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(A) Given the differential equation $\frac{dy}{dx} = 2(x + 1)$.Integrating both sides with respect to $x$,we get:$y(x) = \int 2(x + 1) dx = (x + 1)^2

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$2$

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(A) Given the differential equation $\frac{dy}{dx} = 2(x + 1)$.Integrating both sides with respect to $x$,we get:$y(x) = \int 2(x + 1) dx = (x + 1)^2 + C = x^2 + 2x + 1 + C$.Let $K = 1 + C$,so $y(x) = (x + 1)^2 + C$.The curve is a parabola opening upwards with vertex at $(-1, C)$. For the curve to bound an area with the $x-$axis,it must lie below the $x-$axis,so $C  0$.The roots of $y(x) = 0$ are $(x + 1)^2 = -C$,so $x = -1 \pm \sqrt{-C}$.The area bounded by the curve and the $x-$axis is given by:$A = \int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} (0 - ((x + 1)^2 + C)) dx = -\int_{-1-\sqrt{-C}}^{-1+\sqrt{-C}} ((x + 1)^2 + C) dx$.Let $u = x + 1$,then $du = dx$. The limits change to $-\sqrt{-C}$ to $\sqrt{-C}$.$A = -\int_{-\sqrt{-C}}^{\sqrt{-C}} (u^2 + C) du = -[\frac{u^3}{3} + Cu]_{-\sqrt{-C}}^{\sqrt{-C}} = -[(\frac{(-C)^{3/2}}{3} + C\sqrt{-C}) - (\frac{-(-C)^{3/2}}{3} - C\sqrt{-C})]$.$A = -[\frac{2}{3}(-C)^{3/2} + 2C\sqrt{-C}] = -[\frac{2}{3}(-C)\sqrt{-C} - 2(-C)\sqrt{-C}] = -[-\frac{4}{3}(-C)^{3/2}] = \frac{4}{3}(-C)^{3/2}$.Given $A = \frac{4\sqrt{8}}{3} = \frac{4(2\sqrt{2})}{3} = \frac{8\sqrt{2}}{3}$.So,$\frac{4}{3}(-C)^{3/2} = \frac{8\sqrt{2}}{3} \Rightarrow (-C)^{3/2} = 2\sqrt{2} = (\sqrt{2})^3$.Thus,$-C = 2$,which means $C = -2$.The function is $y(x) = (x + 1)^2 - 2 = x^2 + 2x - 1$.Therefore,$y(1) = (1)^2 + 2(1) - 1 = 1 + 2 - 1 = 2$.

Let the curve $y = y(x)$ be the solution of the differential equation,$\frac{dy}{dx} = 2(x + 1)$. If the numerical value of the area bounded by the curve $y = y(x)$ and the $x-$axis is $\frac{4\sqrt{8}}{3}$,then the value of $y(1)$ is equal to

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