Let $f\$be a$\$positive function. Let $I_{1} = \int_{1 - k}^{k}{xf\left\{ x(1 - x) \right\},}\ I_{2} = \int_{1 - k}^{k}{f\left\{ x(1 - x) \right\}}$ $dx\$where $2k - 1 0.$Then$,\ \frac{I_{1}}{I_{2}}$ is

Question

Let $f\$be a$\$positive function. Let $I_{1} = \int_{1 - k}^{k}{xf\left\{ x(1 - x) \right\},}\ I_{2} = \int_{1 - k}^{k}{f\left\{ x(1 - x) \right\}}$ $dx\$where $2k - 1 > 0.$Then$,\ \frac{I_{1}}{I_{2}}$ is

Dr. Abubacker · Accepted Answer

(c)
$I_{1} = \int_{1 - k}^{k}{xf\left\{ x(1 - x) ight\} dx}$
$= \int_{1 - k}^{k}{(1 - x)f\left\lbrack (1 - x)\left\{ 1 - (1 - x) ight\} ightbrack dx(putx = 1 - x)}$
$= \int_{1 - k}^{k}{(1 - x)f\left\{ x(1 - x) ight\} dx}$
$= \int_{1 - k}^{k}{f\left\{ x(1 - x) ight\} dx - \int_{1 - k}^{k}{xf\left\{ x(1 - x) ight\} dx}}$
$= I_{2} - I_{1}$
$	herefore 2I_{1} = I_{2}$
$\Rightarrow \ \ \frac{I_{1}}{I_{2}} = \frac{1}{2}$

Q. Let $f\$be a$\$positive function. Let $I_{1} = \int_{1 - k}^{k}{xf\left\{ x(1 - x) \right\},}\ I_{2} = \int_{1 - k}^{k}{f\left\{ x(1 - x) \right\}}$ $dx\$where $2k - 1 > 0.$Then$,\ \frac{I_{1}}{I_{2}}$ is

Explanation

Reviewed by Dr. Abubacker

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