{1 + (y')^2}$. Squaring both sides,$y^2 = 1 + (y')^2$,which implies $(y')^2 = y^2 - 1$. Thus,$y' = \pm \sqrt{y^2 - 1}$. Separating variables,$\frac{dy}{\sqrt{y^2 - 1}} = \pm dx$. Integrating both sides,$\ln|y + \sqrt{y^2 - 1}| = \pm x + C$. Given the curve passes through $(0, 1)$,at $x=0$,$y=1$. Substituting these values: $\ln|1 + \sqrt{1-1}| = \pm 0 + C \implies \ln(1) = C \implies C = 0$. So,$\ln|y + \sqrt{y^2 - 1}| = x$ or $-x$. This simplifies to $y + \sqrt{y^2 - 1} = e^x$ or $y + \sqrt{y^2 - 1} = e^{-x}$. Solving for $y$,we get $y = \frac{e^x + e^{-x}}{2} = \cosh(x)$. Also,if $y=1$,then $y'=0$,which satisfies $y = \sqrt{1+(y')^2} \implies 1 = \sqrt{1+0} = 1$. Thus,both $y = \cosh(x)$ and $y=1$ are solutions. Therefore,the correct option is $(D)$.

The area bounded by a curve,the $x$-axis,and the ordinate of some point $(x, y)$ of the curve is equal to the length of the corresponding arc of the curve from a fixed point. If the curve passes through the point $P(0, 1)$,then the equation of this curve is:

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