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) = $$\int\limits_{ - 2}^4 {\left( {y + 4 - {{{y^2}} \over 2}} \right)dy} $$

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

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

= 30 - 12 sq. unit

= 18 sq. unit