If $$f\left( {\frac{1}{x}} \right) + {x^2}f\left( x \right) = 0,\,x > 0,$$       and $$I = \int_{\frac{1}{x}}^x {f\left( z \right)dz,\,\frac{1}{2} \leqslant x \leqslant 2,} $$       then $$I$$ is :

Solution :
$$\eqalign{ & {\text{Put }}z = \frac{1}{t} \cr & {\text{Then }}I = \int_x^{\frac{1}{x}} {f\left( {\frac{1}{t}} \right).\frac{{ - 1}}{{{t^2}}}\,} dt \cr & = \int_x^{\frac{1}{x}} {\left\{ { - {t^2}.f\left( t \right)} \right\}.\frac{{ - 1}}{{{t^2}}}dt} \,\,\left( {{\text{from the question}}} \right) \cr & = \int_x^{\frac{1}{x}} {f\left( t \right)dt} \cr & = - \int_{\frac{1}{x}}^x {f\left( t \right)dt} \cr & = - I \cr & \therefore \,\,2I = 0 \cr} $$