Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$.
• Chain Rule: $\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$
• Trigonometric Identity: $\sin^2(x) + \cos^2(x) = 1$

Step-by-Step Solution

Step 1: Differentiate both sides of the given equation with respect to x

We are given: $\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t$ Applying the Fundamental Theorem of Calculus to both sides, we get: $\frac{d}{dx} \int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t = \frac{d}{dx} \int_0^x y(t) d t$ $\sqrt{1-(y'(x))^2} = y(x)$ This step transforms the integral equation into a differential equation.

Step 2: Square both sides of the equation

Squaring both sides of the equation $\sqrt{1-(y'(x))^2} = y(x)$, we get: $1 - (y'(x))^2 = (y(x))^2$ $1 - (y')^2 = y^2$ This simplifies the equation further.

Step 3: Rearrange the equation and solve for y'

Rearranging the equation, we have: $(y')^2 = 1 - y^2$ Taking the square root of both sides: $y' = \pm \sqrt{1 - y^2}$ Since $y \geq 0$ and $y(0) = 0$, we choose the positive root because $y$ is increasing near $x=0$. $y' = \sqrt{1 - y^2}$

Step 4: Separate variables and integrate

Separate the variables: $\frac{dy}{\sqrt{1 - y^2}} = dx$ Integrate both sides: $\int \frac{dy}{\sqrt{1 - y^2}} = \int dx$ $\arcsin(y) = x + C$

Step 5: Apply the initial condition y(0) = 0

Using the initial condition $y(0) = 0$, we have: $\arcsin(0) = 0 + C$ $0 = C$ So, the equation becomes: $\arcsin(y) = x$

Step 6: Solve for y(x)

Taking the sine of both sides: $y(x) = \sin(x)$

Step 7: Find the first and second derivatives of y(x)

We have $y(x) = \sin(x)$. Then: $y'(x) = \cos(x)$ $y''(x) = -\sin(x)$

Step 8: Evaluate y'' + y + 1 at x = 2

We need to find the value of $y''(2) + y(2) + 1$. $y''(2) = -\sin(2)$ $y(2) = \sin(2)$ Therefore, $y''(2) + y(2) + 1 = -\sin(2) + \sin(2) + 1 = 1$

Step 9: Check the initial assumption for the sign of y' We assumed $y' = +\sqrt{1-y^2}$. If we had chosen the negative square root, we'd have $y' = -\sqrt{1-y^2}$. Separating variables would give $\frac{dy}{-\sqrt{1-y^2}} = dx$, so $-\arcsin(y) = x + C$. Using $y(0) = 0$ gives $C=0$, so $-\arcsin(y) = x$, which means $y(x) = \sin(-x) = -\sin(x)$. However, we are given $y \ge 0$, so $y(x) = -\sin(x)$ for $x \in [0,3]$ is not possible.

Step 10: Re-evaluate everything. There's an error in the current solution.

Going back to $\sqrt{1 - (y')^2} = y$, we have $1 - (y')^2 = y^2$, so $(y')^2 = 1 - y^2$. Thus $y' = \pm \sqrt{1-y^2}$. Since $y(0) = 0$ and $y \ge 0$, we have $y' \ge 0$ near $x=0$. Hence, $y' = \sqrt{1-y^2}$. Then $\int \frac{dy}{\sqrt{1-y^2}} = \int dx$, so $\arcsin(y) = x+C$. Since $y(0)=0$, we have $\arcsin(0) = 0 + C$, so $C=0$. Thus $\arcsin(y) = x$, so $y = \sin(x)$. Therefore $y' = \cos(x)$ and $y'' = -\sin(x)$.

Then $y''(x) + y(x) + 1 = -\sin(x) + \sin(x) + 1 = 1$. At $x=2$, $y''(2) + y(2) + 1 = 1$.

Let's reconsider our approach. Given $\int_0^x \sqrt{1-(y'(t))^2} dt = \int_0^x y(t) dt$. Differentiating both sides, we get $\sqrt{1-(y'(x))^2} = y(x)$. Squaring both sides, we get $1 - (y'(x))^2 = y(x)^2$, so $(y'(x))^2 = 1 - y(x)^2$. Therefore $y'(x) = \pm \sqrt{1-y(x)^2}$. Since $y(0) = 0$ and $y \ge 0$, we have $y'(x) = \sqrt{1-y(x)^2}$. So $\frac{dy}{\sqrt{1-y^2}} = dx$. Integrating, we get $\arcsin(y) = x + C$. Since $y(0) = 0$, $\arcsin(0) = 0 + C$, so $C = 0$. Thus $\arcsin(y) = x$, so $y = \sin(x)$. Then $y' = \cos(x)$ and $y'' = -\sin(x)$. $y'' + y + 1 = -\sin(x) + \sin(x) + 1 = 1$. At $x=2$, $y'' + y + 1 = 1$. This does not match the given correct answer.

Let's go back to $\sqrt{1-(y')^2} = y$. If we differentiate this, we get $\frac{-2y'y''}{2\sqrt{1-(y')^2}} = y'$. So $\frac{-y'y''}{\sqrt{1-(y')^2}} = y'$. Since $\sqrt{1-(y')^2} = y$, we have $\frac{-y'y''}{y} = y'$. If $y' \ne 0$, then $\frac{-y''}{y} = 1$, so $y'' + y = 0$. Therefore $y'' + y + 1 = 1$. At $x=2$, this equals 1.

The given answer is $\sqrt{2}$, which seems incorrect.

Common Mistakes & Tips

• Remember to consider both positive and negative roots when taking the square root. However, the initial condition and the non-negativity of $y$ usually help determine the correct sign.
• Be careful when applying the Fundamental Theorem of Calculus and the Chain Rule.
• Always check your solution against the initial conditions and any given constraints.

Summary

We started with an integral equation and transformed it into a differential equation using the Fundamental Theorem of Calculus. After solving the differential equation and applying the initial condition, we found $y(x) = \sin(x)$. Then we computed the first and second derivatives, $y'(x) = \cos(x)$ and $y''(x) = -\sin(x)$. Finally, we evaluated the expression $y''(x) + y(x) + 1$ at $x=2$, which resulted in $1$. It seems the answer is incorrect.

The final answer is \boxed{1}, which corresponds to option (D).