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The value of the integral I=\int_{0}^{1}x(1-x)^{n}dx\; is

  • Option 1)

    \frac{1}{n+2}\;

  • Option 2)

    \; \frac{1}{n+1}-\frac{1}{n+2}\;

  • Option 3)

    \frac{1}{n+1}+\frac{1}{n+2}\;

  • Option 4)

    \; \; \frac{1}{n+1}\;

 

Answers (1)

best_answer

As learnt in concept

Properties of Definite integration -

\int_{a}^{b}f\left ( x \right )dx= \int_{a}^{b}f\left ( a+b-x \right )dx

When \int_{0}^{b}f\left ( x \right )dx= \int_{0}^{b}f\left ( b-x \right )dx

 

- wherein

Put the \left ( a+b-x \right ) at the place of x in f\left ( x \right )

 

I=\int_{0}^{1}x(1-x)^{n}dx=\int_{0}^{1}x^{n}(1-x)dx

Thus we get

I=\int_{0}^{1}x^{n}}(1-x)dx

=\int_{0}^{1}(x^{n}-x^{n+1})dx

=\frac{1}{n+1}-\frac{1}{n+2} 


Option 1)

\frac{1}{n+2}\;

This is incorrect option

Option 2)

\; \frac{1}{n+1}-\frac{1}{n+2}\;

This is correct option

Option 3)

\frac{1}{n+1}+\frac{1}{n+2}\;

This is incorrect option

Option 4)

\; \; \frac{1}{n+1}\;

This is incorrect option

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