#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 46 Maths Textbook Solution.

$\frac{1}{42}$

Hint: Use indefinite integral formula then put limits to solve this integral

Given:

$\int_{0}^{1}x\left ( 1-x \right )^{5}dx$

Sol:

$\int_{0}^{1} x(1-x)^{5} d x=\int_{0}^{1} x\left(1-5 x+10 x^{2}-10 x^{3}+5 x^{4}-x^{5}\right) d x$

(By Binomial theorem)

$=\int_{0}^{1}\left(x-5 x^{2}+10 x^{3}-10 x^{4}+5 x^{5}-x^{6}\right) d x$

$=\int_{0}^{1} x d x-5 \int_{0}^{1} x^{2} d x+10 \int_{0}^{1} x^{3} d x-10 \int_{0}^{1} x^{4} d x+5 \int_{0}^{1} x^{5} d x-\int_{0}^{1} x^{6} d x$

$\inline =\left[\frac{x^{1+1}}{1+1}\right]_{0}^{1}-5\left[\frac{x^{2+1}}{2+1}\right]_{0}^{1}+10\left[\frac{x^{3+1}}{3+1}\right]_{0}^{1}-10\left[\frac{x^{4+1}}{4+1}\right]_{0}^{1}+5\left[\frac{x^{5+1}}{5+1}\right]_{0}^{1}-\left[\frac{x^{6+1}}{6+1}\right]_{0}^{1}$

$\inline =\left[\frac{x^{2}}{2}\right]_{0}^{1}-5\left[\frac{x^{3}}{3}\right]_{0}^{1}+10\left[\frac{x^{4}}{4}\right]_{0}^{1}-10\left[\frac{x^{5}}{5}\right]_{0}^{1}+5\left[\frac{x^{6}}{6}\right]_{0}^{1}-\left[\frac{x^{7}}{7}\right]_{0}^{1}$

$\inline =\frac{1}{2}[1-0]-\frac{5}{3}\left[1^{3}-0^{3}\right]+\frac{10}{4}\left[1^{4}-0^{4}\right]-\frac{10}{5}\left[1^{5}-0^{5}\right]+\frac{5}{6}\left[1^{6}-0^{6}\right]-\frac{1}{7}\left[1^{7}-0^{7}\right]$

$\inline =\frac{1}{2}-\frac{5}{3}+\frac{5}{2}-2+\frac{5}{6}-\frac{1}{7}$

$\inline =\frac{3-10+15-12+5}{6}-\frac{1}{7}$

$\inline =\frac{1}{6}(23-22)-\frac{1}{7}$

$=\frac{7-6}{42}$

$=\frac{1}{42}$