<s>[INST] 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
 [/INST] 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. 
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