Evaluate $\int\limits_0^1 x(1-x)^n dx$; (when $n \in \mathbb{N}$).

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

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

$\int\limits_0^1 x(1-x)^n dx = \int_0^1 (1-x)(1-(1-x))^n dx$

$\left[ \text{as, } \int\limits_0^a f(x) dx = \int\limits_0^a f(a-x) dx \right]$

$= \int\limits_0^1 (1-x)x^n dx$

$= \int\limits_0^1 x^n dx - \int\limits_0^1 x^{n+1} dx$

$= \frac{1}{n+1} [x^{n+1}]_0^1 - \frac{1}{n+2} [x^{n+2}]_0^1$

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

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