A) \[\frac{1}{n+1}\]
B) \[\frac{1}{n+2}\]
C) \[\frac{1}{n+1}-\frac{1}{n+2}\]
D) \[\frac{1}{n+1}+\frac{1}{n+2}\]
Correct Answer: C
Solution :
Given, \[l=\int_{0}^{1}{x\,{{(1-x)}^{n}}dx}\] Put \[1-x=z\Rightarrow -dx=dz\] \[l-\int_{1}^{0}{(1-z)\,{{z}^{n}}(-dz)=\int_{0}^{1}{(1-z)\,{{z}^{n}}dz}}\] \[=\int_{0}^{1}{({{z}^{n}}}-{{z}^{n+1}})dz=\left[ \frac{{{z}^{n+1}}}{n+1}-\frac{{{z}^{n+2}}}{n+2} \right]_{0}^{1}\] \[=\frac{1}{n+1}-\frac{1}{n+2}\]
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