Question

The value of the integral I = ${\int }_0^{1}x(1-x{)}^{n}dx$ is

• A. $\frac{1}{n+1}$
• B. $\frac{1}{n+2}$
• C. $\frac{1}{n+1}-\frac{1}{n+2}$
• D. $\frac{1}{n+1}+\frac{1}{n+2}$

Answer 1

The correct option is C $\frac{1}{n+1}-\frac{1}{n+2}$

$l={\int }_0^{1}x(1-x)6ndx$
$l={\int }_0^1(1-x)x^{n}dx$
$(\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$
$l={(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2})}_0^1$
$l=\frac{1}{n+1}-\frac{1}{n+2}$