For more information, please see full course syllabus of AP Calculus AB

For more information, please see full course syllabus of AP Calculus AB

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## Transcription

### 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

### AP Calculus AB Online Prep Course

### 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 + 5.7*0585

*+ 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 hope that made sense.*0940

*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 and*0945

*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 sum*1547

*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/ 2²*1792

*= 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

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