If $I_{1}=\int_{0}^{1}\left(1-x^{50}\right)^{100} d x$ and $I_{2}=\int_{0}^{1}\left(1-x^{50}\right)^{101} d x$ such that $I_{2}=\alpha I_{1}$ then $\alpha$ equals to:

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Definite Integrals

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

and

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

such that

I_{2} = α I_{1}

then

α

equals to:

Answer: (C)

Integrate

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

by parts, differentiating

(1 - x^{50})^{101}

and integrating

d x

: the boundary term vanishes and

I_{2} = 5050 \int_{0}^{1} x^{50} (1 - x^{50})^{100} d x

. Writing

x^{50} = 1 - (1 - x^{50})

in that integral splits it as

I_{1} - I_{2}

, so

I_{2} = 5050 (I_{1} - I_{2})

. Thus

5051 I_{2} = 5050 I_{1}

and

α = \frac{I _{2}}{I _{1}} = \frac{5050}{5051}

JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Single correct option

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