f '(x) = $\frac{192{\mathrm{x}}^3}{2+{\sin }^4\mathrm{\pi x}}\forall \mathrm{x}\in \mathrm{R}$

Clearly f '(x) is increasing in $[\frac{1}{2},1]$

f '(x) $\le$ f '(1) = 96

$\text{f }\text{'}\left(\text{x}\right)\ge \mathrm{f}\text{'}\left(\frac{1}{2}\right)=\text{ 8}$

for maximum, f (x) = y = 96 $(\mathrm{x}-\frac{1}{2})$

for minimum, f (x) = y = 8 $(\mathrm{x}-\frac{1}{2})$

$\underset{1/2}{\overset{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\le \underset{1/2}{\overset{1}{\int }}96(\mathrm{x}-\frac{1}{2})\mathrm{dx}=\text{ 12}$

$=\underset{1/2}{\overset{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\ge \underset{1/2}{\overset{1}{\int }}8(\mathrm{x}-\frac{1}{2})\mathrm{dx}={4{\mathrm{x}}^2-4\mathrm{x}|}_{1/2}^1=0–(\text{1 }–\text{ 2})=\text{ 1}$

$\therefore 1\le \underset{1/2}{\overset{1}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\le 12.$