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 $f(f(x)).$

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: $I=\frac {5}{14}$.
