If $f\left(\frac{1}{x}\right)+x^2 f(x)=0, x>0$ and $I=\int\limits_{1 / x}^x f(t) d t, \frac{1}{2} \leq x \leq 2$, then $I$ is equal to
Answer & explanation
Correct answer: option 3
We have,
$I=\int\limits_{1 / x}^x f(t) d t$
$\Rightarrow I =-\int\limits_x^{1 / x} f\left(\frac{1}{u}\right) \frac{1}{u^2} d u$, where $t=\frac{1}{u}$
$\Rightarrow I =-\int\limits_x^{1 / x}-\frac{u^2 f(u)}{u^2} d u \left[∵ f\left(\frac{1}{x}\right)=-x^2 f(x)\right]$
$\Rightarrow I =\int\limits_x^{1 / x} f(u) d u$