A) \[\frac{1}{6}\]
B) \[\frac{1}{7}\]
C) \[\frac{6}{7}\]
D) \[\frac{5}{6}\]
E) \[\frac{1}{42}\]
Correct Answer: E
Solution :
\[\int_{0}^{1}{x{{(1-x)}^{5}}dx}\] \[=\int_{0}^{1}{{{x}^{(2-1)}}}.{{(1-x)}^{(6-1)}}dx\] \[=B(2,6)\] \[\left[ \begin{align} & \because B(m,n)=\int_{0}^{1}{{{x}^{m-1}}{{(1-x)}^{n-1}}dx} \\ & \because B(m,n)=\frac{\left| \!{\overline {\, m \,}} \right. \,\left| \!{\overline {\, n \,}} \right. }{\left| \!{\overline {\, m \,}} \right. +\left| \!{\overline {\, n \,}} \right. } \\ \end{align} \right]\] \[=\frac{\left| \!{\overline {\, 2 \,}} \right. \,\,\left| \!{\overline {\, 6 \,}} \right. }{\left| \!{\overline {\, (2+6) \,}} \right. }=\frac{\left| \!{\overline {\, 2 \,}} \right. \,\,\left| \!{\overline {\, 6 \,}} \right. }{\left| \!{\overline {\, 8 \,}} \right. }\] \[=\frac{1.5.4.3.2.1}{7.6.5.4.3.2.1}\] \[=\frac{1}{42}\]
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