Key Concepts and Formulas

• Solving a first-order differential equation by integration: $\int \frac{dy}{dx} dx = y(x) + C$
• Discriminant of a quadratic equation $ax^2 + bx + c = 0$: $D = b^2 - 4ac$. For two distinct real roots, $D > 0$.
• Area bounded by a parabola $y = ax^2 + bx + c$ and the x-axis (when $D > 0$): Area $= \frac{|D|^{3/2}}{6a^2}$

Step-by-Step Solution

Step 1: Solving the Differential Equation

We are given the differential equation $\frac{dy}{dx} = 2(x + 1)$. To find $y(x)$, we integrate both sides with respect to $x$.

$\int dy = \int 2(x + 1) dx$

$y = 2 \int (x + 1) dx$

$y = 2 \left( \frac{x^2}{2} + x \right) + C$

$y = x^2 + 2x + C$

Here, $C$ is the constant of integration. This represents a family of parabolas.

Step 2: Analyzing the Intersection with the x-axis

For the parabola $y = x^2 + 2x + C$ to enclose a finite area with the x-axis, it must intersect the x-axis at two distinct points. This means the equation $x^2 + 2x + C = 0$ must have two distinct real roots. We determine the condition for this by examining the discriminant.

For $x^2 + 2x + C = 0$, we have $a = 1$, $b = 2$, and $c = C$. The discriminant is:

$D = b^2 - 4ac = 2^2 - 4(1)(C) = 4 - 4C$

For two distinct real roots, we need $D > 0$:

$4 - 4C > 0$

$4 > 4C$

$C < 1$

This condition ensures a bounded area.

Step 3: Applying the Area Formula

We are given that the area bounded by the curve and the x-axis is $\frac{4\sqrt{8}}{3} = \frac{8\sqrt{2}}{3}$. We use the formula for the area bounded by a parabola and the x-axis: Area $= \frac{|D|^{3/2}}{6a^2}$.

We have Area $= \frac{8\sqrt{2}}{3}$, $a = 1$, and $D = 4 - 4C$. Substituting these values into the area formula:

$\frac{8\sqrt{2}}{3} = \frac{|4 - 4C|^{3/2}}{6(1)^2}$

$\frac{8\sqrt{2}}{3} = \frac{|4 - 4C|^{3/2}}{6}$

Multiply both sides by 6:

$16\sqrt{2} = |4 - 4C|^{3/2}$

Since $C < 1$, $4 - 4C > 0$, so we can drop the absolute value:

$16\sqrt{2} = (4 - 4C)^{3/2}$

Express $16\sqrt{2}$ as a power of 2: $16\sqrt{2} = 2^4 \cdot 2^{1/2} = 2^{9/2}$. Thus,

$2^{9/2} = (4 - 4C)^{3/2}$

Raise both sides to the power of $2/3$:

$(2^{9/2})^{2/3} = ((4 - 4C)^{3/2})^{2/3}$

$2^3 = 4 - 4C$

$8 = 4 - 4C$

Step 4: Solving for C

Now we solve for $C$:

$8 = 4 - 4C$

$4 = -4C$

$C = -1$

This value of $C$ satisfies the condition $C < 1$.

Step 5: Finding y(1)

Substitute $C = -1$ into the equation $y = x^2 + 2x + C$:

$y(x) = x^2 + 2x - 1$

Now, we evaluate $y(1)$:

$y(1) = (1)^2 + 2(1) - 1 = 1 + 2 - 1 = 2$

Common Mistakes & Tips

• Always remember the constant of integration, $C$, when solving differential equations.
• Be careful with algebraic manipulations, especially when dealing with fractional exponents.
• Memorize the formula for the area bounded by a parabola and the x-axis to save time.

Summary

We solved the differential equation to find the general form of the curve, then used the area condition and the discriminant to find the specific value of the constant of integration, $C$. Finally, we substituted this value back into the equation and evaluated $y(1)$ to get the final answer.

The final answer is \boxed{2}.