1. Given differential equation

$\frac{dy}{dx}=2(x+1)$

Integrate with respect to $x$:

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

So the family of curves is

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

2. Find where the curve meets the x-axis

For the area bounded by the curve and the $x$-axis to be finite, the parabola must intersect the $x$-axis at two points.

Set $y=0$:

$x^2+2x+C=0$

Equivalently,

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

Let

$a^2=1-C \qquad (a>0)$

Then

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

The roots are at

$x=-1-a,\quad x=-1+a$

Between these roots, $y\le 0$, so the bounded area with the $x$-axis is

$A=\int_{-1-a}^{-1+a} |y|\,dx=\int_{-1-a}^{-1+a} \left(a^2-(x+1)^2\right)dx$

3. Compute the area

Put $u=x+1$. Then limits become $-a$ to $a$:

$A=\int_{-a}^{a} (a^2-u^2)\,du$

$A=\left[a^2u-\frac{u^3}{3}\right]_{-a}^{a}$

$A=\frac{4a^3}{3}$

Given

$\frac{4a^3}{3}=\frac{4\sqrt{8}}{3}$

So,

$a^3=\sqrt{8}=2\sqrt{2}=(\sqrt{2})^3$

Hence,

$a=\sqrt{2}$

4. Find the constant $C$

Since

$a^2=1-C$

we get

$2=1-C$

$C=-1$

Therefore,

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

5. Compute $y(1)$

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

Final Answer

$\boxed{2}$