Official solution
For point of intersection
4x2
= 8x + 12
\[ \Rightarrow \] x2 - 2x - 3 = 0
\[ \Rightarrow \] x = --1, 3
Required area = area of the shaded region
= \[\int\limits_{ - 1}^3 {\left( {8x + 12 - 4{x^2}} \right)dx} \]
= \[4\left[ {2.{{{x^2}} \over 2} + 3x - {{{x^3}} \over 3}} \right]_{ - 1}^3\]
= (36 + 36 -- 36) -- (4 -- 12 + \[{4 \over 3}\])
= \[{{128} \over 3}\]