Let the curve y = y(x) be the solution of the differential...

$\dfrac{dy}{dx} = 2\left(x + 1\right)$

$⇒ ∫ dy = ∫ 2\left(x + 1\right)dx$

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

Area $= \dfrac{4\sqrt{8}}{3}$

$⇒ 2{∫}_{- 1}^{- 1 + \sqrt{1 - C}}\left(- {\left(x + 1\right)}^{2} - C + 1\right)dx = \dfrac{4\sqrt{8}}{3}$

$⇒ 2{\left[- \dfrac{{\left(x + 1\right)}^{3}}{3} - Cx + x\right]}_{- 1}^{- 1 + \sqrt{1 - C}} = \dfrac{4\sqrt{8}}{3}$

$⇒ {\left(- \sqrt{1 - C}\right)}^{3} + 3C - 3C\sqrt{1 - C} - 3C - 3 + 3\sqrt{1 - C} + 3 = 2\sqrt{8}$

$⇒ C = - 1$

$⇒ f\left(x\right) = {x}^{2} + 2x - 1, f\left(1\right) = 2$