this is solved example in adg
Q)
Evaluate ∫01(tx+1−x)ndx where n is a positive integer and 't' is a parameter independent of 'x' .
hence show that ∫01xk(1−x)n−kdx=[nCk(n+1)]−1
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Q)
Evaluate ∫01(tx+1−x)ndx where n is a positive integer and 't' is a parameter independent of 'x' .
hence show that ∫01xk(1−x)n−kdx=[nCk(n+1)]−1