Question

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

• 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 $\frac{1}{n+1}-\frac{1}{n+2}$

${\int }_0^{1}x{(1-x)}^{n}dx$
Substituting $x={\sin }^2\theta$
$dx=2\sin \theta \cos \theta d\theta$ and $x=0$, $\theta =0$, $x=1$, $\theta =\pi /2$
$\therefore$ ${\int }_0^{1}x{(1-x)}^{n}dx={\int }_0^{\pi /2}{\sin }^2\theta {\cos }^{2n}\left(2\sin \theta \cos \theta \right)d\theta$
$2{\int }_0^{\frac{\pi }{2}}{\sin }^3\theta {\cos }^{2n+1}\theta d\theta$
Using ${\int }_0^{\pi /2}{\sin }^{2n+1}\theta {\cos }^{2n+1}d\theta =\frac{[\left(2n\right)(2n-2)...2][\left(2n\right)(2n-2)...2]}{(4n+2)\left(4n\right)(4n-2)...2}$
$\therefore$ $2{\int }_0^{\pi /2}{\sin }^3\theta {\cos }^{2n+1}\theta d\theta =\frac{2[2\times \left(2n\right)(2n-2)(2n-4)...4.2]}{(2n+4)(2n+2)\left(2n\right)(2n-2)...4.2}$
$=\frac{2\times 2\times 1}{(2n+4)(2n+2)}=\frac{1}{(n+2)(n+1)}$
$=\frac{1}{n+1}-\frac{1}{n+2}$ (by partial fraction)