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A continuously differentiable function $f:[0,1]\to[0,1]$ has the properties

(a) $f(0)=f(1)=0$

(b) $f'(x)$ is a non-increasing function of $x$.

Prove that the arc-length of the graph does not exceed $3$.

As I understand from the question that we want to prove that $\int_{0}^{1}f(x)dx <3$.

The first property give the conditions of Rolle's theorem implies that $f'(c)=0$, $c \in(0,1)$.

The second property gives the clue of the max value exists at $c$.

I tried to use first mean value theorem of integral, but no conclusion find.

Is there is any other technique to solve this question?

topeik
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user786
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    Arc-length would be $\int_{0}^{1}\sqrt{1+(f'(x))^2}dx$ – topeik Dec 10 '19 at 03:43
  • How I calculate the upper bound of of $f^{'}(x)$. – user786 Dec 10 '19 at 03:55
  • By Rolle's theorem, $\exists c\in (0,1)$ such that $f'(c)=0$. $f'(x)$ is non-increasing, so $f'(x)\le f'(c)=0 ;\forall x \in (0,1), x\ge c; f'(0)\ge f'(x)\ge f'(c)=0 ;\forall x \in (0,1), x\le c$.

    $\int_0^1 \sqrt{1+(f'(x))^2}dx=\int_0^c \sqrt{1+(f'(x))^2}dx+\int_c^1 \sqrt{1+(f'(x))^2}dx\le \int_0^c \sqrt{1+(f'(0))^2}dx+\int_c^1 dx=c\sqrt{1+(f'(0))^2}+(1-c)$

     – Divide1918 Dec 10 '19 at 04:24
  • I could not think of how to properly bound f'(0) yet, but once I (or anyone else) figured out that part we're done. – Divide1918 Dec 10 '19 at 04:25

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