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### Example Problems for 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
• Example I: Using Left Endpoint & Right Endpoint to Approximate Area Under a Curve 0:10
• Example II: Using 5 Rectangles, Approximate the Area Under the Curve 11:32
• Example III: Find the True Area by Evaluating the Limit Expression 16:07
• Example IV: Find the True Area by Evaluating the Limit Expression 24:52

### Transcription: Example Problems for Area Under a Curve

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

Today, we are going to do some example problems for the area under a curve.0004

Let us get started.0009

Use the graphs on the following pages to find left endpoint,0012

right endpoint approximations for the area under the curve from 0 to 12.0018

First use n = 6, we are going to use 6 rectangles then do the same for n = 12.0022

Draw all the appropriate rectangles for each case.0030

There are 4 copies of the graph, one for each case, do the work, write on the graph, find areas, things like that.0033

Let us get started.0041

For here, we have n = 6 and we are going to do left endpoints.0043

We know the a is equal to 0, we know that b is equal to 12.0052

We have our 0 here and we have our 12 over here.0057

We know that Δx, that is going to be b - a/n, which is 12 - 0/ 6, which is equal to 2.0063

Our Δx is going to equal 2.0076

Left endpoints, left endpoint, go up to the graph, draw your rectangle.0080

This is the endpoint, go up to the graph, draw your rectangle.0090

Go up to the graph, draw your rectangle.0096

Go up to the graph, go to the right to the next x value, draw your rectangle.0103

Go up to the graph to your next x point, draw your rectangle.0108

Go up to the graph, rectangle.0112

There you go, 1, 2, 3, 4, 5, 6, Δx is 2.0116

Therefore, our area, left, 6 rectangles, is equal to 2 which is the Δx.0124

It is going to be f of this + f of that + f of this + f of that + f of this + f of that.0133

1, 2, 3, 4, 5, 6, read it right off the graph.0141

This x value is going to be 9.0147

I read it right off the graph, it is 8.8.0153

It is going to be 8.4, the next one is 7.6 + 6.4.0158

My last one is going to be 5, there we go.0168

I get a value of 90.4 using these.0173

My left endpoint approximation using 6 rectangles gives me 90.4.0181

We are going to do again n = 6.0190

We are going to do right endpoints.0195

Once again, a is equal to 0, b is equal to 12.0199

It is the area I’m looking for, which means my Δx is going to be 12 - 0/6.0206

Again, we are looking at 2 right endpoints.0213

I start from the right, go up to the graph, and I go to the left, until I hit the next point.0217

It is that one, 12, this is 2.0225

That is what this means, the base of my rectangles have a length of 2.0229

From here, I go up to the graph, and I go to the left to the next x endpoint.0232

That is my second rectangle.0238

I go up to the graph, go to the left, that is my 3rd rectangle.0241

Up the graph, to the left, that is my 4th rectangle.0247

Up to the graph, to the left, that is my 5th rectangle.0253

Up to the graph, left, that is my 6th rectangle.0257

Let us go ahead and do 1, 2, 3, 4, 5, 6.0266

You can number them anyway you want.0270

You can go from the left to the right, it does not really matter.0271

My area with right endpoints using 6 rectangles is 2 Δx ×, we are starting over here, or I can start over here.0275

It just depends on how you want to do it.0288

We are using right endpoints, I just going to go ahead and start over here.0293

When I do that, it is going to be 3.2 + 5 + 6.4 + 7.6 + 8.4 + 8.8.0299

3.2, 5, 6.4, 7.6, 8.4, 8.8.0321

When I do that, I get 78.8.0331

Now we are going to do 12 rectangles.0342

This time, n = 12.0345

Once again, Δx is equal to 12 - 0/12.0348

Now the Δx is 1, I have 12 rectangles.0355

We are going to do the left endpoints.0359

From the left endpoint 0, right endpoint is 12.0365

Up, go across, that is one rectangle.0371

Up, go across, 2nd rectangle.0376

Up, go across, 3rd rectangle.0379

Up to the graph, go across, 4th rectangle.0383

Up, go across, up to the graph, go across.0386

Up to the graph, go across, up to the graph, go across.0393

Up to the graph, go across, up to the graph, go across.0398

Up to the graph, go across, up to the graph, go across.0404

There you go, the points 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.0409

Those are the points that I'm going to be looking at the area.0419

Using left endpoints 12, it is going to equal, Δx is 1.0422

That height, that height, that height, 9.0428

I have got 9 + 8.9 + 8.8 + 8.6 + 8.4 + 8 + 7.6 + 7 + 6.4 + 5.7 + 5 + 4.2.0440

I get a value of 87.6.0468

These are my f values.0475

Those numbers that I have read off the graph are these right here.0481

Left endpoints, right endpoints, n = 12, Δx = 12 - 0/12 = 1.0485

Starting from the right, starting from here, go up to the graph, go across, that is my first rectangle.0504

Go up, go across, 2nd rectangle.0512

Go up, go across, 3rd rectangle.0516

Go up, go across, 4th rectangle.0520

Go up to the graph, go across, 5th rectangle.0524

Go up to the graph, go across, 6th rectangle, 7th rectangle, 8th rectangle, 9th rectangle, 10th rectangle, 11th rectangle, 12th rectangle.0528

4, 5, 6, 7, 8, 9, 10, 11, 12.0554

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.0564

The heights are going to be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.0570

The area that you get is area using right endpoints.0581

12 rectangles is 1 × we are going to have 3.2 + 4.2 + 5 + 5.70585

+ 6.4 + 7 + 7.6 + 8 + 8.4 + 8.6 + 8.8 + 8.9.0600

We are going to get a value of 81. 8.0617

Here is what we have, a left 6 was 90.4.0621

A left 12 was 87.6, a right 6 was 78.8.0632

A right 12 was 81.8, this is dropping as I went from 6 to 12.0651

As I made more rectangles, smaller rectangles, this one went up.0659

There is going to be some value that they are going to hit, some a true value.0665

The true value actually happens to be 84.96.0671

You will find out how to get that, in a couple of lessons from now.0675

This drops, this rises, lower sum comes up, upper sum comes down.0681

There is some point where they meet, that point is the true area.0686

It turns out to be 84.96.0689

Using 5 rectangles, approximate the area under the graph of f(x) = x² + 2, from x = -2 to 3 using midpoints.0694

That is use the point midway between the left and right endpoints of a given approximating rectangle.0704

We have got n = 5, we are using midpoints Δx = b - a/ n which is 3, - and - 2/ 5 which is 1.0711

The base length of the rectangle is equal to 1.0736

We are using midpoints.0740

The points I’m interested are, this is one rectangle right here, midpoint between the x values.0747

This is going to be a rectangle but I’m using the midpoint of the rectangle, not the left endpoint, not the right endpoint.0755

I'm using midpoints, that is another point, that is another point, that is another point, that is another point.0762

These x values, that is going to be my x.0768

x sub 1,x sub 2, x sub 4, x sub 5.0775

Now from those points, I go up to the graph and I go to the left endpoint, to the right, the right endpoint, that is my rectangle.0780

I’m going to go ahead and draw this as a dotted line.0794

From whatever point that I choose, that is what I go up to the graph.0802

That is going to be my f value.0807

From this midpoint, I go up to the graph, that is going to be my f value.0808

From there, I go to the right endpoint, to the left endpoint which is 1.0813

1, that is the width of my rectangle, I just happen to be taking points in between.0819

From the midpoint, I go up to the graph, I go to the right, to one x value, to the left x value.0824

That is my other rectangle, 1 rectangle, 2 rectangle, 3 rectangle, midpoint.0831

I go up to that point, I go to the right.0839

To one x, I go to the left to one x, that is my rectangle.0847

Midpoint, I go up there, I go to the right, to one, to left, to one, that is my other rectangle.0852

1, 2, 3, 4, 5, and these are my rectangles.0864

My area, midpoints using 5 rectangles is going to be 1 ×, it is going to be f(-1.5) + f(-0.5)0870

+ f(0.5) + f(1.5) + f(2.5).0889

These are the points that I'm going to take the f value of, that I’m going to put into my function x² + 2 to get the y value.0902

The area of the midpoints 5 is equal to 1 ×, it is going to be 4.25 + 2.25 + 2.25 + 4.25 + 7.0625.0914

The area is going to equal 20.0625, the approximate area.0931

I have my points but I’m using midpoints.0941

From my rectangle, from that midpoint, that is where I go up to the graph and0945

I go left and right to the x value below it, the x value above it.0949

This is the height of my rectangle now.0954

Not left endpoint, up to the graph, or right endpoint, up to the graph, but a midpoint up to the graph, and go left and right.0956

Let us see, given the function f(x) = 3√x from 0 to 27, write an expression for the true area under the graph using limits in sigma notation.0969

Let us find Δx first, let me go to blue.0981

I have got Δx = b - a/ n, that is going to be 27 – 0/ n, which is going to be 27/n, that is my Δx.0987

My area is the limit as n goes to infinity of the sum i = 1 to n of, we said Δx.0999

The equation was the limit as n goes to infinity of the sum of Δx × f(x) sub i.1017

Δx is 27/ n.1031

F is this, so it is going to be 3√x sub i.1038

We need to find what x sub i is, in terms of n.1047

We know what the x sub i are going to be, we have a beginning point, we have an ending point.1055

We actually can find some expression for x sub I, to put into this.1061

Let us list them, we have x sub 1, that is just 0, that is the left endpoint.1067

x sub 2, that is 0 + Δx, that is going to be 27/n.1074

x sub 3, that is just x sub 2 which is 27/n + another Δx, that is going to be 54/n.1083

x sub 4 equal, it is going to be 54/n + another Δx, that is going to be 81/n.1092

x sub i is going to be some, we are going to be looking for some function of i/n.1103

Let me not write this just yet, I'm looking for some sort of a pattern I can find here.1110

Some relationship between, I see that there are n’s in the denominator and I see a 27, 54, and 81.1116

I need to find a relationship between 27, 2, 54, 3, 81, and 4.1124

In other words, is there a relationship between the I and some f(i) which will give me these numbers.1129

I'm looking for an expression that I can substitute.1143

I’m looking for an x sub i that I can substitute into here, that I can substitute for x sub i.1164

I need some relation between i and n.1179

In other words, I'm looking for some x sub i which is going to be some function of i/n.1192

Based on this pattern that I elucidated from the first x sub 1, the 2nd point, the 3rd point, the 4th point.1199

I saw this pattern, I’m looking for something here.1208

I have got it, this is 2 and this is 27.1214

This is 3 and this is 54.1219

This is 4 and this is 81.1222

27 is 1 ×, if i is 2, then this is 2 - 1 × 27.1224

If i is 3, then this is 3 - 1 which is 2 × 27.1237

If i is 4, 81 is equal to 3 × 27.1243

Therefore, my x sub i is going to equal i - 1 × 27/n.1249

This relationship, as i runs 1, 2, 3, 4, 5, 6, 7, gives me these numbers.1260

Now I can put this into here and what I end up with is the following.1272

A = the limit as n goes to infinity, the sum as i runs from 1 to n of 27/n × 3√27 × i - 1/n.1281

Those are perfectly acceptable, I can absolutely leave it like this.1305

However, there is another way to simplify this thing, to simplify the function.1311

I can adjust the index as follows.1330

Now I had x sub 1 = 0, x sub 2 = 27/n, I had x sub 3 = 54/n, I had x sub 4 = 81/n.1340

Therefore, I had x sub i is equal to i - 1 × the 27/n.1357

I can do this, we can do, we can set a equal to 0.1365

And then, we can call x sub 1 27/n, x sub 2 54/n, x sub 3 is 81/n.1378

Therefore, x sub i is just 27i/ n.1390

It does not have that i - 1 quality.1396

This is just 1, 27, 2, 2 × 27 is 54, 3, 3 × 27 is 81.1398

i × 27, it makes it a little bit easier.1405

All I have done is instead of calling x1 0, I just shifted the index down.1410

I just called a = 0, I have called x1 the 27/n, I shifted the index.1415

I end up with a is equal to the limit as n goes to infinity, the sum from 1 to n of 27/n × 3√27 i/ n.1424

It is a lot simpler and I can actually pull out the 3√27 here.1443

I can write a = the limit as n goes to infinity of the sum i = 1 to n.1447

3√27 is 3, 3 × 27 is 81.1457

We get 81/n × 3√i/n.1460

When I actually evaluate the sum which I will do in the next problem, a version of that,1472

I just pull out all the n, all that I'm left with under the summation sign is an i.1476

Everything else comes out as a constant.1481

It is really nice.1484

That is our final answer, or this one, either one is fine.1486

Given the function f(x) = 4x³ from 0 to 2, find the true area by actually evaluating the limit expression that you obtain.1495

Use the following fact about the sum of cubes in the first n integers.1504

When I end up with a sum of cubes, 1³ + 2³ + 3³, all the way to whatever number n³,1508

I have a closed form expression for that, that is this.1517

If I have a sum that I can express as this, I can express in sigma notation, I can replace that sigma notation with this.1520

That is what this is saying.1528

Let us go ahead and do this problem.1530

Let us start with our Δx, I’m going to start over here.1533

Δx = b - a/ n, it is just going to be 2/n.1536

I know, therefore, that my area is going to equal the limit as n goes to infinity of the sum1547

as i goes from 1 to n of Δx, which is my 2/n × 4 × x sub i³.1558

That is my definition.1571

My question is, what can we put for this x sub i in there?1575

Let us see what we have got.1587

Let us come over here.1589

I think I’m going to do, I’m going to do what I did last time.1595

I’m going to set a equal to 0, and I’m going to set x sub 1 equal to, it is 0 + Δx which is 2/n.1599

x sub 2 = 2/n + 2/n which is 4/n.1612

x sub 3 = 4/n + 2/n which is 6/n.1619

I got my relationship 1, 2, 2, 4, 3, 6, x sub i is equal to 2 sub i/n.1624

A is equal to the limit as n goes to infinity, i goes from 1 to n of 2/n × 4 x sub i³.1649

x sub i is this, 4 × 2i/ n³.1665

This is going to equal the limit as n goes to infinity of the sum i = 1 to n.1676

2 × 4 is 8, this is 8/n × 8 i³/ n³.1686

We are just doing everything that we need to do.1698

This is a function of i, the index is what the variable is under a summation sign.1707

8 × 8 is 64, n × n³ = n⁴.1714

I’m just going to pull all of that and take it outside of the summation sign, which I can actually do.1722

This = the limit as n goes to infinity, 8 × 8 is 64, n × n³ is n⁴ × the summation of i goes from 1 to n of i³.1728

The index that I choose, that is the variable.1748

Everything else, I can pull out is a constant.1750

This is this, now I have something already, I know what this is.1753

The sum as i goes from 1 to n of i³, that is just equal to 1³ + 2³ + all the way to n³.1760

They gave me an expression for that, they said it is this.1772

That = n × n + 1/ 2².1776

I’m going to put this in for that and I get the following.1784

I get a = the limit as n goes to infinity of 64/ n⁴ × n × n + 1/ 2²1792

= the limit as n goes to infinity of 64/ n⁴ × n² × n² + 2n + 1/ 4,1811

= the limit as n goes to infinity of 64/ n⁴ × n⁴ + 2n³ + n²/ 4.1832

64 divided by 4, this is going to equal the limit as n goes to infinity of 16 × n⁴/ n⁴ is 1.1865

2n³/ n⁴ is 2/n.1879

N²/ n⁴ is n².1885

When I take n to infinity, this term goes to 0, this term goes to 0.1892

That is it, nice and simple.1902

I have my area = the limit as n goes to infinity of the sum i to n of my Δx × my f(x sub i).1904

I found an expression for Δx, I put it in.1917

I used my x sub 1, x sub 2, x sub 3, x sub 4, to elucidate some relationship,1921

in terms of n and I that I can put in for f(x sub i).1927

I put it in, I solved this, pulled out what I needed leaving just i.1932

As much as I can, i, underneath the summation symbol.1939

I had a closed form expression for that summation symbol.1942

I solved that, combined by simplifying.1945

And then, my final step, I just took the limit is n goes to infinity.1948

This goes to 0, this goes to 0, the limit is 16.1952

This is a long way of doing this.1958

In a couple of lessons we will find a very quick way of doing,1960

but it is really important that we understand where this comes from.1963

We did not just pull it out of nowhere.1967

Clearly, a lot of intricate detailed work goes into defining some of these higher mathematical concepts,1968

like finding area under a curve which we are going to call integration,1975

which is going to be essentially anti-differentiation.1979

Thank you so much for joining us here at www.educator.com.1983

We will see you next time, bye.1986