Question

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

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

Answer 1

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

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

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

$I={\int }_0^1(1-x)x^{n}dx={\int }_0^1{x}^{n}dx-{\int }_0^1{x}^{n+1}dx$

$=\frac{\left(x^{n+1}\right)!}{n+1}-\frac{\left(x^{n+1}\right)!}{n+2}=\frac{1}{n+1}-\frac{1}{n+2}$