Let $f(x)$ be a positive function and $I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x$ and $I_2=\int_{-1}^2 f(x(1-x)) d x$. Then the value of $\frac{I_2}{I_1}$ is equal to ________

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{I}_1=\int_{-\frac{1}{2}}^1 2 \mathrm{xf}(2 \mathrm{x}(1-2 \mathrm{x})) \mathrm{dx} \ & \Rightarrow 2 \mathrm{x}=\mathrm{t} \Rightarrow 2 \mathrm{dx}=\mathrm{dt} \quad \Rightarrow \mathrm{I}_1=\frac{1}{2} \int_{-1}^2 \mathrm{tf}(\mathrm{t}(1-\mathrm{t})) \mathrm{dt} \ & \Rightarrow 2 \mathrm{I}_1=\int_{-1}^2(1-\mathrm{t}) \mathrm{f}(1-\mathrm{t})(1-(1-\mathrm{t})) \mathrm{dt} \ & \Rightarrow 2 \mathrm{I}_1=\int_{-1}^2 \mathrm{f}\left(\mathrm{t}(1-\mathrm{t}) \mathrm{dt}-\int_{-1}^2 \mathrm{tf}(\mathrm{t}(1-\mathrm{t}) \mathrm{dt}ight. \ & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2-2 \mathrm{I}_1 \ & \Rightarrow 4 \mathrm{I}_1=\mathrm{I}_2 \ & \Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=4\end{aligned}$

Let $f(x)$ be a positive function and $I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x$ and $I_2=\int_{-1}^2…

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