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 :
\[I=\int_{0}^{1}{x{{(1-x)}^{n}}dx}\] \[-I=\int_{0}^{1}{-x{{(1-x)}^{n}}dx=\int_{0}^{1}{(1-x-1){{(1-x)}^{n}}dx}}\] \[=\int_{0}^{1}{{{(1-x)}^{n+1}}dx-\int_{0}^{1}{{{(1-x)}^{n}}dx}}\] \[=\left[ \frac{{{(1-x)}^{n+2}}}{-(n+2)} \right]_{0}^{1}-\left[ \frac{{{(1-x)}^{n+1}}}{-(n+1)} \right]_{0}^{1}=\frac{1}{n+2}-\frac{1}{n+1}\] \[\Rightarrow I=\frac{1}{n+1}-\frac{1}{n+2}.\]
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