# 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.

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 $$\int_0^2f(-x^3+3x^2+3x+1)dx\le \frac{1}{3} \int_0^2f(g(x))g'(x)dx=\frac{1}{3}\int_1^{11} f(t)dt=2$$
• 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 $g'(0)=g'(2)=3$ Dec 4 '20 at 19:05