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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### 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

### AP Calculus AB Online Prep Course

### 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 and*0158

*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 and*2410

*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

1 answer

Last reply by: Professor Hovasapian

Fri Mar 23, 2018 5:15 AM

Post by Magic Fu on March 20 at 03:30:15 PM

Hi, Professor Hovasapian.

I am going to ask a question that is completely unrelated to Calc AB.

How do you find bounds of polar functions when you try to find the area?