Let $y= f(x)$ be a monotonic function then area bounded between $y=f(x),x=a,x=b$ and $y=c$ is least for $c=f\left(\frac{a+b}{2}\right)$ where $a<b$
My working:
WLOG let $f$ be increasing and $c=f(t)$
$A(t)=\int_a^tf(t)-f(x)dx+\int_t^bf(x)-f(t)dx$ $\implies A'(t)=2f'(t)\left(t-\frac{a+b}{2}\right)$ $\implies \text{for } t=\frac{a+b}{2}, A(t)$ is minimum.
Here I used $f$ is differentiable. Which is not given the problem statement. Can we do it in some other way or differentiability is required to prove it?
There is not need for $f$ to be differentiable. If you look at the picture (sorry for the ugly picture) you can notice that, for example, if $t>\frac{a+b}{2}$ then when you calculate $A(t)-A\left(\frac{a+b}{2}\right)$, you add the red and yellow areas and you subtract the blue and green ones. And obviously:
$$\text{Red}+\text{Yellow}\ge\text{Red}=\frac{b-a}{2}\left(f(t)-f\left(\frac{a+b}{2}\right)\right)$$
$$\text{Blue}+\text{Green}\le\text{Blue}+\text{Green}+\text{Yellow}=\frac{b-a}{2}\left(f(t)-f\left(\frac{a+b}{2}\right)\right)$$
So $A(t)-A\left(\frac{a+b}{2}\right)\ge 0$. The case $t<\frac{a+b}{2}$ is similar.