Question

The value of the integral $I=\underset{0}{\overset{1}{\int }}{(1-x)}^{n}dx$ is

Answer 1

${\int }_0^1{\left(1-x\right)}^{n}dx$

$=-{\left[\frac{{\left(1-x\right)}^{n+1}}{n+1}\right]}_0^1$

$=-\left[0-\frac{1}{n+1}\right]$

$=\frac{1}{n+1}$