In the reaction with I$_2$, the oxidation state of sulphur changes from $+2$ to $+2.5$.
In the reaction with Br$_2$, the oxidation state of sulphur changes from $+2$ to $+6$.
Thus, both I$_2$ and Br$_2$ are oxidants, but Br$_2$ is stronger as it increases the oxidation state of sulphur further.
Final Answer:
Bromine is a stronger oxidant than iodine.
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to: