The area enclosed by the curve $y = -x^2$ and the line $x + y + 2 = 0$ is

Find the points of intersection of the curve $y = -x^2$ and the line $x + y + 2 = 0$.

Substituting $y = -x^2$ into the line equation:

$x - x^2 + 2 = 0$

$x^2 - x - 2 = 0$

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

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

In the interval $[-1, 2]$, the parabola $y = -x^2$ lies above the line $y = -x - 2$.

The required area is given by:

$A = \int_{-1}^{2} [(-x^2) - (-x - 2)] dx$

$A = \int_{-1}^{2} (-x^2 + x + 2) dx$

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

$A = \left( -\dfrac{8}{3} + \dfrac{4}{2} + 4 \right) - \left( \dfrac{1}{3} + \dfrac{1}{2} - 2 \right)$

$A = \left( 6 - \dfrac{8}{3} \right) - \left( \dfrac{5}{6} - 2 \right)$

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

$A = \dfrac{20}{6} + \dfrac{7}{6} = \dfrac{27}{6} = 4.5$ sq units.

Answer: $4.5$ sq units