What Are Riemann Sums?

Riemann Sums are very similar to Darboux Sums, but defined slightly differently.

The basic idea for Riemann Sums is the same as that for Darboux sums: fitting rectangles under or over a curve.

However, the definition uses not the maximum or minimum value within each interval, but any arbitrary constant rule.

Formally, when we partition the interval $[a,b]$ (just like we did last time) into $\{x_0,x_1,\ldots x_n\}$, the Riemann sum is defined as

$$\sum_{i=1}^{n}f(c_i)(x_i-x_{i-1})$$where $c_i$ is an arbitrary constant that satisfies $x_{i-1}\le c_i\le x_i$.

For example, when $c_i=x_{i-1}$, we obtain the Left Riemann Sum. The above diagram is an example of this type of sum. When $c_i=x_i$, we obtain the Right Riemann Sum. This sum, when used on the same curve as in our previous example, looks like this:

Do you see why these are called the Left Riemann Sum and Right Riemann Sum, respectively?

Usually, we partition the interval $[a,b]$ into equal parts. That is, $$x_i-x_{i-1}=x_j-x_{j-1}$$ for all possible integers $i,j$. However, this is not necessary.

See if you can figure out how $c_i$ is defined for these two Riemann sums:

Middle Riemann Sum

$ci$ is defined as $ci=\dfrac{xi+x{i-1}}{2}$. Note that this is because $c_i$ is taken to be the middle of each interval.

Trapezoidal Sum

This one is defined as $ci=\dfrac{f(xi)+f(x{i-1})}{2}$. This can be figured out by recalling the formula for the area of a trapezoid: average of the bases times the height. We already have the height as $(xi-x{i-1})$, so we need $ci$ to be the average of the bases. This is clearly the expression above.

To illustrate the subtle difference between the Riemann sum and the Darboux sum, observe the diagram below, comparing the upper Darboux sum and the Left Riemann sum:

Note that the upper Darboux sum is guaranteed to be greater than the actual integral and the lower Darboux sum is guaranteed to be less, while the types of Riemann sums depend on the shape of the curve.

In the next topic you will learn more about the Left, Right, and Middle Riemann Sums

Here's an exercise about finding Riemann sums. We're going to do Riemann sums on the function $f(x) = x$ for $0 \le x \le 1$.

The triangle is the area that we want to approximate. What is the actual area here?

Good job! Now let's compute a Right Riemann Sum of the area. First let's only draw 2 rectangles.

Good job! We have 2 rectangles, and their heights are $1/2$ and $1$. Their bases are both $1/2$. Therefore, the answer is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot 1 = \frac{3}{4}$.

Note that $3/4 > 1/2$, the actual area. This is expected by our diagram.

Let's divide it into 3 rectangles now.

Great! Since the heights are $1/2, 2/3, 1$ in that order, the total area is $\frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{2}{3} + \frac{1}{3} \cdot {1} = \frac{2}{3}.$

Still, $2/3 > 1/2$, our actual area, but it is a closer estimate that $3/4$ at least. (Because $2/3 - 1/2 < 3/4 - 1/2$.)

The heights would be $1/4, 2/4 = 1/2, 3/4, 1$, so the total area would be $\frac{1}{4} \cdot \frac{1}{4} + \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{4} \cdot \frac{3}{4} + \frac{1}{4} \cdot 1 = \frac{5}{8}.$

Once again, $5/8 > 1/2$ still, but it is a better estimate than either of the 2 above estimates.

In fact, with some series summation techniques you can show that the Right Riemann Sum with $n$ rectangles gives total area $\frac{n+1}{2n}$ (you can check that this holds for $n = 2, 3, 4$ above). This quantity is always greater than $1/2$. But...

What is $\lim_{n \rightarrow \infty} \frac{n+1}{2n}?$ It's just $1/2$, which just happens to be the actual area of the region. See, Riemann sums don't need to have fancy summation notation to work. Instead, just dividing into rectangles and adding areas works very well. :)