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If [x] stands for the greatest function, the value of  \int_{4}^{10}\frac{\left [ x^{2} \right ]dx}{\left [x^{2}-28x+196 \right ]+\left [x^{2} \right ]}  is

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    3

  • Option 4)

    none of these

 

Answers (1)

As we learnt

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 )

 

 Applying the property

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

we have

I=\int_{4}^{10}\frac{\left [ x^{2} \right ]}{\left [ \left ( 14-x \right )^{2} \right ]+\left [ x^{2} \right ]}dx=\int_{4}^{10}\frac{\left [ 14-x \right ]^{2}}{\left [ x^{2} \right ]+\left [ \left ( 14-x \right )^{2} \right ]}dx

\therefore 2I=\int_{4}^{10}dx=6\: \: or\: \: I=3


Option 1)

0

Option 2)

1

Option 3)

3

Option 4)

none of these

Posted by

Vakul

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