Official solution
x2 \[ \le \] y \[ \le \] -- 2x + 3
\[ \Rightarrow \] x2
= -- 2x + 3
\[ \Rightarrow \] x2
+ 2x -- 3 = 0
\[ \Rightarrow \] (x + 3) (x -- 1) = 0
\[ \Rightarrow \] x = -- 3, x = 1
Area = \[\int\limits_{ - 3}^1 {\left( { - 2x + 3 - {x^2}} \right)dx} \]
= \[\left( { - {x^2} + 3x - {{{x^3}} \over 3}} \right)_{ - 3}^1\]
= \[\left( { - 1 + 3 - {1 \over 3} + 9 + 9 - 9} \right)\]
= 11 - \[{{1 \over 3}}\]
= \[{{{32} \over 3}}\] sq. unit