Official solution
y - x = 2, x2 = y
Now, x2 = 2 + x
\[\Rightarrow\] x2 - x - 2 = 0
\[\Rightarrow\] (x + 1)(x - 2) = 0
Area = \[\int\limits_{ - 1}^2 {(2 + x - {x^2})} \]
\[ = \left| {2x + {{{x^2}} \over 2} - {{{x^3}} \over 3}} \right|_{ - 1}^2\]
\[ = \left( {4 + 2 - {8 \over 3}} \right) - \left( { - 2 + {1 \over 2} + {1 \over 3}} \right)\]
\[ = 6 - 3 + 2 - {1 \over 2} = {9 \over 2}\]