Official solution
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