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

# If and ${I}_{2}=\alpha {I}_{1}$, then $\alpha$ equals to

1. A

$\frac{5049}{5050}$

2. B

$\frac{5051}{5050}$

3. C

$\frac{5050}{5051}$

4. D

$\frac{5050}{5049}$

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### Solution:

We have,

${i}_{1}={\int }_{0}^{1} {\left(1-{x}^{50}\right)}^{100}dx$ and ${I}_{2}={\int }_{0}^{1} {\left(1-{x}^{50}\right)}^{101}dx$

Now,

$\therefore ={\int }_{0}^{1} {\left(1-{x}^{50}\right)}^{101}dx$

${I}_{2}={\int }_{0}^{1} {\left(1-{x}^{50}\right)}^{100}\left(1-{x}^{50}\right)dx$

$⇒{I}_{2}={\int }_{0}^{1} {\left(1-{x}^{50}\right)}^{100}dx-{\int }_{0}^{1} {x}^{50}{\left(1-{x}^{50}\right)}^{100}dx$

$\begin{array}{l}⇒ {I}_{2}={I}_{1}+\frac{1}{50}{\int }_{0}^{1} {\left(1-{x}^{50}\right)}^{100}\left(-50{x}^{49}\right)dx\\ ⇒ {I}_{2}={I}_{1}+\frac{1}{50}{\int }_{0}^{1} x{\left(1-{x}^{50}\right)}^{100}\frac{d}{dx}\left(1-{x}^{50}\right)dx\\ ⇒ {I}_{2}={I}_{1}+\frac{1}{50}\left[{\left[x\frac{{\left(1-{x}^{50}\right)}^{101}}{101}\right]}_{0}^{1}-{\int }_{0}^{1} \frac{{\left(1-{x}^{50}\right)}^{101}}{101}dx\right\\ ⇒ {I}_{2}={I}_{1}-\frac{1}{5050}{I}_{2}\\ ⇒ \frac{5051}{5050}{I}_{2}={I}_{1}⇒\frac{{I}_{2}}{{I}_{1}}=\frac{5050}{5051}.\end{array}$

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