3
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $$\int_0^2f(-x^3+3x^2+3x+1)dx\le \frac{1}{3} \int_0^2f(g(x))g'(x)dx=\frac{1}{3}\int_1^{11} f(t)dt=2 $$

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.