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?

• Well... Not really. The numerator is a bunch of Riemann sums which approximate the integral. – Sean Roberson Oct 27 '18 at 16:39
• You are suppose to find the value of this as a function in $n$? Or you need to find the limit as $n$ gets large? – Mason Oct 27 '18 at 16:40
• @Mason The question doesn't mention anything about $n$ which confuses me as well – Jasmine Oct 27 '18 at 16:42
• Where is the problem from? We can dream up functions $f$ where your expression will take different values for different values of $n$. But I think it limits to $1$ as $n$ grows big. – Mason Oct 27 '18 at 16:47
• @Mason The problem was asked in a test – Jasmine Oct 27 '18 at 16:49

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.
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.