Definite Integrals

Definite integrals are best introduced by first presenting the problem they are intended to solve: Finding the area under a curve. Consider the diagram below.

Here, we're visualizing the graph of some function ${y = f(x).}$ Along the graph are some points ${x = a}$ and ${x = b.}$ The area under the curve is shaded pink, and is denoted:

The entire expression above is called a definite integral. The function ${f(x)}$ is called the integrand, the points ${a}$ and ${b}$ are called the limits or bounds of the integral, and the ${dx,}$ a differential, indicates that the variable of integration is ${x.}$ Given the bounds ${a}$ and ${b,}$ we say that the integral is over the interval ${[a, b].}$ The interval ${[a,b]}$ is called the interval of integration.

To compute the area, we use the following algorithm:

1. Divide the area into rectangles.
2. Add the areas of the rectangles.
3. Take the limit as the rectangles get infinitely thin.

For example, suppose we had the following curve:

To find the area beneath the curve over the interval ${[a,b],}$ we start by dividing it into rectangles:

Notice that some of the rectangles go beyond the curve. If we summate the area for each of these rectangles as is, the area under the curve would be over the actual area. This is where taking the limit comes in. If we make the rectangles thinner, we reduce any over and under estimation:

At some point, the rectangles are so infinitesimally thin that any excess area is neglibile. With this idea in mind, let's consider an example. Say we had the function:

We'll start at ${a = 0}$ and let ${b}$ be arbitrary. Visualizing the area beneath the curve as consisting of rectangles:

Suppose there are ${n}$ rectangles. We know that the area of rectangle is given by the formula:

If there are ${n}$ rectangles, then the base length is given by:

Because the base has a length ${b/n,}$ we can determine the height by passing ${b/n}$ into the function:

The sum of the areas is thus:

Notice the pattern of the terms. The coefficients go ${1,}$ ${2,}$ ${3,}$ and so on, up to ${n.}$ The formula reduces to:

Just to ensure the pattern is clear, we'll write ${1}$ as ${1^2:}$

Question: How big is this quantity:

Since we're working with rectangles, a good starting point is to try and build a pyramid with a base of ${n \times n.}$

In the diagram above, we have two views of the pyramid, a top view and a profile view. In the profile view, the slope of the dashed red line is ${2.}$ It follows that the next layer is ${(n-1) \times (n-1).}$ This means the total number of blocks at the base is ${n^2,}$ the rightmost term in our long expression:

If we examine the term a little closer, we see that the term before ${n^2}$ is ${(n-1)^2:}$

Corresponding to the total number of blocks for the next layer, ${(n-1)^2.}$ The volume of a pyramid with a base of side lengths ${n}$ and a height of ${n}$ is given by the formula:

This tells us the following fact:

This new fact (in red) corresponds to the inner pyrmaid, and in turn corresponds to a lower bound. We can also capture the outer upper bound (the volume of the pyramid denoted by the green dashed line in the diagram) with the formula:

Thus, we have:

At this point, we're ready to take the limit. The original expression:

We want to rewrite this expression as:

Going back to our inequality, we divide by ${n^3:}$

Reducing, we get:

As ${n}$ goes to infinity, we see that:

Hence:

It follows that the total area under ${x^2}$ is:

The graph of ${f(x) = x^2:}$

Summation Notation

With our first example done, we introduce another new bit of notation. Writing:

$$1 + 2 + 3 + \ldots + n$$

is tedious. Accordingly, mathematicians use summation notation to express the idea of summing the terms in a sequence of numbers:

$$\sum\limits_{i = 1}^{n} a_i = a_1 + a_2 + \ldots + a_n$$

Thus, the inequality:

$$\dfrac{1}{3} < \dfrac{1^2 + 2^2 + 3^2 + \ldots + n^2}{n^3} < \dfrac{1}{3} \left( 1 + \dfrac{1}{n} \right)^3$$

can be more concisely written as:

$$\dfrac{1}{3} < \dfrac{1}{n^3} \sum\limits_{i = 1}^{n} i^2 < \dfrac{1}{3} \left( 1 + \dfrac{1}{n} \right)^3$$

What we showed, then, was that:

$$\lim\limits_{n \to \infty} \dfrac{1}{n^3} \sum\limits_{i = 1}^{n} i^2 = \dfrac{1}{3}$$

Similarly, we can write the expression:

$$\left( \dfrac{b}{n} \right) \left( \dfrac{b}{n} \right)^2 + \left( \dfrac{b}{n} \right) \left( \dfrac{2b}{n} \right)^2 + \ldots + \left( \dfrac{b}{n} \right) \left( \dfrac{nb}{n} \right)^2$$

as:

$$\sum\limits_{i = 1}^{n} \dfrac{b}{n} \left( \dfrac{ib}{n} \right)^2$$

This yields the expression:

$$\begin{aligned} \sum\limits_{i = 1}^{n} \dfrac{b}{n} \left( \dfrac{ib}{n} \right)^2 = \dfrac{b^3}{n^3} \sum\limits_{i = 1}^{n} i^2 \end{aligned}$$

Let's consider another example: ${f(x) = x.}$ This function is fairly straightforward in terms of a graph:

From the graph above, we see that the area beneath the curve is a triangle, with sides of length ${b.}$ The area is thus:

$$A = \dfrac{1}{2} b \cdot h = \dfrac{1}{2}b^2$$

Notice that we didn't have to do any of the complex calculations we saw previously. There is, however, an even simpler example: ${f(x) = 1.}$

The area is simply:

$$A = b \cdot h = b \cdot 1 = b$$

With these simpler examples, we now have enough to see a pattern:

${f(x)}$	${\int_0^b f(x)~dx}$
${x^2}$	${\dfrac{b^3}{3}}$
${x}$	${\dfrac{b^2}{2}}$
${1}$	${b}$

Perhaps the pattern isn't so obvious. Let's rewrite the functions with powers:

${f(x)}$	${\int_0^b f(x)~dx}$
${x^2}$	${\dfrac{b^3}{3}}$
${x^1}$	${\dfrac{b^2}{2}}$
${x^0}$	${\dfrac{b^1}{1}}$

Examining this pattern, we might reasonably guess that:

$$\int_0^b x^3~dx = \dfrac{b^4}{4}$$

In a later section, we'll see that this guess is correct.