Let $f(x)$ be a positive function and $I_{1}=\int_{-\tfrac{1}{2}}^{1} 2x,f\left(2x(1-2x)\right),dx$ and $I_{2}=\int_{-1}^{2} f\left(x(1-x)\right),dx$. Then the value of $\dfrac{I_{2}}{I_{1}}$ is equal to