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Lecture Comments (6)

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?

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 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?

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 this0753

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 and1521

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 because2134

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