Let  $g\left(x\right)=f_n\left(x\right),$ where  $f_2\left(x\right)=f\left(f\left(x\right)\right),f_3\left(x\right)=f\left(f\left(f\left(x\right)\right)\right)\mathrm{....}$ and  $f\left(x\right)=x^3-\frac{3}{2}{x}^2+x+\frac{1}{4}$ then  $\underset{1/4}{\overset{3/4}{\int }}g\left(x\right)dx$ is 

$$\begin{gathered} f\left(x\right)={(x-\frac{1}{2})}^3+\frac{1}{4}(x-\frac{1}{2})+\frac{1}{2} \\ I=\underset{1/4}{\overset{3/4}{\int }}g\left(x\right)dx let x=t+\frac{1}{2}\Rightarrow I=\underset{-1/4}{\overset{1/4}{\int }}g(t+\frac{1}{2})dt \\ f(t+\frac{1}{2})=t^3=\frac{1}{4}t+\frac{1}{2}\Rightarrow f\left(f(t+\frac{1}{2})\right)={(t^3+\frac{1}{4}t)}^3+\frac{1}{4}(t^3+\frac{1}{4}t)+\frac{1}{2} \\ \Rightarrow g(t+\frac{1}{2})=p\left(t\right)+\frac{1}{2} where p(-t)+p\left(t\right)=0 \\ \Rightarrow I=\underset{-1/4}{\overset{1/4}{\int }}\frac{1}{2}dt=\frac{1}{4} \end{gathered}$$