Defining The Integral and Riemann Sums

The Area Problem

Suppose that we have a continuous function $f$ on the interval $[a, b]$. We are going to define the definite integral of $f$ on $[a, b]$ in the following manner intuitively through solving the "Area Problem":

\begin{align} \int_a^b f(x) \: dx = \lim_{n \to \infty} \sum_{k=1}^n \: f(x_k^*) \: \Delta x \end{align}

Suppose we want to find the area under the continuous function $f$, on the interval $[a, b]$. We denote this with a symbol known as the integral $\int dx$, and we say we integrate a function $f$ from the lower bound $a$ to the upper bound $b$ with the notation $\int_a^b f(x) \: dx$. So we want to calculate $A = \int_a^b f(x) \: dx$ for a function $f$ such as this one:


The pink region signifies the area which we are trying to calculate. Of course, calculating the area under the curve would be rather difficult geometrically. Calculating areas of shapes such as squares, triangles, or even circles is rather easy as we have simply formulas for them. For curves it is more difficult because the curvature of the curve is not necessarily constant. We can try to estimate the area under the curve though using rectangles though. For example, let's estimate the area under the curve using two rectangles. We will take the interval $[a, b]$, and divide it into two chunks, that is, each rectangle will have a width of $\frac{b -a}{2}$. For the heights of the rectangles, we will use use the right endpoints of these divided intervals and evaluate them under the function. We will represent these divisions by the letter $k$. The first division is $k = 1$, and the second division is $k = 2$. Hence the function $f$ evaluated at $a + \frac{1(b - a)}{2}$ and $a + \frac{2(b - a)}{2}$ will give us the heights of our rectangles. The picture below describes the formation of our rectangles:


From our diagram, our estimation is clearly quite off. However, suppose we increase the number of rectangles that we divide our interval up into. For example, let's create four rectangles. The width of each rectangle will now be $\frac{b - a}{4}$ while the heights of each rectangle will be $f$ evaluated at $k =1, 2, 3, 4$ or more appropriately, $f$ evaluated at $a + \frac{1(b-a)}{4}$, $a + \frac{2(b-a)}{4}$, $a + \frac{3(b-a)}{4}$, and $a + \frac{4(b-a)}{4}$. Rather, we will say that the height of the $k^{\mathrm{th}}$ rectangle will be $a + \frac{k(b-a)}{4}$ where $k = 1, 2, 3, 4$. Thus our new picture is as follows:


This picture looks a tiny bit more accurate, but let's try to make our estimate even more accurate! Suppose we construct eight rectangles instead with the same criteria. The widths of the rectangles will be $\frac{b-a}{8}$, while the heights will be $f \left (a + \frac{k(b-a)}{8} \right )$ where $k$ is an integer and $1 ≤ k ≤ 8$. Thus our picture is as follows:


Once again the approximation becomes a more accurate estimation of the area under the curve $f$. Now suppose that we had infinite rectangles whose width were thus, infinitely small. We would thus have the exact area under the curve $f$ on the interval $[a, b]$.

Why did we use right endpoints? (Click the link to see additional explanation):

Defining the Definite Integral

We will now construct a formula to determine the area under the function f on the interval $[a, b]$ given $f$ to be a continuous function and $f(x) ≥ 0$ on $[a, b]$.

First, let's determine a general equation for the area of each individual rectangle. The width of each rectangle is specifically:

\begin{align} w = \frac{(b - a)}{n} \end{align}

The start point of each rectangle is therefore $a + \frac{k(b - a)}{n}$ where $k$ is an integer such that $1 ≤ k ≤ n$, and $n$ is the number of rectangles (or number of divisions). Notice that as $k$ increases, the the position of the width of the rectangle changes by the same amount:


So we can expression this as the change in $x$ such that $\Delta x = \frac{b - a}{n}$. We will call this the $x$ incrementation.

We're also going to need to determine an equation for the heights of the rectangles which is simply the function evaluated at each subsequent increment:

\begin{align} h = f \left ( a + \frac{k(b-a)}{n} \right) \quad \mathrm{or} \quad f(a + k \Delta x) \end{align}

Recall that it doesn't matter what point we choose on our subintervals. Hence, we denote the choice of any $k^{th}$ rectangles' $x$-value on the interval containing the rectangle as $x_k^*$. More specifically, $x_k^* \in [x_{k-1}, x_{k+1}]$. Thus we have an equation for the area of any specific rectangle:

\begin{align} \mathbf{Area \: of \: any \: rectangle} = f(x_k^*) \: \Delta x \end{align}

Now we're not looking for the area of just any rectangle. We're looking for the area of a sum of rectangles. This is known as a Riemann Sum. Hence, we can say that an estimation of the area under the curve of $f$ on the interval $[a, b]$ is:

\begin{align} \sum_{k=1}^n f(x_k^*) \: \Delta x \end{align}

We're almost done. Remember, the exact area under the curve $f$ on the interval $[a, b]$ will be defined when we have an infinite number of rectangles. Plugging in infinity for n won't help us, as infinity is indeterminate. Instead, we can use a limit to represent this. As $n \to \infty$, $\sum_{k=1}^n f(x_k^*) \: \Delta x \to \mathbf{Total \: Area}$. Thus we obtain:

\begin{align} \mathbf{Total \: Area} = \lim_{n \to \infty} \sum_{k=1}^n f(x_k^*) \: \Delta x \end{align}

But our integral we defined earlier represents the area under the function on the interval $[a, b]$, hence:

\begin{align} \int_a^b f(x) \: dx = \lim_{n \to \infty} \sum_{k=1}^n \: f(x_k^*) \: \Delta x \end{align}

Example 1

Using $\int_a^b f(x) \: dx = \lim_{n \to \infty} \sum_{k=1}^n \: f(x_k^*) \: \Delta x$, determine the area under the curve for the function $f(x) = (2 - x^2)$ on the interval [0, 2], more precisely: $\int_0^2 (2 - x^2) \: dx$.

Recall that $\Delta x = \frac{b - a}{n}$. Hence:

\begin{align} \Delta x = \frac{[2 - 0]}{n} = \frac{2}{n} \end{align}

We now need to determine which endpoints we will use to define the height of each rectangle. Let's use the right endpoints for all $n$-rectangles for simplicity. We will evaluate our function at $k \Delta x$, or rather, $\frac{2k}{n}$. That is, the heights of our rectangles will be:

\begin{align} f \left (\frac{2k}{n} \right ) = 2 - \left (\frac{2k}{n} \right )^2 \end{align}

Hence if follows that since $\int_a^b f(x) \: dx = \lim_{n \to \infty} \sum_{k=1}^n \: f(x_k^*) \: \Delta x$, then:

\begin{align} \int_0^2 (2 - x^2) \: dx = \lim_{n \to \infty} \sum_{k=1}^n \: \left ( 2 - \left (\frac{2k}{n} \right )^2 \right ) \: \cdot \frac{2}{n} \end{align}

All we have to do now is simplify:

\begin{align} \int_0^2 (2 - x^2) \: dx = \lim_{n \to \infty} \sum_{k=1}^n \: \left ( 2 - \left (\frac{2k}{n} \right )^2 \right ) \: \cdot \frac{2}{n} \\ = \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \left (2 - \left (\frac{2k}{n} \right )^2 \right) \\ = \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \left (2 - \frac{4k^2}{n^2} \right ) \\ = \lim_{n \to \infty} \frac{2}{n} \left (\sum_{k=1}^n 2 - \sum_{k=1}^n \frac{4k^2}{n^2} \right ) \\ = \lim_{n \to \infty} \frac{2}{n} \left ( 2n - \sum_{k=1}^n \frac{4k^2}{n^2} \right ) \\ = \lim_{n \to \infty} \frac{2}{n} \left ( 2n - \frac{4}{n^2}\sum_{k=1}^n k^2 \right ) \\ = \lim_{n \to \infty} \frac{2}{n} \left ( 2n - \frac{4}{n^2} [1^2 + 2^2 + 3^2 + ... + n^2] \right ) \\ = \lim_{n \to \infty} \frac{2}{n} \left (2n - \frac{4}{n^2} \cdot \frac{n(n-1)(2n +1)}{6} \right ) \\ = \lim_{n \to \infty} 4 - \frac{8}{n^3} \cdot \frac{(n-1)(2n +1)}{n^2} \\ = \lim_{n \to \infty} 4 -\frac{8n(n-1)(2n +1)}{6n^3} \\ = \lim_{n \to \infty} 4 - \frac{(8n^2 - 8n)(2n + 1)}{6n^3} \\ = \lim_{n \to \infty} 4 - \frac{16n^3 + 8n^2 - 16n^2 - 8n}{6n^3} \\ = \lim_{n \to \infty} 4 - \lim_{n \to \infty} \frac{16n^3 - 8n^2 - 8n}{6n^3} \\ = 4 - \lim_{n \to \infty} \frac{16n^3 - 8n^2 - 8n}{6n^3} \\ = 4 - \lim_{n \to \infty} \frac{\frac{16n^3}{n^3} - \frac{8n^2}{n^3} - \frac{8n}{n^3}}{\frac{6n^3}{n^3}} \\ = 4 - \lim_{n \to \infty} \frac{16 - \frac{8}{n} - \frac{8}{n^2}}{6} \\ = 4 - \frac{16}{6} \\ = 4 - \frac{8}{3} \\ = \frac{4}{3} \end{align}

In our simplification, we used a little rule such that:

\begin{align} 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n-1)(2n+1)}{6n} \end{align}

This can be derived in many ways and can also be proven with induction. There are many other rules that appear and we will not state them here.

Nevertheless, we see that the area on the interval $[0, 2]$ for the function $f(x) = (2 - x^2)$ is $\frac{4}{3}$.

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