Let $ I_1 = \int_{\frac{1}{2}}^{1} 2x \cdot f(2x(1 - 2x)) \, dx $ and $ I_2 = \int_{-1}^{1} f(x(1 - x)) \, dx \; \text{then} \frac{I_2}{I_1} \text{ equals to:} $