I=∫01x(1−x)ndx −I=∫01−x(1−x)ndx=∫01(1−x−1)(1−x)ndx =∫01(1−x)n+1dx−∫01(1−x)ndx =[−(n+2)(1−x)n+2]01−[−(n+1)(1−x)n+1]01=n+21−n+11 I=n+11−n+21
JEE Main 2003 — Mathematics Calculus
The value of the integral I=∫01x(1−x)ndx is
Held on 30 Apr 2003 · Verified 6 Jul 2026.
n+11+n+21
n+11
n+21
n+11−n+21
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