JEE Main 2024 (Online) 8th April Morning ShiftQuestion 1
The value of $k\in\mathbb{N}$ for which the integral $I_n=\int_{0}^{1}\left(1-x^{k}\right)^{n}dx$, $n\in\mathbb{N}$, satisfies $147I_{20}=148I_{21}$ is
Definite Integration
Solution
Concept. Integrate $I_n$ by parts to get a relation between $I_n$ and $I_{n-1}$. Formulas used. (1) $I_n=nk\big(I_{n-1}-I_n\big)$, i.e. $I_n=\dfrac{nk}{1+nk}\,I_{n-1}$Step 1 - Derive (1). By parts with $dv=dx$: $$I_n=\Big[x(1-x^k)^n\Big]_0^1+nk\int_0^1x^k(1-x^k)^{n-1}dx.$$ The bracket is $0$; write $x^k=1-(1-x^k)$ to get $nk\big(I_{n-1}-I_n\big)$. Step 2 - Use it for $n=21$. $I_{21}=\dfrac{21k}{1+21k}\,I_{20}$. Step 3 - Solve. $$147I_{20}=148\cdot\frac{21k}{1+21k}I_{20}\ \Longrightarrow\ 147+147\cdot21k=148\cdot21k\ \Longrightarrow\ 21k=147.$$ So $k=7$. Answer: (D)