TITLE: Integral equation
QUESTION [10 upvotes]: For any $f:\mathbb{R}\rightarrow \mathbb{R}$ is a continuous function, prove that $\int_0^4f(x(x-3)^2)dx=2\int_1^3f(x(x-3)^2)dx$. I tried many things, but nothing worked out. I observed that $g(x)+g(4-x)=4$ where $g(x)=x(x-3)^2$. Hence this action $x \rightarrow 4-x$ acts in some peculiar way. Please, any help would be great.

REPLY [2 votes]: This is just a follow-up on @DavidK's answer. Since it seems very hard to compute the inverses of $g(x)=x(x-3)^2$, I will provide a solution where we don't need to compute them. 
Let $h_1(x),h_2(x),h_3(x)$ be inverses of $g(x)$ on the intervals $[0,1],[1,3],[3,4].$ We know by the monotonicity of $g(x)$ that $g:[0,1]\rightarrow [0,4]$,  $g:[1,3]\rightarrow [0,4]$, $g:[3,4]\rightarrow [0,4]$ are surjective functions, also proving that the domains of definition of $h_1(x),h_2(x),h_3(x)$ is $[0,4]$ for all three of them. 
By David K's answer, it is sufficient to prove that $h_1(y)+h_2(y)+h_3(y)=6$ for all $y \in [0,4]$.
Take $y \in [0,4]$. By the above mentioned surjectivity of $g(x)$, there exists $x_1 \in [0,1], x_2 \in [1,3],x_3\in[3,4]$ with $g(x_1)=g(x_2)=g(x_3)=y$. By the property of inverse functions, we know that $h_i(g(x_i))=x_i$, for all $i=1,2,3.$ Hence $$h_1(y)+h_2(y)+h_3(y)=h_1(g(x_1))+h_2(g(x_2))+h_3(g(x_3))=x_1+x_2+x_3$$
the statement to prove is equivalent of proving that: If $x_1 \in [0,1], x_2 \in [1,3],x_3\in[3,4]$ with $g(x_1)=g(x_2)=g(x_3)$, then $x_1+x_2+x_3=6$. 
Suppose $g(x_1)=g(x_2)=g(x_3)=y$ for some $y\in [0,4]$. Let $P(x)=g(x)-y=x^3-6x^2+9x-y$, a cubic polynomial in $x$. We know by hypothesis that $P(x_1)=P(x_2)=P(x_3)=0$, hence $x_1,x_2,x_3$ are the roots of $P$. By the Vieta's formula, we get $x_1+x_2+x_3=6$, and we are done.