Given $f$ is continuous and nonnegative on $[1,11]$ with $\int_{1}^{11}f(x)dx=6$. Show that $\int_{0}^{2}f(-x^3+3x^2+3x+1)dx$ is less or equal to 2.

My Approach
My first instinct was to show that if there exists a partition $P$ such that the lower sum $L(f(-x^3+3x^2+3x+1),P )>2 $, then as $−x^3+3x^2+3x+1$ is one-to-one, we can use the same minimum points to create a $P'$ for $f(x)$, with each closed interval "stretched". If I can show that this lower sum is greater than 6, then I would have a contradiction but I am kinda stuck here. Any hints on how to continue or maybe a new direction? I also noticed that the two functions have the same end point and midpoint, but I haven't figured out if I can use this fact.


Let $g(x):=-x^3+3x^2+3x+1$. Then $g'(x)=-3x^2+6x+3$ and it is easy to see that $g'(x)>0$ on $[0,2]$. So $g(x)$ is strictly increasing there and $g([0,2])=[g(0),g(2)]=[1,11]$. Moreover, It is not dificult to see that $g'(x)\ge 3$ on $[0,2]$ (indeed, $g'(x)$ is a parabola with negative leading coefficient, $g'(0)=g'(2)=3$ so the vertex is on $g(1)=6$). $$\int_0^2f(-x^3+3x^2+3x+1)dx=\int_0^2 f(g(x))dx \le \frac{1}{3} \int_0^2f(g(x))g'(x)dx$$ Now, making substitution $g(x)=t$, one gets $$\int_0^2f(-x^3+3x^2+3x+1)dx\le \frac{1}{3} \int_0^2f(g(x))g'(x)dx=\frac{1}{3}\int_1^{11} f(t)dt=2 $$

  • 1
    $\begingroup$ Thanks a lot! I really like this approach of kind of reversing the chain rule! There are a couple of typos, the more important one being $g'(0)=g'(2)=3$ $\endgroup$
    – David Zhu
    Dec 4 '20 at 19:05
  • $\begingroup$ I'll correct in a moment. $\endgroup$ Dec 4 '20 at 19:08

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