By using the properties of definite integrals, evaluate the integ… - Vidyadip

Question

By using the properties of definite integrals, evaluate the integral $\int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $

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Answer

Let $I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $
$= \int\limits_0^1 {\left( {1 - x} \right)\left\{ {1 - {{\left( {1 - x} \right)}}} \right\}^ndx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx } } } \right]$
$\Rightarrow I = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $
$ = \int\limits_0^1 {\left( {1 - x} \right){x^n}dx} $
$= \int\limits_0^1 {\left( {{x^n} - {x^{n + 1}}} \right)dx} $
$\Rightarrow I = \left( {\frac{{{x^{n + 1}}}}{{n + 1}} - \frac{{{x^{n + 2}}}}{{n + 2}}} \right)_0^1$
$ = \frac{1}{{n + 1}} - \frac{1}{{n + 2}} - \left( {0 - 0} \right)$
$= \frac{{n + 2 - n - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$

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