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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 $f(f(x)).$

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    $\begingroup$ The answer is 5/14 $\endgroup$
    – Ferca
    Jun 10 at 11:30
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    $\begingroup$ That's the right answer but how did you get it $\endgroup$ Jun 10 at 11:32
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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 $\dfrac{5}{14}$.

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    $\begingroup$ Elegant answer ! Nobody tiried the chnage of variable $f(x)=y$; it would have interesting (joke). Cheers and $\to +1$ $\endgroup$ Jun 10 at 13:06
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    $\begingroup$ I made the change of variable $f(x)=y$. Tedious but funny. $\endgroup$ Jun 10 at 14:16
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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.

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    $\begingroup$ I doubt I would have ever gotten to the right answer with this working, 1.5 hours seems pretty quick actually :) +1 for that! $\endgroup$
    – Math Lover
    Jun 10 at 14:21
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    $\begingroup$ (+1) for this approach! I mean, it is beautiful how you can do this thing totally legit and still get the answer! $\endgroup$ Jun 10 at 18:39
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    $\begingroup$ You saw an opportunity to solve for a polynomial and couldn't resist :) I would expect nothing less. (+1) $\endgroup$ Jun 11 at 4:45
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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: $I=\frac {5}{14}$.

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  • $\begingroup$ Shouldn't the lower limit in the the 4th equation be 0. $\endgroup$
    – Z Ahmed
    Jun 12 at 7:37
  • $\begingroup$ Since I have multiplied by $\frac 12$, it is correct. Otherwise, what you are saying is correct too. In your method, we would not multiply by $\frac 12$. $\endgroup$ Jun 12 at 7:42
  • $\begingroup$ Oh! yes you are right. $\endgroup$
    – Z Ahmed
    Jun 12 at 7:54

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