Welcome back! If you took MAT 1475 at CityTech, the definite integral and the fundamental theorem(s) of calculus were the last two topics that you saw. The fundamental theorems—sometimes people talk about the fundamental theorem, but there are really two theorems and you need both—tell you how indefinite integrals (which you saw in Lesson 1; see link here) and definite integrals (which you’ll see today). This lesson is a refresher.

Lesson 2: The Definite Integral & the Fundamental Theorem(s) of Calculus

Learning goals:

  • Explain the terms integrand, limits of integration, and variable of integration.
  • Describe the relationship between the definite integral and net area.
  • Use geometry and the properties of definite integrals to evaluate them.
  • State the meaning of and use the Fundamental Theorems of Calculus.
  • Explain the relationship between differentiation and integration.

Topic:

WeBWorK:

  • Integration – Definite Integrals
  • Integration – Fundamental Theorem constant bounds
  • Integration – Fundamental Theorem variable bounds

Motivating question

How can we use integrals to find the area of an irregular shape in the plane?

Warmup question 1

Consider the semicircle centered at the point (5,0) and with radius 5 which lies above the x-axis.

  1. Sketch this region in the x-y plane.
  2. Determine the area enclosed by this semicircle.

Show answer 1

Since the area enclosed by a circle of radius r is \pi r^2, the area of a semicircle of radius r is \frac{1}{2} \pi r^2. So we don’t need to know the center to answer the question. We know the radius is 5, so the area enclosed by the semicircle is \frac{1}{2}\pi*5^2 = \frac{25}{2}\pi square units.

Basic definite integrals

First, some important remarks

Before we define what a definite integral is, there are two important things to remember:

  1. For now, you should think of definite integrals and indefinite integrals (defined in Lesson 1, link here) as completely different things. The notation looks similar, but you should not expect them to have anything to do with each other…that is, until we get to the Fundamental Theorems of Calculus later in today’s lesson!
  2. We will define the definite integral differently from how your textbook defines it. Take only a quick look at Definition 1.8 in the text (link here). We will not make sense of that definition until we cover Riemann sums in Lesson 21: Approximating areas (link here).

Let’s get started!

Definition: A definite integral is a signed area. That is, if a function f(x) is defined on a closed interval [a,b], then the definite integral \int_a^b f(x) dx is defined as the signed area of the region bounded by the vertical lines x=a and x=b, the x-axis, and the graph y=f(x); if the region is above the x-axis, then we count its area as positive and if the region is below the x-axis, we count its area as negative.

We call a the lower limit of integration and b the upper limit of integration. The function f(x) is still called the integrand and x is still called the variable of integration (just like for indefinite integrals in Lesson 1).

A picture is worth a thousand words. The following picture, Figure 1, illustrates the definition of the definite integral. Figure 1 shows the graph of a function f(x) in red and three regions between the graph y=f(x) and the x-axis and between x=-2 and x=1. The blue and purple regions are above the x-axis and the green region is below the x-axis. To determine the value of the definite integral \int_{-2}^1 f(x)dx, we would need to know the areas of the three regions. Let’s call the area of the blue region A_{blue}, the area of the green region A_{green}, and the area of the purple region A_{purple}. Then

    \[\int_{-2}^1 f(x)dx = A_{blue} - A_{green} + A_{purple}.\]

Figure 1: A graph of a function f(x) and three shaded regions between it and the x-axis, between x=-2 and x=1

For most irregular shapes, like the ones in Figure 1, we won’t have an easy formula for their areas. We’ll work on that later. For now, we’ll restrict our attention to easier shapes.

Exercise 2

Evaluate the definite integral:

    \[ \int_0^{10} \sqrt{25 - (x-5)^2} dx\]

Show answer 2

If you don’t recognize the shape of the graph of the function right away, this will look more difficult than it actually is. Here’s one way to see why it’s not too bad: write y = \sqrt{25 - (x-5)^2}. Square both sides to get y^2 = 25 - (x-5)^2. Add the last term on the right hand side to both sides to get (x-5)^2+ y^2 = 25. You should recognize this as the equation of a circle with center (5,0) and radius 5. The lowest value of x is 0 and the highest value of x is 5.

Squaring both sides made us forget that our original function is the positive square root, so this means our function encloses the semicircle of radius 5, centered at (5,0), above the x-axis. We saw in the warmup exercise that the area enclosed is \frac{25}{2}\pi. So:

    \[ \int_0^{10} \sqrt{25 - (x-5)^2} dx = \frac{25}{2}\pi\]

Examples of basic definite integrals

Video 1 below shows an example where you can use simple area formulas to evaluate the definite integral.

Video 1

Video 2 below shows two examples where you are not given the formula for the function you’re integrating, but you’re given enough information to evaluate the integral.

Video 2

In the examples in Video 2, you are implicitly using some definite integration properties. If you understand the definite integral as a signed area, you can interpret the rules 1.9 to 1.14 in your text (link here) by drawing representative regions. Video 3 below walks you through one of these properties.

Video 3

The Fundamental Theorems of Calculus

While most calculus students have heard of the Fundamental Theorem of Calculus, many forget that there are actually two of them. Different textbooks will refer to one or the other theorem as the First Fundamental Theorem or the Second Fundamental Theorem. It’s not too important which theorem you think is the first one and which theorem you think is the second one, but it is important for you to remember that there are two theorems. We’ll follow the numbering of the two theorems in your text.

Fundamental Theorem of Calculus Part 1 (FTC 1)

We’ll start with the fundamental theorem that relates definite integration and differentiation.

Fundamental Theorem of Calculus Part 1 (FTC 1): Let f(x) be a function which is defined and continuous on the interval [a,b]. Let F(x) = \int_a^x f(t) dt. Then F'(x) = f(x).

This says that F(x) is an antiderivative of f(x)! So you can build an antiderivative of f(x) using this definite integral. Notice that since the variable x is being used as the upper limit of integration, we had to use a different variable of integration, so we chose the variable t.

Another picture is worth another thousand words. Click here to see a Desmos graph of a function f(x) and a shaded region under the graph. The variables in the Desmos graph don’t match our notation in the definition above: instead of t, Desmos uses x; instead of x, Desmos uses b. Drag the b slider back and forth to see how the shaded region changes. In the definition, F(x) is defined as a definite integral, so it represents a signed area, as we learned earlier in today’s lesson. The function F(x) represents the shaded area in the graph, which changes as you drag the b slider. How fast is the area changing? Well, that’s the instantaneous rate of change of F(x)…which we know from Calculus I is F'(x)…which we know from FTC 1 is just f(x)!

Examples using FTC 1

Video 4 below shows a straightforward application of FTC 1.

Video 4

Students sometimes forget FTC 1 because it makes taking derivatives so quick, once you see that FTC 1 applies. You don’t actually have to integrate or differentiate in straightforward examples like the one in Video 4. One way to make a more complicated example is to make one (or both) of the limits of integration a function of x (instead of just x itself). Because you’re differentiating a composition, you end up having to use the chain rule and FTC 1 together. Video 5 below shows such an example.

Video 5

Fundamental Theorem of Calculus Part 2 (FTC 2)

This is the fundamental theorem that most students remember because they use it over and over and over and over again in their Calculus II class. This theorem relates indefinite integrals from Lesson 1 and definite integrals from earlier in today’s lesson.

Fundamental Theorem of Calculus Part 2 (FTC 2): Let f(x) be a function which is defined and continuous on the interval [a,b]. Let F(x) be any antiderivative of f(x). Then \int_a^b f(x) dx = F(a) - F(b).

Why is this a useful theorem? Well, the left hand side is \int_a^bf(x)dx, which usually represents the signed area of an irregular shape, which is usually hard to compute. The right hand side is just the difference of the values of the antiderivative F(x) at the limits of integration. So if you know how to antidifferentiate, you can now find the areas of all kinds of irregular shapes!

Examples using FTC 2

Video 6 below shows a straightforward application of FTC 2 to determine the area under the graph of a quadratic function.

Video 6

Video 7 below shows a straightforward application of FTC 2 to determine the area under the graph of a trigonometric function.

Video 7

Applications

As usual, physics provides us with some great real-world applications of integrals. Video 8 below shows an example of how to find distance and displacement of an object in motion when you know its velocity.

Video 8

Exit Ticket

  1. Differentiate:

        \[\frac{d}{dx}\int_2^x \frac{\sqrt{\cos(t)}}{\ln(t)}dt\]

  2. Evaluate the definite integral:

        \[\int_1^3(3x^2 -2x +1)dx \]