I want to know if my solution for this problem is correct:
Let $f : [0,\infty ) \to [0,\infty)$ with continous first derivative. Suppose that for all $a,b \in \Bbb{R^+}$, the area between the $a,b$ is equal to the arc length of the function (between $a,b$) . Assuming $f$ has a minimum, find the value of $f$ at the minimum.
What I did :
We know that the integral of the function between $a,b$ is equal to the arc length integral formula. Since $f$ is continous, by Mean Value Theorem for integrals :
$f(c)(b-a) = \sqrt{1 + f’(c)^2}(b-a)$ for some $c \in (a,b)$. Then $f(c) = \sqrt{1 + f’(c)^2}$ .
Since the integral is continuous, be can find $a,b$ such that the $c$ is the minimum. We know that $f’(c)=0$ because $f’$ is continuous and $c$ is where $f$ finds its minimum. We conclude that $f(c)=1$
It’s my solution correct? Thanks.