Let $f\;$ be apositive function. Let ${I_1} = \mathop \smallint \nolimits_{1 - k}^k \,\,xf\left\{ {x\left( {1 - x} \right)} \right\},\,\,{I_2} = \mathop \smallint \nolimits_{1 - k}^k \,\,f\left\{ {x\left( {1 - x} \right)} \right\}$ $dx\;$ where $2k\,\, - \,\,1 > 0$. Then $\;\frac{{{I_1}}}{{{I_2}}}$ is

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

Let

f\;

be apositive function. Let

{I_1} = \int \nolimits_{1 - k}^k \,\,xf\left\{ {x\left( {1 - x} \right)} \right\},\,\,{I_2} = \int \nolimits_{1 - k}^k \,\,f\left\{ {x\left( {1 - x} \right)} \right\}

dx\;

where

2k\,\, - \,\,1 > 0

. Then

\;\frac{{{I_1}}}{{{I_2}}}

View answer and explanation

Correct option: C

\frac{1}{2}

Option C (1/2) is correct because by using the property ∫abf(x)dx = ∫abf(a+b-x)dx, we can show that I1 = 1/2 * I2, thus I1/I2 = 1/2.