Question

The value of \(\displaystyle \int_0^1 x(1-x)^9\,dx\) is

• A. \(\frac{1}{110}\)
• B. \(\frac{1}{120}\)
• C. \(-\frac{1}{110}\)
• D. \(-\frac{1}{120}\)

Hint:

Use beta integral: \(\int_0^1 x^{m-1}(1-x)^{n-1}dx=\frac{(m-1)!(n-1)!}{(m+n-1)!}\).

Answer 1

The Correct Option is A

We need to evaluate: \[ \int_0^1 x(1-x)^9\,dx. \] This is a standard beta function integral. The beta function formula is: \[ \int_0^1 x^{m-1}(1-x)^{n-1}\,dx = \frac{(m-1)!(n-1)!}{(m+n-1)!}. \] Now compare: \[ x(1-x)^9=x^1(1-x)^9. \] So, \[ m-1=1 \] and \[ n-1=9. \] Therefore, \[ m=2,\qquad n=10. \] Using the beta function formula: \[ \int_0^1 x(1-x)^9\,dx = \frac{(2-1)!(10-1)!}{(2+10-1)!}. \] \[ = \frac{1!9!}{11!}. \] Now, \[ 11!=11\cdot 10\cdot 9!. \] So, \[ \frac{1!9!}{11!} = \frac{9!}{11\cdot 10\cdot 9!}. \] Cancel \(9!\): \[ =\frac{1}{110}. \] Hence, \[ \int_0^1 x(1-x)^9\,dx=\frac{1}{110}. \]