Let $$f$$ be a positive function. If $${I_1} = \int_{1 - k}^k {xf\left\{ {x\left( {1 - x} \right)} \right\}dx,\,{I_2}} = \int_{1 - k}^k {f\left\{ {x\left( {1 - x} \right)} \right\}dx,} $$           where $$2k - 1 > 0,$$   then $$\frac{{{I_1}}}{{{I_2}}}$$  is :

Solution :
$$\eqalign{ & {I_1} = \int_{1 - k}^k {xf\left\{ {x\left( {1 - x} \right)} \right\}dx} \cr & {\text{Put }}x = 1 - z \cr & {I_1} = \int_k^{1 - k} {\left( {1 - z} \right)f\left\{ {\left( {1 - z} \right)z} \right\}d\left( {1 - z} \right)} \cr & \,\,\,\,\, = \int_{1 - k}^k {f\left\{ {z\left( {1 - z} \right)} \right\}dz} - \int_{1 - k}^k {zf\left\{ {z\left( {1 - z} \right)} \right\}dz} \cr & \,\,\,\,\, = {I_2} - {I_1} \cr & \therefore 2{I_1} = {I_2} \cr} $$