Let $\mathrm{f}(x)$ be positive for all real $x$. If $\mathrm{I}_1=\int_{1-\mathrm{h}}^{\mathrm{h}} x…
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
$2$
$h$
$\frac{1}{2}$
$1$
Solution
$\begin{aligned} \mathrm{I}_1 & =\int_{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x \text { and } \\ \mathrm{I}_2 & =\int_{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x \\ \mathrm{I}_1 & =\int_{1-\mathrm{h}}^{\mathrm{h}}(1-x) \mathrm{f}[(1-x)(1-1+x)] \mathrm{d} x \\ \therefore \quad \ldots\left[\because \int_{\mathrm{a}} \mathrm{f}(x) \mathrm{d} x=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}-x) \mathrm{d} x\right] & \mathrm{I}_1=\int_{1-\mathrm{h}}^{\mathrm{h}}(1-x) \mathrm{f}(x(1-x)) \mathrm{d} x \\ & =\int_{1-\mathrm{h}}^{\mathrm{h}} \mathrm{f}(x(1-x)) \mathrm{d} x-\int_{1-\mathrm{h}}^{\mathrm{h}} x \mathrm{f}(x(1-x)) \mathrm{d} x \\ & \Rightarrow \mathrm{I}_1=\mathrm{I}_2-\mathrm{I}_1 \\ & \Rightarrow 2 \mathrm{I}_1=\mathrm{I}_2 \\ & \Rightarrow \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{1}{2}\end{aligned}$