$$\therefore$$ Area of shaded region

$$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {\left( {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}} + x} \right)dx + \int\limits_0^1 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx} } $$

$$ = \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx + \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {xdx} } $$

Let $$x = {\sin ^3}\theta $$

$$\therefore$$ $$dx = 3{\sin ^2}\theta \cos \theta d\theta $$

$$ = \int\limits_{ - {\pi \over 4}}^{{\pi \over 2}} {3{{\sin }^2}\theta {{\cos }^4}\theta d\theta + \left( {0 + {1 \over {16}}} \right)} $$

$$ = {{9\pi } \over {64}} + {1 \over {16}} - {1 \over {16}} = {{36\pi } \over {256}} = A$$

$$\therefore$$ $${{256A} \over \pi } = 36$$