Let f(x) be positive for all real x. If I_1 = | vedclass

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(C) Given $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b

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$\frac{1}{2}$

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(C) Given $I_1 = \int_{1-h}^{h} x f(x(1-x)) dx$ and $I_2 = \int_{1-h}^{h} f(x(1-x)) dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we have $a+b = (1-h) + h = 1$. Applying this to $I_1$: $I_1 = \int_{1-h}^{h} (1-x) f((1-x)(1-(1-x))) dx$ $I_1 = \int_{1-h}^{h} (1-x) f((1-x)x) dx$ $I_1 = \int_{1-h}^{h} f(x(1-x)) dx - \int_{1-h}^{h} x f(x(1-x)) dx$ $I_1 = I_2 - I_1$ $2I_1 = I_2$ Therefore,$\frac{I_1}{I_2} = \frac{1}{2}$.

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

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