Let f(x) be a positive function,I_1 = int_{-f | vedclass

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(D) Given $I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$. Let $2x = t$,then $2dx = dt$,so $dx = \frac{1}{2} dt$. When $x = -\frac{1}{2}$,$t = -1$. Wh

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$4$

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(D) Given $I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$. Let $2x = t$,then $2dx = dt$,so $dx = \frac{1}{2} dt$. When $x = -\frac{1}{2}$,$t = -1$. When $x = 1$,$t = 2$. Substituting these,$I_1 = \int_{-1}^2 t f(t(1-t)) \frac{1}{2} dt = \frac{1}{2} \int_{-1}^2 t f(t(1-t)) dt$. Thus,$2I_1 = \int_{-1}^2 t f(t(1-t)) dt$. Using the property $\int_a^b g(t) dt = \int_a^b g(a+b-t) dt$,we have $2I_1 = \int_{-1}^2 (1-t) f((1-t)(1-(1-t))) dt = \int_{-1}^2 (1-t) f((1-t)t) dt$. Expanding this,$2I_1 = \int_{-1}^2 f(t(1-t)) dt - \int_{-1}^2 t f(t(1-t)) dt$. Substituting $I_2 = \int_{-1}^2 f(t(1-t)) dt$ and $2I_1 = \int_{-1}^2 t f(t(1-t)) dt$,we get $2I_1 = I_2 - 2I_1$. Therefore,$4I_1 = I_2$,which implies $\frac{I_2}{I_1} = 4$.

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

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