Integration III: Definite Integrals As Upper and Lower Sums

Let's assume that we are asked to solve the definite integral

<Sign in to see all the formulas>

In other words, we are given a function $f(x)$ and are asked to find the area under it between $x=a$ and $x=b$. Let's also assume that the function we are given, $f(x)$, is a bounded function.

Let's follow the recipe below to approximate the area under the curve by rectangles:

1. Let's break up the $x$-axis between $a$ and $b$ into smaller, nonoverlapping closed intervals. This is discussed fully in this lesson. Let's assume that we do this with $N$ closed intervals denoted by
   

   <Sign in to see all the formulas>

   
   Furthermore, recall that each closed interval has a length of
   

   <Sign in to see all the formulas>

   
   respectively. In the figure below we break the x-axis between the limits of integration $a$ and $b$ into three parts. The partition for this would be <Sign in to see all the formulas>.
   
2. Because the function is bounded, we can determine the minimum value of the function $f(x)$ within each closed interval <Sign in to see all the formulas>. Let's denote the minimum value by $m_i$.
   
3. Because the function is bounded, we can determine a maximum value of the function $f(x)$ within each closed interval <Sign in to see all the formulas>. Let's denote the maximum value by $M_i$.
   
4. For each closed interval <Sign in to see all the formulas> we can find the area of two rectangles. The rectangle from step 2 has an area of <Sign in to see all the formulas> and the rectangle from step 3 has an area of <Sign in to see all the formulas>. Also, note that the rectangle from step 2 has a smaller area than the rectangle from step 3. In other words, <Sign in to see all the formulas>.
   
5. We can add up all the small rectangles from step 4 and call that number
   

   <Sign in to see all the formulas>

   
   and we can add up all the large rectangles from step 4 and call that number
   

   <Sign in to see all the formulas>

   
   We call $U(P)$ the upper sum and $L(P)$ the lower sum.
   

The important thing to note is that no matter what <Sign in to see all the formulas> is (keep in mind it is a number), we know that

<Sign in to see all the formulas>

In other words, the actual area will be somewhere between $L(P)$ and $U(P)$.

It is important to note that the first and easiest thing we can do is to approximate the integral with one rectangle. In this case our partition would be <Sign in to see all the formulas> and we would have

<Sign in to see all the formulas>

where $m$ is the minimum value of the function on $[a,b]$ and $M$ is the maximum value of the function on $[a,b]$. Since $m$ and $M$ are the extreme values of $f(x)$ over $[a,b]$, <Sign in to see all the formulas> is the smallest approximation of the integral and <Sign in to see all the formulas> is the largest approximation of the integral.

Now let's state the definition of the definite integral.

The Definite Integral
The definite integral is a unique number denoted by <Sign in to see all the formulas> such that the inequality <Sign in to see all the formulas> is true for every possible partition of $[a,b]$.

We are sure that you were following this lesson quite nicely until this actual definition. You are probably thinking: "What do I do with this?, "What does this mean?", and "Huh?" We agree that this definition is abstract. The key to understanding the definite integral is in the statement 'every possible partition.' There are two ways to use 'every possible partition':

• Visualize breaking $[a,b]$ into more and more rectangles that get smaller and smaller. As we do this $L(P)$ and $U(P)$ will get closer to the real area. In fact, 'every possible partition' means we can get as close to the real number as we want by making our rectangles smaller and smaller.
• Formulaically create every possible partition and then do some algebra to see if we can find <Sign in to see all the formulas>.

The first bullet point above is a good visual to keep in our heads and the second point is one method for actually finding the integral.

Let's tackle the first bullet point now. Let's say we following the recipe above for a partition that we will call $P_1$. Then we following the recipe again for a partition we call $P_2$. But this time the number of closed intervals we break $[a,b]$ into is greater. For example, $P_1$ could be <Sign in to see all the formulas> and $P_2$ could be <Sign in to see all the formulas>. Let's keep doing this until we have 100 different partitions. For example, <Sign in to see all the formulas>. (Note that our subscript denotes the number of rectangles. $P_2$ has two rectangles and <Sign in to see all the formulas> has one hundred rectangles.) Here is the important point. As we break $[a,b]$ into more and more rectangles, within each interval <Sign in to see all the formulas>, $m_i$ will get closer and closer to $M_i$. As such, for each consecutive partition <Sign in to see all the formulas> we have

<Sign in to see all the formulas>

where $m$ and $M$ are the extreme value of the function over $[a,b]$ defined above. Here is a proof that adding even only one point to a partition can improve our knowledge of the what the number <Sign in to see all the formulas> is. Also, a complete logical discussion of when a definite integral exists can be found here.

Finally, let's formulaically create every possible partition to find the actual integral. Let's immediately jump into an example. Let's try to solve

<Sign in to see all the formulas>

where $C$ is a constant. We already know the answer:

So let's find $L(P)$ and $U(P)$ for any arbitrary partition. Since the minimum and maximum along the function is the same, i.e. <Sign in to see all the formulas> in our recipe above, we find that

<Sign in to see all the formulas>

Since

<Sign in to see all the formulas>

we know that

<Sign in to see all the formulas>

Recall that <Sign in to see all the formulas> so that

<Sign in to see all the formulas>

Notice that you have an <Sign in to see all the formulas> and then <Sign in to see all the formulas> and so on. In fact the sum turns out to be <Sign in to see all the formulas>. But remember our limits of integration are $x_0=1$ and $x_N=4$. Therefore,

<Sign in to see all the formulas>

The point is that we have found the answer to <Sign in to see all the formulas> by choosing an arbitrary partition. We only knew $x_0=1$ and $x_N=4$. We did not specify how many rectangles in the partition or exactly where $x_1$ through <Sign in to see all the formulas> were located on the x-axis. Many integrals problems are simple enough to solve this way, but you quickly get to functions that are too complicated. Don't worry about doing this method all the time because the main technique to solve integrals comes from the Fundamental Theorem of Integral Calculus. In fact, if you are in a business Calculus class or something similar you probably wont' be tested on this. However, being able to visualize an integral as the area of more and more rectangles is extremely important to keep in your head.

Finally, it is important to note that our above definition of an integral does not tell us what <Sign in to see all the formulas> is. Once we have a number we think is <Sign in to see all the formulas> then the definition will allow us to prove it or not. Don't worry too much about this though. As we said in the previous paragraph, we will develop a lot of techniques to help us find what the number <Sign in to see all the formulas> is.