Let $f\text{'}\left(x\right)=\frac{192x^3}{2+{\sin }^4\pi x}$ for all $x\in ℝ$with  $f\left(\frac{1}{2}\right)=0.$If $m\le \underset{1/2}{\overset{1}{\int }}f\left(x\right)dx\le M,$ then the possible values of  $m$ and $M$ are

$\begin{array}{l}2\le 2+{\sin }^4\pi x\le 3\\ \frac{1}{3}\le \frac{1}{2+{\sin }^4\pi }\le \frac{1}{2}\\ \frac{192}{3}{x}^3\le f\text{'}\left(x\right)\le \frac{192}{2}{x}^3\end{array}$

$\begin{array}{l}\Rightarrow 64\underset{\frac{1}{2}}{\overset{t}{\int }}{x}^{3}dx\le f\left(t\right)-f\left(\frac{1}{2}\right)\le 96\underset{\frac{1}{2}}{\overset{t}{\int }}{x}^{3}dx\\ \Rightarrow 16\left(t^4-\frac{1}{16}\right)\le f\left(t\right)\le 24\left(t^4-\frac{1}{16}\right)\\ \Rightarrow \underset{\frac{1}{2}}{\overset{1}{\int }}\left(16t^4-1\right)dt\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(t\right)dt\le \underset{\frac{1}{2}}{\overset{1}{\int }}\left(24t^4-\frac{3}{2}\right)dt\\ \Rightarrow {\left[\frac{16}{5}{t}^5-t\right]}_{\frac{1}{2}}^1\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(t\right)dt\le {\left[\hspace{0.17em}24\frac{t^5}{5}-\frac{3t}{2}\right]}_{\frac{1}{2}}^1\\ \Rightarrow \frac{16}{5}\left(1-\frac{1}{32}\right)-\frac{1}{2}\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(t\right)dt\le \frac{24}{5}\left(1-\frac{1}{32}\right)-\frac{3}{2}.\frac{1}{2}\\ \Rightarrow \frac{31}{10}-\frac{1}{2}\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(t\right)dt\le \frac{93}{20}-\frac{3}{4}\\ \Rightarrow 2.6\le \underset{\frac{1}{2}}{\overset{1}{\int }}f\left(t\right)dt\le 3.9\\ \Rightarrow m=1,M=12\end{array}$