I=∫011⋅(1−x10)20dx x10=t x=t1/10 dx=101(t)−9/10dt I=∫01(1−t)20101(t)−9/10dt I=101∫01t−9/10(1−t)20dt a=101b=101c=21
JEE Main 2024 — Mathematics Calculus
Let β(m,n)=∫01xm−1(1−x)n−1 dx, m,n>0. If ∫01(1−x10)20 dx=a×β(b,c), then 100(a+b+c) equals____
Held on 5 Apr 2024 · Verified 6 Jul 2026.
1021
2120
2012
1120
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