Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If  $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$
could some help me with this, thanks
 A: As @dxiv says in the comments, we can write $$f (x) = (x-\frac {1}{2})^3 +\frac {1}{4}x +\frac {3}{8} $$ Thus, substituting $u=x- \frac {1}{2}$ gives us $f (u)=u^3+\frac {1}{4}(u+\frac {1}{2}) +\frac {3}{8} = u^3+\frac {1}{4}u +\frac {1}{2} $. Thus, $$f (f (u)) =(u^3+\frac {1}{4}u +\frac {1}{2})^3 +\frac {1}{4}[u^3 +\frac {1}{4}u +\frac {1}{2}] +\frac {1}{2}$$ As $x $ goes from $\frac {1}{4} $ to $\frac {3}{4} $,. $u $ goes from $-\frac {1}{4} $ to $\frac {1}{4} $. 
Hope you can take it from here.
A: Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75) 
Add the two 2I =  int(1) 
I= 0.25. 
A: You can use the property given here:
$$\color{blue}{\int_a^b f(x)dx=\int_a^{(a+b)/2}[f(x)+f(a+b-x)]dx \Rightarrow \\  \int_a^b f(f(x))dx=\int_a^{(a+b)/2}[f(f(x))+f(f(a+b-x))]dx};\\
\int_{1/4}^{3/4} f(f(x))dx=\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\
\int_{1/4}^{1/2}[f(f(x))+f(f(1-x))]dx=\\
\int_{1/4}^{1/2}[1]dx=\frac14,$$
because:
$$f(f(x))=x^3-\frac32x^2+x+\frac14;\\
f(f(1-x))=(1-x)^3-\frac32(1-x)^2+(1-x)+\frac14=
-x^3+\frac32x^2-x+\frac34.$$
Also, the property mentioned by RedFloyd is given in the above source:
$$\color{blue}{\int_a^b f(x)dx=\int_a^b f(a+b-x)dx \Rightarrow \\
\int_a^b f(f(x))dx=\int_a^b f(f(a+b-x))dx}.$$
