The area of the region bounded by the line $y = x + 2$ and the curve $x = -y^2$ is

The given curves are $x = -y^2$ and $y = x + 2$.

To find the points of intersection, substitute $x = y - 2$ into the equation of the parabola:

$y - 2 = -y^2$

$y^2 + y - 2 = 0$

$(y + 2)(y - 1) = 0$

The points of intersection are at $y = -2$ and $y = 1$.

The area of the bounded region is given by integrating the difference of the right curve $x = -y^2$ and the left curve $x = y - 2$ with respect to $y$ from $y = -2$ to $y = 1$:

$A = \int_{-2}^{1} (-y^2 - (y - 2)) \, dy$

$A = \int_{-2}^{1} (-y^2 - y + 2) \, dy$

Integrating the expression:

$A = \left[ -\dfrac{y^3}{3} - \dfrac{y^2}{2} + 2y \right]_{-2}^{1}$

Substituting the upper limit $y = 1$:

$-\dfrac{1}{3} - \dfrac{1}{2} + 2 = \dfrac{-2 - 3 + 12}{6} = \dfrac{7}{6}$

Substituting the lower limit $y = -2$:

$-\dfrac{(-2)^3}{3} - \dfrac{(-2)^2}{2} + 2(-2) = \dfrac{8}{3} - 2 - 4 = -\dfrac{10}{3}$

Calculating the total area:

$A = \dfrac{7}{6} - \left( -\dfrac{10}{3} \right) = \dfrac{7}{6} + \dfrac{20}{6} = \dfrac{27}{6} = 4.5$

Answer: $4.5\ \text{sq units}$