Question

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

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int_0^1\rm x {(1 - x)}^{n} \: dx$

We know,

$\boxed{ \rm{ \:\displaystyle\int_0^a\rm f(x)dx \: = \: \displaystyle\int_0^a\rm f(a - x) \: dx \: }}$

So, using this property, we get

$\rm \: = \: \displaystyle\int_0^1\rm (1 - x) {[1 - (1 - x)]}^{n} \: dx$

$\rm \: = \: \displaystyle\int_0^1\rm (1 - x) {x}^{n} \: dx$

$\rm \: = \: \displaystyle\int_0^1\rm ({x}^{n} - {x}^{n + 1} ) \: dx$

$\rm \: = \: \bigg[\dfrac{ {x}^{n + 1} }{n + 1} - \dfrac{ {x}^{n + 2} }{n + 2} \bigg]_0^1$

$\rm \: = \: \dfrac{1}{n + 1} - \dfrac{1}{n + 2}$

$\rm \: = \: \dfrac{n + 2 - n - 1}{(n + 1)(n + 2)}$

$\rm \: = \: \dfrac{1}{(n + 1)(n + 2)}$

Hence,

$\rm\implies \:\boxed{ \rm{ \: \displaystyle\int_0^1\rm x {(1 - x)}^{n} \: dx = \frac{1}{(n + 1)(n + 2)} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:\displaystyle\int_a^b\rm f(x)dx \: = \: \displaystyle\int_a^b\rm f(y)dy \: }}$

$\boxed{ \rm{ \:\displaystyle\int_a^b\rm f(x)dx \: = \: - \: \displaystyle\int_b^a\rm f(x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_a^b\rm f(x)dx \: = \: \displaystyle\int_a^b\rm f(a + b - x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^a\rm f(x)dx \: = \:2 \displaystyle\int_0^a\rm f(x)dx \: if \: f( - x) = f(x)}}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^a\rm f(x)dx \: = \:0\: \: if \: f( - x) = - f(x)}}$

Answer 2

Step-by-step explanation:

$\begin{array}{l} \text{Let } \displaystyle\rm I =\int_{0}^{1} x(1-x)^{n} d x \\ \\ \boxed{\pmb {\text{ Using } \displaystyle\rm P_{4}: \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x}} \\ \\ \displaystyle\rm \therefore \quad I =\int_{0}^{1}(1-x)[1-(1-x)]^{n} d x \\ \\ \displaystyle\rm I =\int_{0}^{1}(1-x)[1-1+x]^{n} d x \\ \\ \displaystyle\rm I =\int_{0}^{1}(1-x)[x]^{n} d x \\ \\ \displaystyle\rm I =\int_{0}^{1}(1-x) x^{n} d x \end{array}$

$\[ \begin{array}{l} \displaystyle\rm \[ I =\int_{0}^{1}\left(x^{n}-x^{n+1}\right) d x \] \\ \\ \displaystyle\rm I =\int_{0}^{1} x^{n} d x-\int_{0}^{1} x^{n+1} d x \\\\ \displaystyle\rm I =\left[\frac{x^{n+1}}{n+1}\right]_{0}^{1}-\left[\frac{x^{n+2}}{n+2}\right]_{0}^{1} \\\\ \displaystyle\rm I =\left[\frac{(1)^{n+1}}{n+1}-\frac{(0)^{n+1}}{n+1}\right]-\left[\frac{(1)^{n+2}}{n+2}-\frac{(0)^{n+2}}{n+2}\right] \\\\ \displaystyle\rm I =\frac{1}{n+1}-\frac{1}{n+2} \\\\ \displaystyle\rm I =\frac{n+2-(n+1)}{(n+1)(n+2)} \\\\ \color{darkcyan}\boxed{\pmb{ \displaystyle\rm I =\frac{1}{(n+1)(n+2)}}} \end{array} \]$