Question

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

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

Answer 1

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

$I={\int }_0^{1}x(1-x{)}^{n}dx$
$I={\int }_0^{1}x(1-x{)}^{n}dx={\int }_0^1(1-x\left)\right(x^n)dx$
$={\int }_0^1(x^n-x^{n+1})dx$
$={\int }_0^1{x}^{n}dx-{\int }_0^1{x}^{x+1}dx$
$=\frac{x^n}{n+1}{|}_0^1-\frac{x^{n+1}}{n+2}{|}_0^1$
$=\frac{1}{n+1}-\frac{1}{n+2}$
$=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}$