Question

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

Answer 1

Answer: 11.00

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