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Ishan Dasgupta Samarendra here is the solution
using the property f ( x ) = f ( 2 a − x ) the limit becomes half
2 ∫ 0 2 1 ( x n + ( 1 − x ) n ) n 1 d x ..... ( 1 )
plz see carefully.
for x ∈ ( 0 , 2 1 )
x has smaller value than ( 1 − x ) .
taking common ( 1 − x ) n in ( 1 )
also n → ∞ lim ( 1 − x x ) n → 0
the integral reduces to
2 ∫ 0 2 1 ( 1 − x ) × ( 1 n ) n 1 . d x
2 ∫ 0 2 1 ( 1 − x ) . d x
4 3
hope it helps