Question
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{1}_{0}\text{x}(1-\text{x})^{\text{n}}\ \text{dx}$

Answer

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

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