The value of the integral $I=\int_0^1 x(1-x)^n d x$ is
The value of the integral $I=\int_0^1 x(1-x)^n d x$ is
$\frac{1}{n+1}+\frac{1}{n+2}$
$\frac{1}{n+1}$
$\frac{1}{n+2}$
$\frac{1}{n+1}-\frac{1}{n+2}$
Solution
$I=\int_0^1 x(1-x)^n d x$
$-I=\int_0^1-x(1-x)^n d x=\int_0^1(1-x-1)(1-x)^n d x$
$=\int_0^1(1-x)^{n+1} d x-\int_0^1(1-x)^n d x$
$=\left[\frac{(1-x)^{n+2}}{-(n+2)}\right]_0^1-\left[\frac{(1-x)^{n+1}}{-(n+1)}\right]_0^1=\frac{1}{n+2}-\frac{1}{n+1}$
$I=\frac{1}{n+1}-\frac{1}{n+2}$