Find the area enclosed by the curve $y = - {x^2}$ and the straight lilne $x + y + 2 = 0$.

We have $y = - {x^2}$ and $x + y + 2 = 0$

Solving we get,
${x^2} = x + 2$

$\Rightarrow$ ${x^2} - x - 2 = 0$

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

$\Rightarrow$ $x = 2, - 1$

The graph of above function is downward parabola.

From the figure, area of shaded region,

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

$= \int_{ - 1}^2 {\left( {x + 2 - {x^2}} \right)} dx$
$= \left[ {\frac{{{x^2}}}{2} + 2x - \frac{{{x^3}}}{3}} \right]_{ - 1}^2$

$= \left[ {\frac{{{2^2}}}{2} + 2(2) - \frac{{{2^3}}}{3}} \right] - \left[ {\frac{{{{( - 1)}^2}}}{2} + 2( - 1) - \frac{{{{( - 1)}^3}}}{3}} \right]$

$= \left[ {6 - \frac{8}{3}} \right] - \left[ {\frac{1}{2} - 2 + \frac{1}{3}} \right] = \frac{9}{2}$.sq units