Question

Let $f(x)=2 x^3-3 x^2-x+\frac{3}{2}$ and $\int\limits_{\frac{1}{8}}^{\frac{7}{8}} f(f(x)) d x=\frac{m}{n}$ (where $m, n$ are relative prime) then value of $(m+n)$ is equal to

Answer 1

$ f(x)=x^3+(x-1)^3-4 x+\frac{5}{2} $
$ f(1-x)=(1-x)^3+(-x)^3-4(1-x)+\frac{5}{2} $
$ \Rightarrow f(x)+f(1-x)=1$
Let $I=\int\limits_{\frac{1}{8}}^{\frac{7}{8}} f(f(x)) d x$
Apply king property
$I=\int\limits_{\frac{1}{8}}^{\frac{7}{8}} f(1-f(x)) d x$
Adding (1) and (2)
$ 2 I= \int\limits_{\frac{1}{8}}^{\frac{7}{8}} f(f(x)) d x+\int\limits_{\frac{1}{8}}^{\frac{7}{8}} f(1-f(x)) d x$
$ 2 I= \int\limits_{\frac{1}{8}}^{\frac{7}{8}} \mathrm{dx}=\frac{3}{4}$
$ \Rightarrow I=\frac{3}{8}$
$m = 3$
$n = 8$
$m + n = 11$