Question

Let $f\left(x\right)=x^3-\frac{3x^2}{2}+x+\frac{1}{4}$. Find the value of ${\left({\int }_{1/4}^{3/4}f\left(f\left(x\right)\right)dx\right)}^{-1}$

Answer 1

Given $f\left(x\right)=x^3-\frac{3x^2}{2}+x+\frac{1}{4}=\frac{1}{4}(4x^3-6x^2+4x+1)$

$=\frac{1}{4}\cdot (4x^3-6x^2+4x-1+2)$

$f\left(x\right)=\frac{1}{4}[x^4-(1-x{)}^4]+\frac{2}{4}$

$f(1-x)=\frac{1}{4}\left[\right(1-x{)}^4-x^4]+\frac{2}{4}$

$\therefore f\left(x\right)+f(1-x)=\frac{2}{4}+\frac{2}{4}=1$

Replacing $x$ by $f\left(x\right)$

we have $f\left[f\right(x\left)\right]+f[1-f(x\left)\right]=1$

$\mathrm{Now}I={\int }_{1/4}^{3/4}f\left(f\right(x\left)\right)dx.....\left(3\right)$

Also,

$I={\int }_{1/4}^{3/4}f\left(f\right(1-x\left)\right)dx={\int }_{1/4}^{3/4}f(1-f(x\left)\right)dx........\left(4\right)$

$\mathrm{adding}\left(3\right)$ and $\left(4\right)$

$2I={\int }_{1/4}^{3/4}\left[f\right(f\left(x\right))+f(1-f\left(x\right)\left)\right]dx={\int }_{1/4}^{3/4}dx$

$2I=\frac{1}{2}$

$I=\frac{1}{4}$

$\therefore I^{-1}=4$