Given I1=∫01(1−(1−x4)3)5x3 dx and I2=∫01(1−(1−x4)3)5+1x3 dx then the value of I1I2 is k, then the value of 15−34{k} is,… (Where {.} denotes the fractional part function)……….

Let, 1−x4=t3I1=∫01(1−t3)5dt and 3I2=∫01(1−t3)5+1dt I1I2=3+15+13+15

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Given $I_{1} = \int_{0}^{1} {(1 - {(1 - x^{4})}^{\sqrt{3}})}^{\sqrt{5}} x^{3} d x$ and $I_{2} = \int_{0}^{1} {(1 - {(1 - x^{4})}^{\sqrt{3}})}^{\sqrt{5} + 1} x^{3} d x$ then the value of $\frac{I_{1}}{I_{2}}$ is k, then the value of $\frac{\sqrt{15} - \sqrt{3}}{4 {k}}$ is,… (Where ${.}$ denotes the fractional part function)……….

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answer is 3.

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Detailed Solution

Let, $1 - x^{4} = t$

$3 I_{1} = \int_{0}^{1} {(1 - t^{\sqrt{3}})}^{\sqrt{5}} d t a n d 3 I_{2} = \int_{0}^{1} {(1 - t^{\sqrt{3}})}^{\sqrt{5} + 1} d t \frac{I_{1}}{I_{2}} = \frac{\sqrt{3} + \sqrt{15} + 1}{\sqrt{3} + \sqrt{15}}$

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