Finding value of this expression. 
If $f(x)$ is a differentiable function and $f''(x)<0 ~ \forall x \in
 \mathbb R$ then the value of:   $$\left[\dfrac{
 \displaystyle\sum_{r=0}^{n-1}f\left(\frac r n\right)\dfrac 1n + 
 \displaystyle\sum_{r=1}^{n}f\left(\frac r n\right)\dfrac 1n }{2
\displaystyle\int_0^1 f(x)dx}\right] $$ is equal to ? 
(where $[\cdot]$ is the greatest integer function)

My attempt
I tried solving it for $n\rightarrow \infty$ and got the answer as 1. But the answer is given as 0. Also while discussing on this question with my fellow mates I heard the term 'Reimann sum' which I am clueless about. Please help me understanding what is Reimann sum? How does it answer this question? Is it possible to solve this without using the concept of Reimann sum?
 A: If you are given the additional condition that $f(x) > 0$ for all $x$, then indeed the answer is $0$, and there exists a nice pictoral proof using tools I might expect students in a calculus course to have just learned.
Without that condition (or some other condition imposing some sort of regularity), the question isn't well-formed.
Let's give a simple example of how there are multiple possible answers, using three closely related functions:
$$\begin{align}
f(x) &= -3x^2 + 2x + 2, \\
g(x) &= -3x^2 + 2x - 2, \\
h(x) &= -3x^2 + 2x.
\end{align}$$
For simplicity, let's look at the case when $n = 1$. Then the numerator of your test problem is equal to
$$ F(0) + F(1)$$
and the denominator is equal to
$$ 2 \int_0^1 F(t) dt,$$
(applied to the function $F$). In this case, we find the following values:
$$ \begin{array}{l|ccc}
F &\text{numerator} &\text{denominator} & \lfloor \frac{\text{num}}{\text{den}} \rfloor \\
\hline
f(x) & 3 & 4 & 0 \\
g(x) & -5 & -4 & 1 \\
h(x) & -1 & 0 & N/A
\end{array}$$
We can make the floor of the ratio $0$, $1$, or not defined at all if the denominator integral is exactly $0$. Thus the question is malformed.

Let us now answer the question I think the exam meant to ask, which is with the additional constraint that $f(x) > 0$ for all $x$. Then in this case, I might rewrite the term that you are looking for as
$$ \left \lfloor
\frac{\sum_{r = 0}^{n} \frac{f(r) + f(r+1)}{2} \frac{1}{n}}{\int_0^1 f(t) dt}
\right \rfloor.$$
The point is that the numerator is exactly the $n$-point trapezoidal approximation of the integral in the denominator, formed by taking $n$ intervals and approximating the integral on that interval by the average of the left and right endpoints of that interval, times the width of the interval.
But as $f''(x) < 0$, this means that for all $t \in [a,b]$, we have that $f(t) > \frac{f(a) + f(b)}{2}$, i.e. the function is concave (sometimes called convex down). Thus every trapezoid in the numerator is a slightly smaller approximation of the integral, and since everything is positive (by my added assumption), the numerator is strictly smaller than the denominator --- and the floor is zero.
