JEE Main 2024 (Online) 5th April Evening ShiftQuestion 1
Let $\beta(m,n)=\displaystyle\int_{0}^{1}x^{m-1}(1-x)^{n-1}dx$, $m,n>0$. If $\displaystyle\int_{0}^{1}\left(1-x^{10}\right)^{20}dx=a\times\beta(b,c)$, then $100(a+b+c)$ equals
Definite Integration
Solution
Concept. With $u=x^{10}$ the integral becomes a beta integral. Formulas used. (1) $u=x^{10}$: $x=u^{1/10}$, $dx=\frac1{10}u^{\frac1{10}-1}du$Step 1 - Substitute. By (1), $$\int_0^1(1-x^{10})^{20}dx=\frac1{10}\int_0^1u^{\frac1{10}-1}(1-u)^{20}du=\frac1{10}\,\beta\Big(\frac1{10},21\Big).$$ Step 2 - Read off. $a=\frac1{10}$, $b=\frac1{10}$, $c=21$. Step 3 - Value. $100(a+b+c)=100\cdot21.2=2120$. Answer: (A)