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)

${\int }_0^{1}x{(1-x)}^{5}dx$
$={\int }_0^1{x}^{(2-1)}.{(1-x)}^{(6-1)}dx$
$=B(2,6)$ $\left[\begin{array}{rl}& \because B(m,n)={\int }_0^1{x}^{m-1}{(1-x)}^{n-1}dx\\ & \because B(m,n)=\frac{|\overline{m}|\overline{n}}{|\overline{m}+|\overline{n}}\end{array}\right]$
$=\frac{|\overline{2}|\overline{6}}{|\overline{(2+6)}}=\frac{|\overline{2}|\overline{6}}{|\overline{8}}$
$=\frac{\mathrm{1.5.4.3.2.1}}{\mathrm{7.6.5.4.3.2.1}}$
$=\frac{1}{42}$