Find the area of the region enclosed by the parabola $x^2 = y$ and the line $y = x + 2$.

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

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

We have, $x^2 = y$ and $y = x + 2$.

On solving both equations we get:

$x^2 = x + 2$

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

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

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

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

$∴$ Required area of shaded region $= \int\limits_{-1}^{2} (x + 2) \, dx - \int\limits_{-1}^{2} x^2 \, dx$

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

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

$= 6 + \frac{3}{2} - \frac{9}{3} = \frac{36+9-18}{6}$

$= \frac{27}{6} = \frac{9}{2} \text{ sq. units}$