Mathematics

If $$\displaystyle f\left ( \frac{1}{x} \right )+x^{2}f\left ( x \right )=0$$ for $$x> 0,$$ 
and $$\displaystyle I=\int_{1/x}^{x}f\left ( z \right )dz, \frac{1}{2}\leq x\leq 2$$ 
then $$\displaystyle I$$ is?


ANSWER

$$0$$


SOLUTION
Substitute $$\displaystyle z=\dfrac { 1 }{ t } $$

$$\displaystyle I=\int _{ x }^{ \dfrac { 1 }{ t }  }{ f\left( \dfrac { 1 }{ t }  \right)  } .\left( \dfrac { -1 }{ { t }^{ 2 } }  \right) dt=\int _{ x }^{ \dfrac { 1 }{ x }  }{ \left( -{ t }^{ 2 }.f\left( t \right)  \right)  } \left( \dfrac { -1 }{ { t }^{ 2 } }  \right) $$

$$\displaystyle =\int _{ x }^{ \dfrac { 1 }{ x }  }{ f\left( t \right)  } dt=-\int _{ \dfrac { 1 }{ x }  }^{ x }{ f\left( t \right)  } dt=-I\\ \therefore 2I=0$$
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Single Correct Medium Published on 17th 09, 2020
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