Official solution
For point of intersection
x + y = 1 \[ \Rightarrow \] x = 1 -- y
y2 = \[{x \over 2}\] \[ \Rightarrow \] 2y2
= x
2y2
= 1 -- y \[ \Rightarrow \] 2y2
+ y -- 1 = 0
\[ \Rightarrow \] (2y -- 1) (y + 1) = 0
\[ \Rightarrow \] y = \[{1 \over 2}\] or -1
Total area = \[4\int\limits_0^{{1 \over 2}} {\left[ {\left( {1 - x} \right) - \left( {\sqrt {{x \over 2}} } \right)} \right]} dx\]
= \[4\left[ {x - {{{x^2}} \over 2} - {1 \over {\sqrt 2 }}{{{x^{3/2}}} \over {3/2}}} \right]_0^{{1 \over 2}}\]
= \[4\left[ {{1 \over 2} - {1 \over 8} - {{\sqrt 2 } \over 3}{{\left( {{1 \over 2}} \right)}^{3/2}}} \right]\]
= 4 \[ \times \] \[{5 \over {24}}\] = \[{5 \over 6}\]