Estimating integrals using Riemann sums Let $f$ be a function such that $f''(x)<0$ for $x$ in $(0,1)$.
We are required to find $$\left\lfloor\left(\frac{\sum_{r=1}^n \frac{f(r/n)}{n}+ \sum_{r=0}^{n-1} \frac{f(r/n)}{n}}{2 \int_{0}^{1} f(x) dx }\right)\right\rfloor$$ (i.e. the greatest integer function of this expression).
The answer given is $0$, which implies the numerator is less than the denominator.
My question is: how do we arrive at this answer?
I have two major concerns:


*

*The denominator of the expression gives $2$ times the actual area under the curve,  whereas in the numerator: one of the sums will give an underestimate while the other an overestimate. How do I know about the extent of underestimation/overestimation? If the overestimation compensates for the underestimation we get the answer as $1$, otherwise $0$. Maybe the concavity of the function gives us the clue? I am not very well equipped with this idea.

*Although it doesn't seem relevant to this particular question, but, as far as I know, to know which Riemann sum (left or right) gives the underestimate/overestimate, we need to know about whether the function is increasing or decreasing over the interval. This question, however, only gives us the concavity! And I am certain that a concave down function isn't necessarily decreasing.
 A: Note that
  $$ \frac{\sum_{r=0}^{n-1} \frac{1}{n} f\left(\frac{r}{n}\right) + \sum_{r=1}^{n} \frac{1}{n} f\left( \frac{r}{n} \right)}{2} = \sum_{r=1}^{n} \frac{1}{n}\frac{f\left(\frac{r-1}{n}\right) + f\left(\frac{r}{n}\right)}{2} $$
Now,
  $$ \frac{1}{n}\frac{f\left(\frac{r-1}{n}\right) + f\left(\frac{r}{n}\right)}{2} $$
is the area of the trapezium formed by the vertices
  $$ \left( \frac{r-1}{n}, 0 \right), \left( \frac{r}{n}, 0 \right), \left( \frac{r}{n}, f\left(\frac{r}{n}\right) \right), \left( \frac{r-1}{n}, f\left(\frac{r-1}{n}\right) \right) $$
Crucially, the line connecting $\left( \frac{r-1}{n}, f\left(\frac{r-1}{n}\right) \right)$ to $\left( \frac{r}{n}, f\left(\frac{r}{n}\right) \right)$ lies completely beneath the curve, because the curve is concave. This means that the area, $T_{r}$, of this trapezium has area smaller than
  $$ \int_{\frac{r-1}{n}}^{\frac{r}{n}} f(x) \, dx $$
Therefore
  $$ \sum_{r=1}^{n} T_{r} \leq \sum_{r=1}^{n} \int_{\frac{r-1}{n}}^{\frac{r}{n}} f(x) \, dx = \int_{0}^{1} f(x) \, dx $$
as required.
