QJEEM19Ma356e: JEE Main 2019 Mathematics - Class 12 JEEM19Ma356e Solution: The area (in sq. units) of the region A = {(x, y) : y 2 2 ≤ x ≤ …

y² = 2x ...........(1)

and x = y + 4 .............(2)

Solving (1) and (2)

(x - 4)² = 2x

$$\Rightarrow$$ x² - 10x + 16 = 0

$$\Rightarrow$$ x = 8, 2 and y = 4, -2

Integrating in y direction from A to B

Required area (A) = $$\underset{-2}{\overset{4}{\int }}\left(y+4-\frac{y^2}{2}\right)dy$$

= $${\left[\frac{y^2}{2}+4y-\frac{y^3}{6}\right]}_{-2}^4$$

= $$\left(\frac{16}{2}+16-\frac{64}{6}\right)$$ - $$\left(\frac{4}{2}-8+\frac{8}{6}\right)$$

= 30 - 12 sq. unit

= 18 sq. unit