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The Area Under a Curve

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • The Area Under a Curve 0:13
    • Approximate Using Rectangles
    • Let's Do This Again, Using 4 Different Rectangles
  • Approximate with Rectangles 16:10
    • Left Endpoint
    • Right Endpoint
    • Left Endpoint vs. Right Endpoint
    • Number of Rectangles
  • True Area 37:36
    • True Area
    • Sigma Notation & Limits
    • When You Have to Explicitly Solve Something

Transcription: The Area Under a Curve

Hello, and welcome back to www.educator.com, and welcome back to AP Calculus.0000

Today, we are going to talk about the problem of finding the area under a curve from one point to another point.0005

Let us jump right on in.0012

Let us see, I want to find the area under the curve f(x) = x² from x = 1 to x = 5.0018

That is my task, here is what I want to do.0024

I’m going to draw my x² curve, something like that.0059

I will say that this is my 1.0065

I will say that maybe 5 is over here.0070

I want to find the area under the curve from the x axis up to the curve, that is what I'm looking for.0074

How can I do that?0081

What I’m basically going to do what is I’m going to approximate it first with rectangles 0086

because I have a formula for finding the area of a rectangle.0091

It is based × the height.0094

I'm going to draw a bunch of rectangles, approximating it that way.0097

Let us start and see what happens.0102

The first thing I only to do is to choose how many rectangles I need to start with.0108

In this case, I’m just going to start by approximating it with 4 rectangles.0111

I’m going to break up my domain into 4 sections.0114

This is my halfway mark, that is my halfway mark, that is my halfway mark.0119

I have from 1, 2, 3, and I have 4.0124

Now I’m going to draw my rectangles.0127

Let us say I start with an approximation using 4 rectangles.0132

The number 4 is arbitrary, I could have used 3, I could have used 7, I could have used 10, I could have used 114.0152

It is just 4 gives us a chance to actually, it is not too many and0158

allows us to see what is going on theoretically without burdening us with too much drawing.0163

Here are the rectangles that I’m going to draw.0172

It is going to be one rectangle, two rectangles.0175

This is my first one and I want that area.0186

This is my 2nd, this is my 3rd, and this is my 4th.0190

I’m going to find the areas by multiplying the base × the height, base × the height, base × the height, base × the height.0195

That is going to give me an approximation to the area under the curve.0204

These parts right here, obviously they are not going to be included.0207

It is going to be an underestimate, we will deal with that in just a second.0211

My first area, area 1, that is just going to be 1.0216

This is from 1 to 5, this distance is 4.0220

If I break it up into 4 rectangles, each distance, each base length of a rectangle is 1.0225

The first area is 1 × f(1) because this height right here, it is just f of whatever 1 is.0232

It is up to the graph, this is my graph.0249

F which is equal to x², it is just 1, = 1 × 1 = 1.0253

My area 2, my area 2 = 1 which is the base × f(2).0264

This is 3, this is 4, this is 5.0272

This is going to equal 1 × f(2).0275

The f(2) is 4, it is going to be 1 × 4 which is 4.0279

I take my third area, my third year is going to be this one right here.0286

It is going to be 1 × f(3), because the height of the rectangle is f(3).0290

1 × f(3) which is equal to 1 × 9 which is equal to 9.0296

My 4th area which is going to be 1, which is the base of the rectangle.0304

F(4) is going to be the height of the rectangle.0309

1 × f(4) which is equal to 1 × 16 which is 16.0316

I add these together to get an area l = 3.0326

I will put a little 4 here.0344

This symbolism that I’m using the area l4, 4 stands for the number of rectangles that I actually chose,0346

the number of divisions that I make to the domain.0353

L stands for left the endpoint.0355

I chose left endpoints, the rectangle, the left endpoint, to go up to my graph.0359

Left endpoint for this rectangle, left endpoint for this rectangle, left the endpoint for that rectangle.0367

That is what the symbolism means.0373

My first approximation using left endpoints as my height.0375

The f of the left endpoint to make the height of the rectangle.0384

That gives me an approximation of 30.0388

Clearly, this is an underestimate, not a problem.0390

Let us go over here, let me redraw real quickly.0395

I have got this, I have this one, I have this one, I have this one, and I have this one.0401

This is 1, 2, 3, 4, 5.0415

This area was area 1, this was area 2, this was area 3, and this was area 4.0422

Notice that this al4 = 30 is an underestimate.0429

It is an underestimate because the approximating rectangles are below the graph.0449

The approximating rectangles are below the graph.0461

Our arbitrarily shows 4 rectangles.0476

The left endpoints of each rectangle, they are the ones that correspond to x = 1, x = 2, x = 3, and x = 4.0489

Left, left, left, left of the points that I can choose in my domain.0516

I chose left endpoint starting from the left.0520

The heights of the rectangles corresponded to f(1), f(2), f(3), f(4).0523

It was the distance from 1 to 2 × the height at the left endpoint.0550

This base × that, this base × that, this base × this, this base × that.0555

Let us do the same thing, same graph, I pick x², same number of rectangles 4.0566

But let us use 4 different rectangles, but all over the same domain.0571

Let us do this again, let us go back to blue.0578

Let us do this again but choose 4.0588

Let us choose a different 4 rectangles.0600

A different 4 rectangles but still 4 rectangles.0606

I have got my drawing again, I got my x² graph.0622

I have 1, I have 5, midpoint, midpoint, midpoint.0630

Now I have got 1, 2, 3, 4.0636

This time, I’m going to choose these rectangles.0641

I have area 1, area 2, area 3, and area 4.0661

My area 1 is equal to, the base of the rectangle is still just 1.0670

I will go ahead and do this in red.0676

Is still 1, it is 1 ×, but this time I’m going to take, notice the height of the rectangle is this height right here.0677

That is f(2), it is going to be 1 × 4 = 4.0687

Area 2 is 1 × this time, it is f(3), that is my rectangle right there.0696

That is the height of my rectangle.0705

It is going to be 1 × f(3) which is going to be 1 × 9 which is 9.0709

My 3rd area is going to be 1 × f(4) which is 1 × 16 is 16.0717

My 4th rectangle, the height of the rectangle is going to be 1 × f(5) which is going to be 1 × 25, which is 25.0727

Notice this time, I have taken right endpoints, 4 for each rectangle.0742

From the right endpoint, I went up to the graph.0753

And from there, I drew my rectangle to the next x value, drew my rectangle to the next x value.0756

I add these together to get an area using right endpoints, 4 rectangles, I end up with 54.0761

I will draw a quick representation of it again.0786

This time I have got this and I have got that, 1, 2, 3, 4, 5.0789

This time my rectangles are 1, 2, 3, 4.0800

It is this area, this area, this area, and this area.0807

A right 4 is an overestimate, you can see that it actually goes, here is the graph, the rectangles go above the graph.0818

It is an overestimate, that is clear to see.0831

Again, we have 4 rectangles.0839

The right endpoints, it corresponds to 5, 4, 3, and 2.0858

I will write it in order this way, but I will right it backwards.0872

X = 5, x = 4, x = 3, x = 2, to demonstrate that I actually started from the right working left.0874

The heights of the rectangles ,they correspond to f(5), f(4), f(3), and f(2).0886

The true area is something between the overestimate and the underestimate.0914

It is going to be more than 30 and less than the 54, whatever it is that we got.0927

The true area which we will call a is between these two.0933

In other words, the true area is less than or equal to a right.0946

It is greater than or equal to a left, 4 and 4.0952

This one was 30, I believe, this one was 54.0956

That much we know, we know we have an upper limit and we know we have a lower limit.0960

We know that the area is going to be somewhere in between them.0964

In general, if we want to approximate with rectangles, the area under some f(x),0971

from a point a to a point b, we can do so in two ways.1008

The first way is we can form rectangles whose height is based on a left endpoint.1024

Or we can form our rectangles whose heights are based on right endpoints.1050

Let us do a left endpoint, general scheme.1087

This is all just theory that we are throwing out here.1092

Just want to make sure that this is really solidly understood, before we actually talk about anything specific.1095

Left endpoint, our scheme is going to be as follows.1102

We have some graph, it does not matter what the graph actually looks like.1112

We have an a and we have a b.1118

In this particular case, I’m just going to choose 5.1127

Once I choose the number of rectangles, n is the number of rectangles.1130

In this case, for the sake of argument here, I’m just going to choose n = 5.1140

My Δx which is going to be the length of my base, that is just equal to the right endpoint - the left endpoint.1145

In other words, the length of the domain divided by 5.1155

That is it, I’m just breaking this up into 5 things.1160

I have got 1, 2, 3, 4, there we go.1167

I'm going to call a, x sub 1, I’m going to call this x sub 2, x sub 3, x sub 4, x sub 5.1173

B, I’m going to call x sub 6.1183

The length is just one endpoint + the Δx, whatever this is + the Δ.1188

The Δx is actually the same for all of them.1198

I take the difference between this right endpoint and this left endpoint.1200

I divide by the number of rectangles that I’m going to have under this graph, that gives me the length of the bases.1204

We are just doing left most endpoints and here is how it works.1211

Here is the procedure for doing left most endpoints.1215

You take the left most endpoint which is our a, which is we are going to call x1.1220

And then, you go up to the graph.1236

We start here, we go up to the graph.1245

The next thing we do, you go right to the next x value which is x sub 2.1251

From here, you go up to the graph, and then, you go to the right, to the next x value, that is your first rectangle.1272

Now from x2, starting at x2, you go up to the graph, that puts us here.1285

Then, you go right to the next x value which is x3.1305

You go right which is x3.1320

You keep going this way, from x3 you go up to the graph, you go to the right to the next x value.1329

That is your 3rd rectangle.1336

X4, you go up to the graph, that gives me my next rectangle.1339

It is not going to be that if you use left endpoints, you are going to be an underestimate, right endpoints overestimate.1346

That worked out that way because of the graph that we chose, x², it was that way.1353

The graph itself does not matter.1359

When you use left endpoints, you start to the left endpoint, you go up to the graph, that is your height.1360

You go up to the next x value, that is your rectangle.1366

Sometimes it is going to be above, sometimes it is going to be below, depending on the shape of the graph.1369

Now last one, from x⁵, we go up to the graph.1375

It is where we hit the graph and then we go to the right, to the next x value.1380

That is our area 1, that is our area 2, area 3, area 4, area 5.1386

We have to have this many rectangles as that.1393

We have the left endpoints that we choose because we have that many rectangles.1398

That is how many endpoints we have.1402

We have 5, in this case, x1, x2, x3, x4, x5.1405

Left endpoints, we started with a.1410

The procedures are just the same.1413

Whatever point that you choose, go up to the graph, and then go to the right until you hit the next x value.1414

That is your height, where you hit the graph.1424

That is all we are doing.1428

Now our area is just equal to a1 + a2 + a3 + a4 + a5.1429

This is equal to f(x1) × Δx, f(x1) is this, this is our Δx + f(x2) × Δ.1442

This is f(x2), this is our Δx + f(x3) × Δx + f(x4) × Δx + f(x5) × Δx.1456

I’m going to factor out the Δx, area = Δx × f(x1) + f(x2) + f(x3) + f(x4) + f(x5).1474

That is it, the Δx is the same.1499

It is the right endpoint of the domain - the left endpoint of the domain.1501

The length divided by the number of rectangles I choose.1508

Δx is the same in all cases, I can just factor it out.1511

It just becomes f(x1) + f(x2) + f(x3) + f(x4) + f(x5).1514

That gives me my 5 rectangles.1522

I have to have as many terms as I have n, that is how I keep track.1522

This is left endpoints.1529

Let us do the case with right endpoints.1534

I have a graph, I have a, I have b.1548

1, 2, 3, 4, I have broken it up into 1, 2, 3, 4, 5 sections.1552

N is equal to 5, my Δx which is the length of one section, that is b - a/5, in this case.1560

I call my ax1, this is my second point, my 3rd point, my 4th point, my 5th point.1571

B, I call my 6th point.1579

Now using right endpoints, this is right endpoints.1583

That procedure, you go to the right most endpoint which is b.1591

Nice standard procedure, you go to the rightmost endpoint which is b which is x = 6.1603

You go up to the graph, and then you go left to the next x value below it.1608

The left to the x value below it, which in this case is x sub 5.1621

That right endpoint, you go up to the graph, you go to the left until you hit this x value.1636

That is your first rectangle.1644

Now from x5, starting from x5, you go up to the graph then go left to the next x value.1648

Starting at x5, you go up to the graph, you hit here.1670

You go to the left to the next x value, that is your second area, and so on.1674

I will do one more, this one is going to be x4.1682

Notice we stop at x4.1685

From x4, you go up to where it hits the graph.1687

And then, you go to the left, that is your 3rd rectangle.1697

From x3, you go up to where it hits the graph, and then you go to the left to where your next x value.1701

That is your third rectangle.1711

From x3, you go up to where it hits the graph.1712

You go over to the next x value.1716

This last x value is a, you stop there.1720

I’m going to call this a1, this a2, a3, a4, and a5.1726

I started from the right using right endpoints.1733

For a particular interval, the last one I chose, left endpoints, now I’m choosing right endpoints.1736

The last one we do the left, left, left, left, left.1744

Now it is right, right, right, right, right, starting from the right.1747

Again, my area is equal to a1 + a2 + a3 + a4 + a5.1752

My area = Δx f(x6), Δx(x6) that gives me this area, +Δx f(x5).1765

F(x5) that gives me this area, and so on.1785

+ Δx f(x4) + Δx f(x3) + Δx f(x2).1789

Where do I stop? 1, 2, 3, 4, 5 terms, that is where I stop.1800

My area = Δx × f(x6) f(x5) f(x4), this is all just notation here for something that is clear geometrically.1806

I'm just taking a height × width.1822

F(x4) + f(x3), this is all just algebra because we need to turn geometry into algebra.1825

F(x2), that is that.1833

The rectangles that I got, if I’m choosing right endpoints and going up to the graph, 1841

are not the same as the rectangles that I choose from the left going to the left endpoints.1846

We want to see that, let us superimpose them.1851

Let us go back to blue.1857

These left endpoint rectangles are not the same as the right endpoint rectangles.1861

We want to make sure you see that.1888

Here is how, nice and big.1893

We have our function, we have a, we have b, breaking it up into 5.1903

1, 2, 3, 4, this is 1, 2, 3, 4, 5 left endpoints.1908

I will do those in blue.1915

Starting from the far most left endpoint.1918

I go up to the graph, I go across, that is my first rectangle.1920

This is x1, this is x2, x3, x4, x5.1926

B, of course is x6.1934

From x2, I go up to the graph and I go across to the right, that is my second rectangle.1937

From x3, I go up to the graph, I go to the right, I come down.1944

At x4 I go up to the graph, go there, and I come down.1949

Go up the graph there and I come down.1955

Those are my left endpoint rectangles, 1, 2, 3, 4, 5.1958

Now I begin with the rightmost endpoint, I’m going to do this in red.1964

From the right most endpoint, from here, I go up to the graph, it is here, and I go to the left, that is my first rectangle.1967

I go up to hit the graph, it is my second rectangle.1980

I go up to hit the graph, my 3rd rectangle.1987

Go up to hit the graph, it is my 4th rectangle.1992

I go up to hit the graph, it is my 5th rectangle.1997

The ones in red, those are the right endpoints.2002

The ones in blue, those are my left endpoints.2017

They are different, they are going to give me different numbers.2021

I wanted to make sure that you knew that they were different rectangles.2027

One of them starting with left endpoints going up to the graph going across.2030

The other for the right endpoints, I’m starting from the right going up and cutting across.2035

Let us give the general form, let us go back to blue.2044

In general, for any n, I’m not going to specify what n is.2048

For any n there are n + 1 x values.2059

N is the number of rectangles.2074

If you decide that you are going to do 5 rectangles, you are going to have six points total.2079

x1, x2, x3, x4, x5, x6, that is going to be your a x2, x3, x4, x5, and then b.2087

Let us draw this out.2097

We have our general graph of any function, this is our f(x), no matter what it is.2106

We have our a that is going to be our x1.2112

We go all the way to our b, this is going to be our x n + 1.2117

In other words, if I choose n rectangles, my b is actually my x n + 1 terms.2123

If I choose 10 rectangles, a is my x1, b is going to be my x sub 11.2131

If I choose 30 rectangles, b is going to be my x31.2136

However, I break it up.2149

Δx is equal to b - a divided by n.2152

Final point of the domain, initial point of the domain.2160

I subtract them, that gives me the distance.2164

I divide by the number of rectangles that I choose, that gives me my Δx which is the length of the base of one of my rectangles.2166

Left endpoint, area is equal to Δx × f(x1) f(x2), all the way to f(x sub n).2174

My right endpoint, area is going to be Δx which is the same but this time it is going to be f(x2) + f(x3) f(x) n + 1.2194

If I start from the left, it is going to be this, this, this, this, this, this, this.2216

If I start from the right, it is going to be this, this, this, this, this, this, this.2222

It does not include the x1.2227

If I go from left, it does not include the x n + 1.2230

That is why it ends at x sub n.2233

It ends here from going to the left.2236

Going from the right, it includes everything except it ends here.2238

It does not include that one, that is what is going on.2243

This is the definition of the left endpoint area.2247

This is the definition of the right endpoint area.2250

As you take more rectangles, that is n higher and higher, 10, 20, 30, 40, 50, 2257

clearly your Δx is going to get smaller because you are divided b - a/ Δ n.2273

N higher and higher, you get a better approximation of the true area.2278

That make sense, you are taking thinner rectangles so there is not as much a gap up near the graph.2295

Let us start again here, let us erase this.2312

As you take more and more rectangles, n higher and higher, you get better approximations to the true area.2317

We define the true area as the limit as we take n, the number of rectangles, to infinity.2336

That is what we do in calculus, we always take something to infinity.2358

N to infinity of the left endpoint area or the right endpoint area.2363

Once you form that thing, the left endpoint area, right endpoint area, with Δx and f(x), 2372

once you have a function, that is going to be some function of n, you take n to infinity.2378

Our definition is this, the true area = the limit as n goes to infinity of the right,2384

a true = the limit as n goes to infinity of the left.2401

It can be shown, it can be demonstrated, which you will do if you are a math major and2410

you go on to take some course called analysis,2413

you have to go to demonstrate that these two limits end up being the same.2416

It make sense, remember when we first started this lesson, we had a lower sum of 30.2420

And then, we have an upper sum of 54.2426

If I went a little higher, let us say to 10 rectangles, I will be a little bit more than 30, I will be a little bit less than 54.2429

There is going to be some number that they are going to converge to,2435

as the number of rectangles become smaller and smaller.2438

That is the true area, that is what we have done.2440

We found an upper sum, we found a lower sum.2443

If we take n smaller and smaller rectangles, the lower sum is going to rise, the upper sum is going to drop.2447

There is going to be some point where they are going to meet.2452

It can be shown that these two limits are equal and the limit is the true area.2456

It gets even more interesting than that.2484

In fact, for your x sub 1, x sub 2, x sub 3, and so on, 2486

you can actually choose any point between an x sub n and x sub n + 1.2505

In other words, for any rectangle, you do not have to pick a left endpoint or a right endpoint, 2522

that is just very systematic because it gives us a systematic procedure for doing things.2527

You can actually choose any point you want in there, it absolutely does not matter.2532

For every different rectangle, you can choose a different point.2535

It is better if you just choose the same point consistently, either left or right, or midpoint is often a good one that we use.2539

The truth is that you can choose any point you want.2546

It does not have to be left or right endpoints, it can be midpoints which is used often.2561

Often case is when you take the midpoint, you actually get a better approximation.2591

You get an approximation, and then you take the limit as n goes to infinity.2594

You get your particular area.2599

Let us finish this off with a discussion of something called the sigma notation.2604

Let me go back to blue here.2610

Sigma notation is a shorthand notation, you have seen it before back in algebra, pre calculus, things like that.2612

It is a shorthand notation for long sums.2620

We do not wand to write out, f(x1) Δx, we need a shorthand notation for that.2623

Let us go ahead.2631

The left area of n, we said was equal to Δx × f(x1) f(x2) +…+ f(x sub n).2632

In sigma notation, it looks like this.2649

The sum I to n, i is the index.2653

1, 2, 3, 4, 1, 2, 3, 4, 5, 6, it goes up to however many you want to add.2658

It is going to be Δx f(x sub i).2667

That is it, you put i1 in for here, that is the 1st term.2671

And then, you go 2, that is the 2nd term.2677

The 3, that is the 3rd term.2680

All the way to n, that is the last term.2682

This is where the index starts and this is where the index stops.2688

If right, n is equal to, we said that the right one, the right endpoint was f(x2) + f(x3) + f(x) n + 1.2709

Let us actually do it so you can read it.2729

That sigma notation, the sum i goes from 1 to n, Δx (x) i + 1.2736

We said that the area was the limit as n goes to infinity of ar n.2754

Which was the same as the limit as n goes to infinity of the al n.2768

Area = the limit as n goes to infinity.2774

How is this for a pretty intimidating looking thing, Δx f(x sub i).2781

Let me put the left one first and let me put the right one next.2796

Left are x sub i = the limit as n goes to infinity.2800

I just put this in for this here.2807

1 to n, Δx f(x sub i) + 1.2815

This is how we do it.2828

You form this thing, you sum it up, you get some expression in n, and then, you take n into infinity, that gives you your area.2830

That is what we are ultimately going to do.2839

Again, this is all the theoretical stuff.2840

What is important, what I want you to take from this lesson is the idea of a rectangle.2842

Choose your left endpoints, go up to the graph, go over.2847

Building your rectangles, that is what is important.2850

Or right endpoints, or midpoints, whatever it is.2853

For whatever point that you choose, you are going to go up to the graph and you are going to build your rectangle from there.2857

That is what is really important here.2862

As long as you can understand where this came from, that this is the base of the rectangle and 2866

these are the heights of the rectangles, that is all that is necessary.2870

The rest it just tedious notation.2874

Let us finish it off by saying, when you have to explicitly solve something like,2877

the limit as n goes to infinity of the sum from 1 to n of Δx f(x sub n), 2903

the first thing you are going to do is work from the inside out, like everything else in math.2916

I know it is notation but notation is just a shorthand for telling you what to do, it is just an algorithm.2931

This says, the first thing you want to do is first find an expression for this.2937

First, find an expression for this, for whatever it is.2944

If they give you an f, you find a Δx.2948

Remember, Δx = b - a/ n.2951

Second, if there is a closed form expression for this sum, which are often will be for our purposes,2957

you calculate the sum, calculate this expression.2994

When you calculate it, it is going to make the sum go away.3000

In other words, there is a way of finding what the sum of something is often.3003

Not always, but often.3008

The last thing you do, whatever expression you get, you take the limit as n goes to infinity.3009

You know how to do limits.3017

You take the limit as n goes to infinity of the expression you just got.3024

Do not worry, we are going to be doing example problems.3042

You are left with the answer which is your area.3046

I will go to go ahead end this lesson here.3052

The next lesson is going to be the example problems for this particular discussion.3055

Thank you so much for joining us here at www.educator.com.3059

We will see you next time, bye.3061