How to arrive at this integral? int_0^1 x^{n+2}(1-x)^n dx+2 int_0^1 x^{n+1}(1-x)^{n+1}dx+int_0^1 x^n(1-x)^{n+2}dx=2I(n+1)+2 int_0^1 x^{n+2}(1-x)^n dx

Jared Lowe 2022-11-05 Answered
How to arrive at this integral?
However, I am not sure how to arrive at this:
= 0 1 x n + 2 ( 1 x ) n d x + 2 0 1 x n + 1 ( 1 x ) n + 1 d x + 0 1 x n ( 1 x ) n + 2 d x = 2 I ( n + 1 ) + 2 0 1 x n + 2 ( 1 x ) n d x
and why the highlighted part is equal to each other? What type of integral is that?
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Answers (1)

Stella Andrade
Answered 2022-11-06 Author has 19 answers
Explanation:
Just to clarify, if symmetry is no the first thing you saw:
0 1 x n + 2 ( 1 x ) n d x = [ t = 1 x d t = d x x = 0 , 1 t = 1 , 0 x = 1 t ] = 0 1 ( 1 t ) n + 2 t n d t = 0 1 ( 1 x ) n + 2 x n d x
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