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The value of the integral,  \int_{3}^{6}\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}\; dx\; \; is

  • Option 1)

    1/2

  • Option 2)

    3/2

  • Option 3)

    2

  • Option 4)

    1

 

Answers (1)

best_answer

As leant 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_{3}^{6}\frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}}dx

Also, we know that

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

Thus, I=\int_{3}^{6}\frac{\sqrt{9-x}}{\sqrt{9-(9-x)+\sqrt{9-x})}}dx

I=\int_{3}^{6}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx

2I=\int_{3}^{6}dx=[x]_{3}^{6}

I=\frac{3}{2}


Option 1)

1/2

This option is incorrect

Option 2)

3/2

This is correct

Option 3)

2

This option is incorrect

Option 4)

1

This option is incorrect

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