Let \(\mathrm{f}(x)\) be positive for all real \(x\). If \(\mathrm{I}_1=\int_{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x\) and \(\mathrm{I}_2=\int_{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x\), where \((2 h-1)>0\), then \(\frac{I_1}{I_2}\) is

\(\mathrm{I}_1 =\int_{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x \text { and } \)
\( \mathrm{I}_2 =\int_{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x \)
\( \mathrm{I}_1 =\int_{1-\mathrm{h}}^{\mathrm{h}}(1-x) \mathrm{f}[(1-x)(1-1+x)] \mathrm{d} x \)
\( \therefore \ldots\left[\because \int_{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d} x\right] \mathrm{I}_1=\) \(\int_{1-\mathrm{h}}^{\mathrm{h}}(1-x) \mathrm{f}(x(1-x)) \mathrm{d} x \)
\( =\int_{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x-\int_{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x \)
\( \Rightarrow \mathrm{I}_1=\mathrm{I}_2-\mathrm{I}_1 \)
\( \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2 \)
\( \Rightarrow \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{1}{2}\)