Question

${\int }_0^1{x}^{70}(1-x{)}^{30}dx$ is equal to

• A. $\frac{101}{{}^{100}{C}_{30}}$
• B. $\frac{101}{100\cdot {}^{101}{C}_{30}}$
• C. $\frac{101}{100\cdot {}^{100}{C}_{30}}$
• D. $\frac{1}{101\cdot {}^{100}{C}_{30}}$

Answer 1

The correct option is D $\frac{1}{101\cdot {}^{100}{C}_{30}}$

Let, $x={\sin }^2\theta \Rightarrow dx=2\sin \theta \cos \theta d\theta \therefore I={\int }_0^{\pi /2}\left({\sin }^2\theta {)}^{70}\right(\cos \theta {)}^{60}\cdot 2\sin \theta \cos \theta d\theta =2{\int }_0^{\pi /2}{\sin }^{141}\theta {\cos }^{61}d\theta =2\times \frac{(140\times 138\times \cdots 2)(60\times 58\times \cdots 2)}{(202\times 200\times 198\times \cdots 2)}\times 1\{\because I_{m,n}=\frac{(m-1)(m-3)(m-5)\cdots (n-1)(n-3)(n-5)}{(m+n)(m+n-2)(m+n-4)\cdots }\cdot 1\text{, when}m,n\text{is not even}\}=\frac{2\times (2^{70}\times 70!)\times (2^{30}\times 30!)}{2^{101}\times 101!}=\frac{30!\times 70!}{101\times (100!)}=\frac{1}{101\cdot {}^{100}{C}_{30}}$