If I1=∫01 1−x50100dx,I2=∫01 1−x50101 and I2=αI1, then α equals to

We have, i1=∫01 1−x50100dx and I2=∫01 1−x50101dxNow,∴=∫01 1−x50101dxI2=∫01 1−x501001−x50dx⇒I2=∫01 1−x50100dx−∫01 x501−x50100dx⇒ I2=I1+150∫01 1−x50100−50x49dx⇒ I2=I1+150∫01 x1−x50100ddx1−x50dx⇒ I2=I1+150x1−x5010110101−∫01 1−x50101101dx⇒ I2=I1−15050I2⇒ 50515050I2=I1⇒I2I1=50505051.

If I1=∫01 1−x50100dx,I2=∫01 1−x50101 and I2=αI1, then α equals to

We have, i1=∫01 1−x50100dx and I2=∫01 1−x50101dxNow,∴=∫01 1−x50101dxI2=∫01 1−x501001−x50dx⇒I2=∫01 1−x50100dx−∫01 x501−x50100dx⇒ I2=I1+150∫01 1−x50100−50x49dx⇒ I2=I1+150∫01 x1−x50100ddx1−x50dx⇒ I2=I1+150x1−x5010110101−∫01 1−x50101101dx⇒ I2=I1−15050I2⇒ 50515050I2=I1⇒I2I1=50505051.

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If $I_{1} = \int_{0}^{1} {(1 - x^{50})}^{100} d x, I_{2} = \int_{0}^{1} {(1 - x^{50})}^{101}$ and $I_{2} = α I_{1}$ , then $α$ equals to

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$\frac{5049}{5050}$

$\frac{5051}{5050}$

$\frac{5050}{5051}$

$\frac{5050}{5049}$

answer is C.

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Detailed Solution

We have,

$i_{1} = \int_{0}^{1} {(1 - x^{50})}^{100} d x$ and $I_{2} = \int_{0}^{1} {(1 - x^{50})}^{101} d x$

Now,

$∴= \int_{0}^{1} {(1 - x^{50})}^{101} d x$

$I_{2} = \int_{0}^{1} {(1 - x^{50})}^{100} (1 - x^{50}) d x$

$\Rightarrow I_{2} = \int_{0}^{1} {(1 - x^{50})}^{100} d x - \int_{0}^{1} x^{50} {(1 - x^{50})}^{100} d x$

$\begin{array}{ll} \Rightarrow I_{2} = I_{1} + \frac{1}{50} \int_{0}^{1} {(1 - x^{50})}^{100} (- 50 x^{49}) d x \\ \Rightarrow I_{2} = I_{1} + \frac{1}{50} \int_{0}^{1} x {(1 - x^{50})}^{100} \frac{d}{d x} (1 - x^{50}) d x \\ \Rightarrow I_{2} = I_{1} + \frac{1}{50} [{[x \frac{{(1 - x^{50})}^{101}}{101}]}_{0}^{1} - \int_{0}^{1} \frac{{(1 - x^{50})}^{101}}{101} d x \\ \Rightarrow I_{2} = I_{1} - \frac{1}{5050} I_{2} \\ \Rightarrow \frac{5051}{5050} I_{2} = I_{1} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{5050}{5051} . \end{array}$