Suppose $f(x)$ is concave down and positive on the interval $(a,b)$ and $f(a) = f(b)$. Suppose $\int_a^b f(x)dx = c$ and $k < c$. How can I find the smallest interval $(a',b')$ such that $\int_{a'}^{b'} f(x)dx = k$?
This question has had me stuck for hours. From graphing it out, I'm pretty sure that the interval $(a', b')$ should be such that $f(a') = f(b')$, and I know I can use the Intermediate Value Theorem and concavity to show that the $a'$ and $b'$ here are unique. Is there some kind of argument I can make with the Mean Value Theorem? Do I need something like the Fundamental Lemma of Calculus of Variations? I'm not sure where to go from here.
Any help would be much appreciated! Thank you.
I think the smallest interval $(a',b')$ for which $\int_{a'}^{b'} f(x)dx = k$ can be found as follows. It's not super-slick, but I think it's correct (with a few straightforward details to add). For simplicity, write $\int_{s}^{t} f(x)dx$ as $I(s,t)$.
First, because the function is concave down and positive on $(a,b)$, $a$ and $b$ are both finite. (Not hard to show, I suspect.)
Then, $f$ is non-negative and concave down on $[a,b]$, so it has exactly one maximum on $[a,b]$. Let $f(c)=M$ be the maximum.
Case a: ($c=a$). In this case, the sought interval is $(a,b')$ for which $I(a,b') = k$. (There is a unique $b'$ for which this is true, since $I(a,b')$ is an increasing function of $b'$.) No smaller interval works, because $I(a+\epsilon,b'+\epsilon) < k$ because $f$ is decreasing on $(a,b)$, and a smaller interval with integral $k$ would lie within some interval of the form $(a+\epsilon,b'+\epsilon)$ and have an integral even smaller than $I(a+\epsilon,b'+\epsilon)<k$.
Case b: ($c=b$). By the same argument, the sought interval is $(a',b)$ for which $I(a',b') = k$
Case c: ($a<c<b$). Without loss of generality, suppose that $M=f(c) > f(a) \ge f(b)$.
Now find a friend and meet her at the point $(c,M)$ on the graph. Face away from each other and walk along the graph away from $(c,M)$ in opposite directions with constant rate $dy\over dt$ so your $y$-coordinates at any moment are equal. You walk left, she walks right, and at any moment before either of you hits $a$ or $b$, your $x$-coordinates are $c-Y$ and $c+H$, where $f(c-Y)=f(c+H)$. Pay attention to $I(c-Y,c+H)$, which is an increasing function. If $I(c-Y,c+H)$ reaches $k$ while you are both within $(a,b)$, you have reached the desired solution interval. (No smaller interval works by an argument similar to the one above).
If you reach $x=a$ (this will happen before she reaches $x=b$, because we assumed $f(a)\ge f(b)$ and you won’t both reach the interval endpoints simultaneously because $I(c-Y,c+H)>k$), stop, but let your friend keep walking until $I(a,c+H)=k$.
Since your friend walked longer than you, $f(c+H) < f(a)$, and a “no smaller interval with integral $k$” argument is again similar to above.