If $f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}$, then $\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx =\,$? This is a question from a practice workbook for a college entrance exam.

Let
$$f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}.$$
Find
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx.$$

While I know that computing $f(f(x))$ is an option, it is very time consuming and wouldn't be practical considering the time limit of the exam. I believe there must be a more elegant solution.
Looking at the limits, I tried to find useful things about $f(\frac{1}{7}+\frac{6}{7}-x)$
The relation I obtained was that $f(x) + f(1-x) = 12/12 = 1$. I don't know how to use this for the direct integral of $f(f(x)).$
 A: We know,
$ \displaystyle \int_{a}^{1-a}f(f(x))\,dx = \int_{a}^{1-a}f(f(1-x))\,dx$.
So,
$\displaystyle \int_{a}^{1-a}f(f(x))\,dx = \frac{1}{2} \int_a^{1-a}\left[f(f(x))+f(f(1-x)) \right] \ dx$
Now for the given function, observe that $f(x) + f(1-x) = 1 \implies f(1-x) = 1 - f(x)$
So, $f(f(x)) + f(f(1-x)) = f(f(x)) + f(1-f(x)) = 1$
So we have,
$\displaystyle \int_{a}^{1-a}f(f(x))\,dx = \frac{1}{2} (1-2a)$
Here $a = \dfrac{1}{7}$ and that leads to $\dfrac{5}{14}$.
A: In a comment after @Math Lover's elagant answer, I told that nobody tried the change of variable $f(x)=y$. So, I tried for the fun of it
$$I=\int f[f(x)]\,dx=\int \Big[ \frac 1{12}+f(x)-\frac 12 f^2(x)+\frac 13 f^3(x)\Big]\,dx$$
Let $f(x)=y$. Solving the cubic with the hyperbolic method for only one real root
$$x=\frac{1}{2}-\sqrt{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2-4
   y}{\sqrt{3}}\right)\right)$$
$$dx=\frac{4 \cosh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2-4
   y}{\sqrt{3}}\right)\right)}{\sqrt{48 (y-1) y+21}}\,dy$$ So,
$$20480\,I=-10240 \sqrt{3} \sinh (t)+2700 \cosh (2 t)+1350 \cosh (4 t)+15 \cosh (8 t)+12 \cosh   (10 t)$$ where $$t=\frac{1}{3} \sinh ^{-1}\left(\frac{2-4 y}{\sqrt{3}}\right)$$ Now, for $y$, the upper and lower bounds are $\frac{3223}{4116}$ and $\frac{893}{4116}$ and then for $t$, they are $$\mp \frac{1}{3} \sinh ^{-1}\left(\frac{1165}{1029 \sqrt{3}}\right)$$ Because of the symmetry in $t$, all cosines disappear and the result for the definite integral is just
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f[f(x)]\,dx =\sqrt{3} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{1165}{1029 \sqrt{3}}\right)\right)$$ whic, numerically is $0.3571428571428571428571429$; its reciprocal is $2.8=\frac {14}5$ so the value of $\frac 5{14}$.
For those who are curious, this took me close to one and half hour.
A: Firstly, the result you had arrived at before your edit was incorrect. The correct result, as you've now written, would be $$f(x)+f(1-x)=1$$
Substituting $y\to x-\frac 12$ in this relation, we get:
$$f\left(\frac 12 +y\right)+f\left(\frac 12 -y\right)=1$$
Now, let us substitute $t=x-\frac 12$ in the integral. It becomes equivalent to:
$$I=\int_{-\frac {5}{14}}^{\frac {5}{14}} f\left(f\left(t+\frac 12\right)\right) dt$$
Applying Feynman's substitution here, we arrive at:
$$I=\frac 12 \int_{-\frac {5}{14}}^{\frac {5}{14}} f\left(f\left(t+\frac 12\right)\right)+f\left(f\left(\frac 12 -t\right)\right)dt$$
Now, let, for some $y$, $f\left(\frac 12 +y\right)=k$.
Then we have from our obtained relation, $f\left(\frac 12-y\right)=1-k$
This means, $f\left(f\left(\frac 12+y\right)\right)=f(k)$ and $f\left(f\left(\frac 12-y\right)\right)=f(1-k)$.
Adding, we get $$f\left(f\left(\frac 12+y\right)\right)+f\left(f\left(\frac 12-y\right)\right)=f(k)+f(1-k)=1$$
This means that the given integral is: $I=\frac {5}{14}$.
