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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Erin Monagan*

Erin has been writing and editing for several years and has a master's degree in fiction writing.

Explore how driving backwards takes you where you've already been as we define definite integrals. This lesson will also teach you the relationship between definite integrals and Riemann sums. Then, discover how an integral changes when it is above and below the x-axis.

Let's say you want to find the area between some line and the *x*-axis. You know that you could use a Riemann sum approach. This approach divides the region up into a bunch of different slices, and you estimate the area of each slice. As the width of each slice gets thinner, your estimate of the total area gets better. Eventually, you'll get the total area exactly. We write this as the limit as *delta x* goes to zero of the sum of all of the slices from *k*=1 to *n* of *f*(*x* sub *k*) * *delta x* sub *k*, or the height times the width of each slice.

This limit equals the **definite integral** from *a* to *b* of *f(x)dx*. This is the **integral**, but what does integral mean? We've said that it can mean the area under the curve, but does it always mean that?

Let's consider this example: You're driving and you graph your velocity as a function of time. Let's say your velocity is 20 mph, so let's graph that. You're driving 20 mph from *t*=0 to *t*=.5 hrs. If you're driving for a half an hour at 20 mph, how far have you gone? Well, 20 * .5 means that you've gone 10 miles, because the distance traveled is your velocity * time. On this particular graph, velocity is the height of the box and time is the width of the box, and velocity * time gives you the area of this box. In the case when you're graphing velocity as a function of time, the integral (the area underneath the curve) is also equal to the distance that you've traveled.

Let's look at another case. Let's say that for the first half an hour, you're driving 20 mph because you're stuck in traffic. Then the next half hour you're going 40 mph; traffic has gotten a little bit lighter. Now how far have you traveled? The first half hour you traveled 10 miles. In the second half hour, you've traveled 20 miles, because 40 * .5 = 20). In total, you've traveled 10 + 20 miles, which is 30 miles. Again, this is the area underneath the curve of velocity as a function of time.

In this case, the integral of velocity as a function of time gives you the area under the curve, which is your distance traveled. We can make this a little more specific and say that if your velocity is given as a function, *f(t)*, and you're traveling from time *a* to time *b*, then the distance that you've traveled equals the integral from lower limit *a* to upper limit *b* of *f(t)dt*. In this case, you're integrating the function *f(t)* with respect to *t*.

But what happens in a case like this? Here, your velocity is 30 mph for the first half hour - maybe you're driving away from home - and 30 minutes in, your velocity becomes -30 mph. You stop the car a half hour into your drive, put it in reverse and drive backwards at 30 mph for a half hour. In this case, if you go forward at 30 mph for a half hour and then you go backward at 30 mph for a half hour, you're going to end up in the same place. So here if I integrate from a lower limit of *t*=0 to an upper limit of *t*=1 my function *f(t)dt*, I get zero. If I were just looking at the area, it wouldn't be zero, it would be a positive number. What happened?

This is a very important fact about integrals. Before we said that a Riemann sum is just an area, but an integral can be positive or negative. If you're above the *x*-axis, it's the area, but if you're below the *x*-axis, it's the negative area.

Let's go back to our forward and backward driving. For the first half an hour, you're traveling 30 mph (30 * .5 = 15). That's 15 miles forward, because you're going forward. Let's multiply the width of this times the height. The height is going to be negative, 0.5 * -30 = -15 miles. This isn't the area under the graph, because area cannot be negative. But, this is the integral of this region. When you add these two integrals together, you have 15 -15 = 0.

Let's review. The **definite integral** is the limit as *delta x* goes to zero of the sum from *k*=1 to *n* of *f*(*x* sub *k*) * *delta x* sub *k*. This is just adding up all of your slices in the Riemann sum. This equals the integral from *a* to *b* of *f(x)dx*. An integral above the *x*-axis will be positive, and an integral below the *x-axis* will be negative.

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6 in chapter 12 of the course:

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Math 104: Calculus16 chapters | 135 lessons | 11 flashcard sets

- Go to Continuity

- Go to Series

- Go to Limits

- Summation Notation and Mathematical Series 6:01
- How to Use Riemann Sums for Functions and Graphs 7:25
- How to Identify and Draw Left, Right and Middle Riemann Sums 11:25
- What is the Trapezoid Rule? 10:19
- How to Find the Limits of Riemann Sums 8:04
- Definite Integrals: Definition 6:49
- Linear Properties of Definite Integrals 7:38
- Average Value Theorem 5:17
- The Fundamental Theorem of Calculus 7:52
- Indefinite Integrals as Anti Derivatives 9:57
- How to Find the Arc Length of a Function 7:11
- Go to Area Under the Curve and Integrals

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