Find the area of the following region using integration: $\{(x, y) : y^2 \leq 2x \text{ and } y \geq x - 4\}$

18

The correct answer is Option (1) → 18

Given:

$y^2 = 2x \dots (i)$

$y = x - 4 \dots (ii)$

Required area is $OABCO$ from $(i)$ and $(ii)$:

$(x - 4)^2 = 2x$

$x^2 - 10x + 16 = 0$

$(x - 8)(x - 2) = 0$

$x = 8 \text{ and } x = 2$

$∴\text{ Intersection points } (2, -2) \text{ and } (8, 4)$

Required Area:

$\int_{-2}^{4} \left[ (y + 4) - \frac{y^2}{2} \right] \, dy = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} \right]_{-2}^{4}$

$= \left( 8 + 16 - \frac{32}{3} \right) - \left( 2 - 8 + \frac{4}{3} \right)$

$= 30 - 12 = 18 \text{ unit}^2$