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Math 104: Calculus14 chapters | 116 lessons | 11 flashcard sets

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Lesson Transcript

Instructor:
*Zach Pino*

What would happen if you could draw an infinite number of infinitesimally thin rectangles? You'd get the exact area under a curve! Define the Holy Grail of calculus, the integral, in this lesson.

Let's go back and say you're trying to calculate the amount of land that you have - the area of your property. Your property extends from some road to the River Newton, between a fire hydrant on the west side and a pine tree on the east side. You know that you can estimate the area by measuring how far it is from the road to the river at the fire hydrant and multiplying that number times the distance between the fire hydrant and the pine tree. That gives you a rough estimate of the area, as it's the height of your property times the width of your property.

You know that you could make a better estimate of the amount of property that you have by dividing this up and calculating two separate areas - say the area from the fire hydrant to the middle of your property line and then measuring again how far back the river is and the middle of your property and multiplying that height times the width from that middle point to your pine tree. Once you have these two separate areas, you can add them together and get a better estimate of your total area.

If you divided your property into three regions and measured the area of each one of them, you might get still a better estimate of your overall area. And really, you could do this over 16 regions. Maybe each one is the width of a lawn mower. So as you're riding on your lawn mower, you measure how many feet it is from the road to the river; you turn around and measure how far it is from the river to the road. And you can continue doing this in big stripes.

This is just slicing your property and measuring the area of each stripe separately. This would give you an even better estimate of your area. When you mow the lawn, you get a pretty good idea of how much land you have. Sure, there's the odd patch of grass that's left over, and maybe every now and then you drive over the road a little bit, but it's a pretty good estimate as to how much land you actually have.

All this is is a **Riemann sum**. It's the area between the river and the road, and you've added up the sums of all of these little slices. In Riemann sum notation, we call the river *f(x)*, the road the *x*-axis, the fire hydrant is at *x* = some value *a*, and the pine tree is at the *x* value of *b*. We write this whole Riemann sum as the sum over all *n* slices - so that's *k*=1 to *n* (from the first slice to the *n*th) - the height of that slice, which is *f*(*x* sub *k*), the height of slice *k*, times the width of that slice, that's *delta x* sub *k*. So the height times the width, sum them all up and you get a pretty good estimate for your area.

Let's say we have a really funky looking property. Say the river follows the line *x*^3 - sin(*x*) + 1, and our property goes from *x*=0 to *x*=2. If we take one slice and we estimate this with one Riemann sum, a left-side Riemann sum, we might get an area that is 2. If we use two slices, dividing this in the middle, then we might get an area of 2.16 or so. If we divide this into five slices, our area becomes 3.3. If we divide it into ten slices, it's roughly 3.9, 20 slices is 4.4 or so. If I divide this into 100 slices - say I'm able to take my lawnmower and go up and back 100 times - I might estimate the area to be 4.5.

But let's say that's not good enough, and I want to take smaller slices. So rather than a gigantic lawnmower, I'm going to take the width of a measuring tape and take 1,000 slices. I find that my area is just over 4.5, maybe about 4.6. I march down to city hall and I say, 'Hey, my land is about 4.5 or 4.6.' They say, 'You know what, Erin, your actual land is 4.58. So you're getting closer the more slices you take, but maybe rather than a measuring tape, maybe you need something thinner to take even smaller slices - maybe a hair width.' So I say, 'Okay, if I take a hair width, that's going to make the number of slices that I'm taking even higher, and that'll get me even closer to the actual area. It might make sense that the smaller the width, the closer I'm getting to the actual area.'

This brings up an important point about Riemann sums. Say you take *n* slices between point *a* and point *b* on the *x*-axis. So this is *n* times up with your lawn mower between the fire hydrant and the pine tree. You might estimate your total area to be the sum from *k*=1 to *n* of *f*(*x* sub *k*) - that's the height of that *k* slice - times *delta x* sub *k*, that's the width of that slice. But let's say you're going to take the limit of this as *n* goes to infinity. So this means you're taking thinner and thinner and thinner slices; you're moving from a lawnmower to a measuring tape, to a hair, to a nanotube - thinner and thinner slices. That means that *delta x* is going to zero. Well when you take this limit, the limit as *delta x* goes to zero of the sum *k*=1 to *n* of *f(x) delta x*, you end up defining what's known as the **integral**.

This integral goes from *a* to *b* of *f(x)dx*. So here, *a* is our left limit, and *b* is our right limit. *f(x)* is called the integrand, and *dx* is really the variable we're integrating over, so that's *x*. You can recognize an integral, because it has this line here, which is the integral, and it has something like *dx*.

If you were integrating over the variable *y*, it would be *dy*, and so on and so forth. Another way to remember this is that when you're taking the limit as *delta x* goes to zero (as your slice goes to zero), your sum becomes the integral. This *k*=1 to *n* is really just defining all of your slices, so that's like defining your region as being from *a* (the left side) to *b* (the right side). Your *f*(*x* sub *k*) just becomes *f(x)* and *delta x* sub *k* becomes *dx*, like 'D is for *delta*.'

Let's review. If you take the limit as your slice size goes to zero of some **Riemann sum** over the region *x*=*a* to *x*=*b*, you get the **integral** from *a* to *b* of *f(x)dx*. This is what's known as a **definite integral**.

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Math 104: Calculus14 chapters | 116 lessons | 11 flashcard sets

- Go to Continuity

- 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
- How to Use Riemann Sums to Calculate Integrals 7:21
- 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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