# 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:
$\begin{array}{rl}\phantom{=}& {\int }_{0}^{1}{x}^{n+2}\left(1-x{\right)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx+2{\int }_{0}^{1}{x}^{n+1}\left(1-x{\right)}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dx+{\int }_{0}^{1}{x}^{n}\left(1-x{\right)}^{n+2}\phantom{\rule{thinmathspace}{0ex}}dx\\ =& 2I\left(n+1\right)+2{\int }_{0}^{1}{x}^{n+2}\left(1-x{\right)}^{n}\phantom{\rule{thinmathspace}{0ex}}dx\end{array}$
and why the highlighted part is equal to each other? What type of integral is that?
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Explanation:
Just to clarify, if symmetry is no the first thing you saw:
${\int }_{0}^{1}{x}^{n+2}\left(1-x{\right)}^{n}dx=\left[\begin{array}{rl}t& =1-x\\ -dt& =dx\\ x& =0,1↔t=1,0\\ x& =1-t\end{array}\right]={\int }_{0}^{1}\left(1-t{\right)}^{n+2}{t}^{n}dt={\int }_{0}^{1}\left(1-x{\right)}^{n+2}{x}^{n}dx$