The area is given by the integral,
[tex]\displaystyle\int_{-1/2}^{1/2}(1-2x^2-|x|)\,\mathrm dx[/tex]
The integrand is even, so we can simplify the integral somewhat as
[tex]\displaystyle2\int_0^{1/2}(1-2x^2-|x|)\,\mathrm dx[/tex]
When [tex]x\ge0[/tex], we have [tex]|x|=x[/tex], so this is also the same as
[tex]\displaystyle2\int_0^{1/2}(1-2x^2-x)\,\mathrm dx[/tex]
which has a value of
[tex]2\left(x-\dfrac23x^3-\dfrac12x^2\right)\bigg|_0^{1/2}=2\left(\dfrac12-\dfrac1{12}-\dfrac18\right)=\boxed{\dfrac7{12}}[/tex]
so that A = 7.
