Let f be a positive function. If
I1 = ∫[1−k to k] x f(x(1−x)) dx and I2 = ∫[1−k to k] f(x(1−x)) dx,
where 2k − 1 0, then the ratio I1: I2 is

Question

Let f be a positive function. If
I1 = ∫[1−k to k] x f(x(1−x)) dx and I2 = ∫[1−k to k] f(x(1−x)) dx,
where 2k − 1 > 0, then the ratio I1: I2 is

Accepted Answer

1/2. Substitute x->1-x in I1; the argument x(1-x) is unchanged and the limits swap back, giving I1 = integral of (1-x) f = I2 - I1. Hence 2I1 = I2, so I1:I2 = 1/2.

Let f be a positive function. If I1 = ∫[1−k to k] x f(x(1−x)) dx and I2 = ∫[1−k to k] f(x(1−x)) dx, where 2k − 1 > 0, then the ratio I1: I2 is

Solution

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