Let int_0^x sqrt{1-left(y^{prime}(t)right)^2} | vedclass

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(A) Given the equation $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} dt = \int_0^x y(t) dt$.Applying the Fundamental Theorem of Calculus by differen

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(A) Given the equation $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} dt = \int_0^x y(t) dt$.Applying the Fundamental Theorem of Calculus by differentiating both sides with respect to $x$,we get:$\sqrt{1-\left(y^{\prime}(x)\right)^2} = y(x)$.Squaring both sides:$1-\left(\frac{dy}{dx}\right)^2 = y^2$.Rearranging the terms:$\left(\frac{dy}{dx}\right)^2 = 1-y^2$.Taking the square root:$\frac{dy}{dx} = \pm \sqrt{1-y^2}$.Since $y(0)=0$ and $y \geq 0$,we choose the positive root for $x$ near $0$:$\frac{dy}{\sqrt{1-y^2}} = dx$.Integrating both sides:$\sin^{-1}(y) = x + C$.Using the initial condition $y(0)=0$,we find $C=0$,so $\sin^{-1}(y) = x$,which implies $y = \sin(x)$.Now,find the derivatives:$y^{\prime} = \cos(x)$ and $y^{\prime \prime} = -\sin(x)$.Substituting these into the expression $y^{\prime \prime} + y + 1$:$-\sin(x) + \sin(x) + 1 = 1$.Thus,at $x=2$,the value is $1$.

Let $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} dt = \int_0^x y(t) dt, 0 \leq x \leq 3, y \geq 0$,$y(0)=0$. Then at $x=2, y^{\prime \prime}+y+1$ is equal to :

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