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

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

### Example Problems for Areas Between Curves

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
- Instructions for the Example Problems
- Example I: y = 7x - x² and y=x
- Example II: x=y²-3, x=e^((1/2)y), y=-1, and y=2
- Example III: y=(1/x), y=(1/x³), and x=4
- Example IV: 15-2x² and y=x²-5
- Example V: x=(1/8)y³ and x=6-y²
- Example VI: y=cos x, y=sin(2x), [0,π/2]
- Example VII: y=2x², y=10x², 7x+2y=10
- Example VIII: Velocity vs. Time

- Intro 0:00
- Instructions for the Example Problems 0:10
- Example I: y = 7x - x² and y=x 0:37
- Example II: x=y²-3, x=e^((1/2)y), y=-1, and y=2 6:25
- Example III: y=(1/x), y=(1/x³), and x=4 12:25
- Example IV: 15-2x² and y=x²-5 15:52
- Example V: x=(1/8)y³ and x=6-y² 20:20
- Example VI: y=cos x, y=sin(2x), [0,π/2] 24:34
- Example VII: y=2x², y=10x², 7x+2y=10 29:51
- Example VIII: Velocity vs. Time 33:23
- Part A: At 2.187 Minutes, Which care is Further Ahead?
- Part B: If We Shaded the Region between the Graphs from t=0 to t=2.187, What Would This Shaded Area Represent?
- Part C: At 4 Minutes Which Car is Ahead?
- Part D: At What Time Will the Cars be Side by Side?

### AP Calculus AB Online Prep Course

### Transcription: Example Problems for Areas Between Curves

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

*Today, we are going to be doing some example problems for areas between curves.*0004

*Let us jump right on in.*0008

*In the following examples, we want you to sketch the given curves, identify the region that is enclosed by these curves.*0011

*State whether you are going to be integrating with respect to x or y.*0020

*Draw a representative rectangle, if you can.*0025

*Finally, find the area of this region, or at the very least set up the integral for the area of this region.*0028

*Let us get started.*0035

*Our first one is going to be y = 7x - x² and y = x.*0039

*Let us go ahead and draw this out and see what is it we are actually dealing with.*0043

*I think I will go ahead and work in blue.*0048

*Let me draw it over here.*0054

*How we are going to draw 7x - x²?*0060

*I'm going to go ahead and work over here.*0064

*I’m going to factor this, y = x × 7 - x and that is equal to 0, that gives me x = 0 and that gives me x = 7.*0068

*I know that the graph is going to hit a 0 and I know the graph is going to hit at 7.*0083

*The midpoint between 0 and 7, the midpoint which is where the vertex is going to be,*0089

*I know this is a -x², I know it is a thing that is going to open down.*0101

*Therefore, I know it is going to be going up this way and up this way.*0105

*The midpoint = 3.5.*0111

*When I put 3.5 into the original, I end up with y = 12.25.*0114

*At 3.5, 12.25, that is that graph right there.*0123

*As far as the y = x is concerned, we know that one, that is just a line that way.*0134

*The region that we are interested in is this region right here.*0141

*In this particular case, I’m going to go ahead and draw a representative rectangle.*0149

*The rectangle is vertical, I'm going to be adding the rectangles this way.*0155

*I'm going to be integrating along x.*0159

*What is the x value of this point?*0172

*We are going to be integrating from 0 all the way to the x value of this point, whatever that is.*0176

*We need to find that.*0186

*What is the x value of this point?*0187

*I’m going to set them equal to each other.*0197

*I have got 7x - x² is equal to x.*0198

*I have got x² - 6x = 0.*0208

*I have got x × x - 6 = 0 which gives me x = 0.*0216

*That is one of them.*0221

*x = 6, that is the other point.*0224

*Remember, this one was 7.*0228

*We integrate from 0 to 6, there we go.*0234

*From here to here, we are going to add up all of the individual rectangles.*0244

*That is going to give us the area of the curve.*0250

*Let us go ahead and do that.*0252

*With each of these, I actually rendered the picture itself.*0256

*We have a nice drawing to look at.*0260

*This is the region that we are interested in, in between here and here.*0263

*This is our 0, this is our 6, therefore, the area is going to be the integral from 0 to 6, upper function - the lower function.*0268

*The upper function is 7x - x² - the lower function which is x.*0281

*I’m going to integrate along x so it is dx.*0290

*This is the integral from 0 to 6 of 7x – x.*0294

*I can simplify 6x - x² dx.*0299

*This is going to equal 6x²/ 2 – x³/ 3.*0306

*I'm going to evaluate this from 0 to 6.*0316

*My answer is going to be 36, that is it.*0319

*I put 6 into here.*0324

*I put 6 into here and I end up with 108 – 72, that is for the 6.*0342

*And then, -, 0 – 0 is for the 0, that gives me my 36.*0350

*Again, you can go ahead and actually evaluate it.*0358

*You can go ahead and use your calculator.*0361

*Remember, we looked at that earlier.*0363

*Just enter the function, do second calc.*0366

*Go down to integration, lower limit, upper limit.*0371

*You have got yourself your integral value.*0375

*In this particular case, it is really easy to evaluate without the calculator, but in case you want to do that.*0378

*You got 36.*0382

*Example number 2, we have x = y² – 3, x = e¹/2 y.*0388

*We have y = -1 and y = 2.*0394

*In this particular case, it is y that is the independent variable.*0397

*We are going to be looking at functions which are left to right, instead of up down.*0400

*Let us take a look at what it is that we have got going here.*0404

*x = y² – 3.*0410

*My y² graph looks like this, the -3 part means it is a shift to the left because it is a function of y.*0412

*It is with left, my graph is actually going to be 1, 2, 3.*0423

*It is going to look something like that.*0430

*x = e ^½ y, when y is equal to 0, e⁰ is 1, that means x is 1.*0435

*It is going to be one of my points.*0452

*Let me go ahead and mark 1, 2.*0455

*Let us go 1, 2, let us go 3.*0476

*When y = 2 it is 2/2, it is 1 e ⁺12.718.*0486

*When y = 2, I got to x 0.718, it is going to be somewhere around there.*0496

*Basically, what I’m going to get is this, it is not going to cross, it is asymptotic right there.*0508

*It is going to look something like that.*0518

*y = -1 that this line, y = 2 is this line.*0521

*The region that I'm interested in is going to be this region right here.*0532

*I have a better picture on the next page.*0539

*It is going to be something like that.*0542

*We are going to be integrating from -1 to 2.*0545

*We are going to be integrating along the y axis.*0549

*Our representative rectangle is going to look something like that.*0553

*I’m going to be integrating vertically.*0557

*We will say best to integrate along the y axis from -1 to 2.*0564

*Therefore, the area is going to be the integral from -1 to 2, right function - left function.*0586

*We are integrating along y.*0606

*Let us take a look at a better picture here.*0610

*Yes, this is the region we are interested in.*0612

*That, that, that, there.*0615

*We want this and we are integrating horizontally from -1 to 2.*0624

*Therefore, our area is going to equal the integral from -1 to 2.*0636

*The right function is going to be e ^½ y - the left function y² - 3 dy.*0647

*That is going to equal the integral from -1 to 2 of e ^½ y - y² + 3 dy.*0664

*That is going to equal 2 e ^½ y - y³/ 3 + 3y evaluated from -1 to 2.*0680

*I went ahead and I just did this via my calculator.*0702

*However, if you want to see what it actually looks like, when you plug 2 in here, you are going to get 3 terms.*0704

*-1 in here, you are going to get 3 terms.*0709

*I got 2e – 8/3 + 6, that is when I plug the 2 in, and then I subtract.*0712

*When I plug in the -1, it is going to be 2e⁻¹/2 + 1/3 – 3.*0722

*If I have done everything correctly, and my final answer that I got was 10.224.*0733

*That is it, nice and straightforward.*0740

*Right function - left function, in this case.*0743

*y = 1/x, y = 1/x³, x = 4.*0747

*This one should be reasonably straightforward, let us do this*0753

*Let us go 1, 1, 1/x.*0760

*Let me make this a little bit bigger.*0767

*I’m going to put the 1,1 right there.*0773

*I have got something like that, that is my y = 1/x, my 1/x³.*0775

*Something like that, that was my y = 1/x³.*0789

*If I go to 4, something like that.*0795

*It looks like I’m going to be integrating from 1 to 4.*0800

*I'm going to be integrating, this is one of my representative rectangles.*0806

*I'm going to be adding the rectangles horizontally, I’m integrating along the x axis.*0809

*We integrate along x from 1 to 4.*0818

*Let us take a look at a picture here.*0825

*Nice, better, looking picture.*0828

*This is the region, here is our 1, here is our 4.*0830

*Here is our representative rectangle and we are adding this way.*0837

*What I have got is the area = the integral from 1 of 4 of the upper function - the lower function.*0842

*The upper function is this one, that is the 1/x - 1/x³ dx.*0861

*It is going to be integral from 1 to 4 of 1/x dx - the integral from 1 to 4 of x⁻³ dx.*0863

*It is going to equal, this one is going to be ln(x) evaluated from 1 to 4, -x⁻²/ -2 evaluated from 1 to 4.*0876

*If I have done everything correctly, I get ln 4 - ln 1 - -1/32 - -1/2, this goes to 0.*0896

*When I add everything up, I get 0.9175.*0917

*Again, it all comes down to the same thing.*0924

*It is going to be the upper function – the lower function or right function - left function.*0926

*All you have to do is find the limits of integration.*0932

*The limits of integration are going to come from either ends that are given to you, in this case x = 4.*0935

*The x value or the y value for where the two graphs happens to meet.*0945

*That is the only thing that is going on.*0949

*15 – 2x², y = x² – 5, I mentioned the biggest difficulty here is actually drawing these functions out,*0955

*if you remember them from pre-calculus.*0962

*Let us go ahead and draw this out.*0965

*15 - 2x² - 2x² opens down, let us see where it actually hits the x axis.*0971

*We have got 15 - 2x² is equal to 0.*0979

*You have got 2x² = 15, x² = 15/2, that means x is going to be + or -2.74.*0986

*It is going to be, I have got a point here, a point here.*0999

*I have one parabola that looks like this.*1010

*I have my other parabola which is going to be the x² – 5.*1014

*x² - 5 = 0, x² = 5, x = + or -2.24.*1019

*2.24 is like there and there, down to 5.*1030

*This parabola goes this way, this way.*1035

*I need the area in between those two curves.*1042

*It is going to be upper – lower.*1048

*Here is going to be one of my representative rectangles.*1051

*I'm going to integrate along the x axis.*1053

*My limits of integration, I need to know the x value of that point and the x value of that point, where those two meet.*1056

*Let us find that out, I find that out by setting the two functions equal to each other.*1061

*15 - 2x² = x² – 5.*1068

*I get 20 = 3x², x² = 20/3, x = + or -2.582.*1074

*This is 2.582, this is -2.582.*1089

*Those are going to be my limits of integration.*1093

*I'm going to take the integral, the area is going to be integral from - 2.582 to +2.582.*1096

*The upper function - the lower function dx.*1106

*Let us see what that looks like.*1115

*There you go, here is your -2.582, here is your +2.582.*1118

*I have got my area is equal to the integral -2.582.*1133

*You are definitely going to need to use a calculator for this.*1140

*2.582 of the upper function which is 15 - 2x² - the lower function which is x² - 5 dx, *1143

*which is equal to the integral -2.582 to +2.582.*1158

*It looks like I have got 20 - 3x² dx.*1165

*This is going to equal 20x - x² evaluated from that to that.*1186

*When I put the values in, I get 68.853.*1201

*There you go, that is the area between the curves.*1209

*Nice, simple, straightforward, not a problem.*1218

*x = 1/8 y³, x = 6 - y², let us go ahead and draw this out.*1226

*Again, it is probably best to just go ahead and use your calculator or something like www.desmos.com.*1239

*What you end up getting is, 6 - y² this is going to be a graph that looks like this.*1246

*x = ½ y³ is going to look something like this, something like that.*1254

*We are looking at this region right there.*1263

*It looks like it is best to go horizontal rectangles, which means we are going to integrate from bottom to top.*1269

*We want to integrate along y, we need the y value of that point and we need the y value of that point.*1281

*That is going to be our lower limit to our upper limit of integration.*1292

*Let us go ahead and do that, find that first.*1296

*We set the two functions equal to each other.*1298

*1/8 y³ = 6 - y².*1302

*I get y³ + 8y² - 48 is equal to 0.*1309

*We use our calculators, we use our software, we use Newton’s method, *1319

*whatever it is that we need to do to find the values of y to satisfy that.*1323

*We end up with two values, y = 2.172 and y = -3.144.*1327

*This value is our 2.172, that is our upper limit of integration.*1343

*This value is our -3.144, that is our lower limit of integration.*1348

*Our area is going to equal -3.144 to 2.172, right function - left function.*1356

*We are integrating along y, this is going to be, dy.*1374

*Here is the region that we are discussing.*1381

*In between there, that is our y value of 2.172.*1390

*Here is our y value of -3.144.*1400

*What have we got?*1407

*Let me go ahead and write it down here.*1408

*We have the area is equal to the integral -3.144 to 2.172, right function - left function.*1412

*I have got 6 - y² – 1/8 y³ dy.*1424

*I get 6y - y³/ 3 – y⁴/ 32.*1436

*I evaluate this from -3.144 to 2.172.*1452

*When I put it into a calculator, I get 20.479.*1459

*Nice and straightforward.*1468

*Let us do something that involves some sine and cosine.*1475

*We have y = cos x, y = sin 2x.*1478

*I would like you to integrate this or find the area the region between those two curves, between 0 and π/2.*1481

*Let us go ahead and draw this out.*1490

*Let us stay in the first quadrant here.*1494

*That is okay, I do not need to make it quite so big.*1497

*y = cos x, period is 2π, it is going to look something like this.*1503

*This is π/2.*1512

*y = sin(2x), the period is π.*1517

*If the period is π, that means it is going to start at 0 and*1521

*it is going to hit 0 again at π/2 that means at π/4, it is going to hit a high point.*1526

*Let us go ahead and call this 1.*1532

*We are going to get something like that, the region that we are interested in, this region right here.*1535

*What is the area of that region?*1550

*Notice, here this function, this is our cos x.*1554

*This function is our sin 2x, from 0 to some value which I will call a for now, cos x is above sin x, that is upper – lower.*1561

*Pass this point, it is the sin 2x that is actually the upper function and the cos x is the lower function.*1573

*I'm going to have to break this up into two integrals.*1581

*Integrate from 0 to a, cos x - sin 2x, from a to π/2, sin 2x - cos x.*1583

*We need to find what a is first.*1592

*How do you find what a is equal to?*1596

*You set the two functions equal to each other and you solve for x.*1598

*I have got cos(x) = sin(x), cos(x) -, sin(2x) there is an identity.*1602

*Sin(2x) = 2 sin x cos - 2 sin x cos x = 0.*1615

*I have moved it to the left and used my identity.*1624

*I’m going to factor out a cos x.*1626

*Cos x, 1 – 2, sin x = 0 that gives me two equations.*1629

*Cos x = 0 and sin x = ½.*1637

*Cos x = 0 that is going to be my π/2.*1643

*I already know that they meet there.*1647

*This value right here, sin(x) = ½.*1649

*Between 0 and π/2, my x = π/ 6, that is equal to my a.*1653

*My a is equal to π/6.*1660

*Therefore, I’m going to be integrating from 0 to a, 0 to π/6.*1664

*I’m going to be integrating from π/6 to π/2.*1674

*This first integral, second integral.*1681

*Let us go ahead and see what this looks like.*1684

*The region that I’m interested in is, a that is going to be my first integral.*1688

*This is going to be my second integral.*1699

*The red one is my cos(x), the blue one is my sin(2x).*1703

*Therefore, what I have got is area is equal to the integral from 0 to π/6 cos x - sin(2x) dx *1711

*+ the integral from π/6 to π/2.*1726

*Now it is sin 2x - cos x.*1732

*Nice and straightforward, the rest is just integration.*1741

*This is going to be sin x + ½ cos 2x evaluated from 0 to π/6 + *1743

*this is going to be -1/2 cos(2x) - sin x evaluated from π/6 to π/2.*1761

*When you actually evaluate this which I would not do here, you are going to get ½.*1776

*If I have done all of my arithmetic correctly, or if my calculator do the arithmetic correctly.*1781

*y = 2x² 10 x² 7x + 2y = 10.*1795

*You know what, this example, I think I'm actually just going to go through really quickly,*1800

*just run through it because I think we get the idea now.*1805

*Do not want to spend too much time hammering the point.*1809

*This is what the graph looks like.*1812

*Here is your graph of the y = 10x².*1815

*This is the 2x².*1822

*This time right here is the 7x + 2y = 10 which I have actually written as y = -7/2 x + 5.*1825

*You are going to solve it for y = mx + b, in order to graph it.*1836

*The region we are interested in is this region right here.*1840

*I’m going to go ahead and do this by breaking this up.*1849

*I’m going to do this is in two integrals.*1853

*I’m going to integrate along x.*1854

*I’m going to take the representative rectangle there for my first region and my second region.*1857

*I’m going to go from 0 to whatever this number is, which I will find in just a minute.*1864

*I'm going to go from this number to whatever this number is, for my second integral.*1870

*When I set y = 10x² equal to y = -7/2 x + 5, I’m going to end up this value right here.*1879

*My x value is going to be 0.553.*1893

*This point, I'm going to set my 2x² equal to my -7/2 x + 5.*1898

*This value I'm going to get is going to be 0.932.*1905

*Once I have that, the integral is really simple.*1917

*It is just going to be the area is equal to the integral from 0 to 0.553, upper function - lower function, *1920

*10x² - 2x² dx + 0.553 to 0.932 of the upper function which is -7/2 x + 5.*1934

*Because we are integrating with respect to x, it has to be a function of x - the lower function.*1955

*From here to here, that is the lower function which is the 2x² dx.*1963

*When I evaluate this integral and solve, I get 0.9341 as my area, that is it.*1968

*This is region 1, this is region 2, I just broke it up.*1980

*I have to find the x value for there, the x value for there, integrate from 0 to that point first *1983

*and that point first, upper – lower, upper – lower.*1990

*That is it, just do what is exactly what you think you should do.*1993

*It is very intuitive.*2000

*Let us go ahead and talk about this problem now.*2006

*Cars A and B start from rest and they accelerate.*2008

*The graphs below show their respective velocity vs. time graph × along the x axis, *2012

*the velocity of the cars is along the y axis.*2018

*Car A is the blue graph, this is A.*2022

*The graph, the value of the function is, this function right here is √2x.*2025

*B is the red graph, this is car B.*2035

*Its function, when expressed as a function of x is 1/5 x³.*2039

*First question we are going to ask, at 2.187 minutes where the graphs meet,*2047

*this point right here, this is our 2.187, which car is further ahead?*2052

*This is a velocity vs. time graph, when you integrate a velocity vs. time graph,*2061

*in other words, when you find the area under the curve up to a certain time, that gives you the total distance traveled.*2067

*Because again, velocity is, let us say meters per second, the differential time element is dt, meters per second.*2075

*The integral of v dt, velocity is expressed in, let us just say it is meters per second.*2088

*Time is expressed in second.*2096

*When you multiply those and add them all up, in other words integrate from 0 to 2.187,*2099

*you are going to get the area under the curve for car A.*2105

*You are going to get the area under the curve for car B, whichever area is bigger, that has gone further.*2108

*Clearly, car A has the bigger area, the area under all of the blue graph is a lot more than the area under the red graph.*2117

*Therefore, part A is really simple, it is car A is further ahead.*2127

*Car A is further ahead and this is further ahead because*2134

*the area under its graph is a lot more than area of the graph for car B which is just that right there.*2140

*If we shaded the region between the graphs from t = 0 to t = 187, what would the shaded area represent?*2155

*Let me go to black.*2162

*If I shaded in the area between the graphs, in other words this area, if I shaded that area, what does that represent?*2164

*The area on the blue graph is the total distance the blue has gone.*2175

*The area under the red graph is the total distance the red car has gone.*2179

*The difference between them is just how much further car A is then car B.*2185

*Part B, I will just write it up here.*2193

*Part B is what does it represent?*2198

*The shaded region represents the distance A is farther from the farther from B.*2204

*That is all, area under the graph for A, area under the graph for B,*2218

*the difference between them is the area between the two graphs.*2225

*It is how much father A is than B.*2228

*Part C, at 4 minutes, which car is ahead, and D, at what time will the cars be side by side?*2233

*When we look at this graph, we see that A, it accelerates faster and steadies out.*2241

*B, accelerates slower, it is further behind but at some point it really starts to accelerate.*2248

*Eventually, it is going to catch up.*2255

*A is going to be ahead of be but at some point B, is going to pass A.*2256

*At 4 minutes, which car is ahead?*2261

*The question we are asking is at 4 minutes which has a greater area under its graph? 4.*2264

*Let us go ahead and work that one out on the next page with a graph.*2272

*We are going to integrate, we are going to find the area under the blue graph from 0 to 4,*2277

*and we are going to find the area under the red graph from 0 to 4.*2283

*We said that the blue graph was √2x and we said this one was 1/5 x³.*2288

*From car A which is the blue graph, we have the area is equal to the integral from 0 to 4 of √2x dx.*2299

*When I solve that, I get 7.543.*2319

*For car B, the area = the integral from 0 to 4 of 1/5 x³ dx.*2324

*When I solve that, I get 12.8.*2336

*Car B, at 4 minutes or 4 seconds, whatever the time unit is, now car B has actually past car A.*2340

*Car B is farther forward.*2347

*The last question asked, at what time are they side by side?*2355

*Side by side means it is going the same distance.*2361

*At what time are the two areas equal?*2364

*We need the area of car A to equal the area of car B.*2368

*The area of car A is going to be the integral from 0 to l.*2379

*L is the time that we are looking for of √2x dx is equal 0 to l, that is the time period.*2384

*From 0 to whatever time we are looking for, we are looking for l, 1/5 x³ dx.*2395

*This is going to be √2 × the integral from 0 to l of x ^½ dx.*2407

*It is going to equal 1/5, the integral from 0 to l of x³ dx.*2415

*This is going to be, I’m going to come up here, 2√2/ 3 x³/2 from 0 to l is going to equal x⁴/ 20 from 0 to l.*2426

*When I put l in for here, when I put l in for here, 0 on for here, 0 on for here, set them equal to each other and solve.*2451

*I get something like this, I get 2√2/ 3 l³/2 = l⁴/20.*2457

*I need to solve for l.*2472

*The equation that this gives me is l⁴/20, let us bring this over this side, set it equal to 0 – 2√2/ 3 l³/2 = 0.*2475

*When I use my mathematical software or my calculator, or whatever it is to find what l is, I get l = 3.24 seconds.*2498

*I hope that made sense.*2512

*I want to know, when are they going to be side by side?*2513

*Side by side means they have gone the same distance.*2517

*Distance is the area underneath their respective graphs.*2520

*I set the areas equal to each other, the areas are the integrals.*2523

*L is my time, the upper limit integration.*2528

*I’m solving for l, I get an equation in l.*2531

*I solve for l, I get 3.24 seconds.*2534

*I did this by using a particular graph.*2537

*Here is what the graph looked like.*2540

*That function that we ended up getting, that this function is our l⁴/20 – 2√2/ 3 l³/2 = 0.*2542

*I just want to know where it is equal 0, 3.24 seconds.*2556

*You can use a calculator, you can use a graph, find where it crosses the x axis, anything you want.*2560

*I hope that make sense, and that is areas between curves.*2567

*Thank you so much for joining us here at www.educator.com.*2572

*We will see you next time, bye.*2574

1 answer

Last reply by: Professor Hovasapian

Wed May 11, 2016 2:55 AM

Post by Acme Wang on April 27 at 03:02:36 AM

In Example II, why the antiderivative of e^(1/2y)equals 2e^(1/2y)?

PS: I am a super fan of your calculus class! It does help me a lot!!! Thank you!

1 answer

Last reply by: Professor Hovasapian

Sat Apr 23, 2016 6:45 PM

Post by Cam-Tuoi Dinh on April 22 at 08:53:12 AM

In example 8, why the unit of the answer for problem d has to be in second, not in minutes?

1 answer

Last reply by: Professor Hovasapian

Sat Mar 26, 2016 4:49 AM

Post by Sazzadur Khan on March 6 at 05:00:24 PM

On example 3, shouldn't the evaluated integral read 20x-x^3?