Question

The value of the integral $ \int_{0}^{1}{x{{(1-x)}^{5}}dx} $ is equal to

• A. $ \frac{1}{6} $
• B. $ \frac{1}{7} $
• C. $ \frac{6}{7} $
• D. $ \frac{5}{6} $
• E. $ \frac{1}{42} $

Answer 1

Answer: (E)

$ \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} $