CUET Preparation Today
The integral $\int\limits_0^1 x(1-x)^{n} d x$ is equal to : |
$\frac{1}{(n+2)(n+3)}$ $\frac{1}{(n+1)(n+2)}$ $\frac{1}{n(n+1)}$ $\frac{1}{(n-1)(n-2)}$ |
$\frac{1}{(n+1)(n+2)}$ |
The correct answer is Option (2) - $\frac{1}{(n+1)(n+2)}$ $I=\int\limits_0^1 x(1-x)^{n} dx$ $=\int\limits_0^1 (1-x)(x)^ndx$ $=\int\limits_0^1x^n-x^{n+1}dx=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_0^1$ $=\frac{1}{(n+1)(n+2)}$ |
