$\int\limits_0^2x(2-x)^ndx$ is equal to

$\frac{2^{n+2}}{(n+1)(n+2)}$

The correct answer is Option (3) → $\frac{2^{n+2}}{(n+1)(n+2)}$

$I=\int_{0}^{2} x(2-x)^n dx$

Put $x=2-t$ so that $dx=-dt$

When $x=0,t=2$ and when $x=2,t=0$

$I=\int_{2}^{0} (2-t)t^n(-dt)=\int_{0}^{2} (2-t)t^n dt$

$I=\int_{0}^{2} (2t^n-t^{n+1})dt$

$I=2\int_{0}^{2} t^n dt-\int_{0}^{2} t^{n+1}dt$

$I=2\left[\frac{t^{n+1}}{n+1}\right]_{0}^{2}-\left[\frac{t^{n+2}}{n+2}\right]_{0}^{2}$

$I=2\cdot\frac{2^{n+1}}{n+1}-\frac{2^{n+2}}{n+2}$

$I=\frac{2^{n+2}}{n+1}-\frac{2^{n+2}}{n+2}$

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

$I=2^{n+2}\cdot\frac{(n+2)-(n+1)}{(n+1)(n+2)}$

$I=\frac{2^{n+2}}{(n+1)(n+2)}$

The value of the integral is $\frac{2^{n+2}}{(n+1)(n+2)}$.