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

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

Answers (1)

Answer:

\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

=\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}

=\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}

=\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]

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

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

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

=\frac{7-6}{42}

=\frac{1}{42}

Posted by

infoexpert21

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