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

$\frac{9}{2} \text{ sq. units}$

The correct answer is Option (1) → $\frac{9}{2} \text{ sq. units}$

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

$\Rightarrow -x - 2 = -x^2$

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

$\Rightarrow x^2 + x - 2x - 2 = 0$

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

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

On solving both equations, we get

Let $A_1 = \text{Area above the line } x + y + 2 = 0$

i.e., $A_1 = \text{Area of CODABC}$

and $A_2 = \text{Area above the curve } y = -x^2$

i.e., $A_2 = \text{Area ODAOBCO}$

$∴$ Area of shaded region $A = A_1 - A_2$

$= \int\limits_{-1}^{2} (-x - 2) \, dx - \int_{-1}^{2} -x^2 \, dx$

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

$= \left| \left[ \frac{x^3}{3} - \frac{x^2}{2} - 2x \right]_{-1}^{2} \right| = \left| \left[ \frac{8}{3} - \frac{4}{2} - 4 + \frac{1}{3} + \frac{1}{2} - 2 \right] \right|$

$= \left| \frac{16 - 12 - 24 + 2 + 3 - 12}{6} \right| = \left| -\frac{27}{6} \right| = \frac{9}{2} \text{ sq. units}$