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For more information, please see full course syllabus of AP Calculus AB
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More Example Problems, Including Net Change Applications

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: Evaluate the Following Indefinite Integral 0:10
  • Example II: Evaluate the Following Definite Integral 0:59
  • Example III: Evaluate the Following Integral 2:59
  • Example IV: Velocity Function 7:46
    • Part A: Net Displacement
    • Part B: Total Distance Travelled
  • Example V: Linear Density Function 20:56
  • Example VI: Acceleration Function 25:10
    • Part A: Velocity Function at Time t
    • Part B: Total Distance Travelled During the Time Interval

Transcription: More Example Problems, Including Net Change Applications

Hello, and welcome back to, and welcome back to AP Calculus.0000

Today, we are going to be doing some more example problems including net change applications.0004

Let us jump right on in.0010

Evaluate the following indefinite integral, straightforward application, anti-differentiation,0012

just add c at the end because we do not have lower and upper limits.0019

It is going to be just nice and straightforward.0025

This going to be x⁴/ 4, it is going to be + x⁻¹/ -1 + c.0030

When we simplify, we get x⁴/ 4.0041

Let us write this as –x⁻¹.0045

You are welcome to write it as 1/x if you like, not a problem, +c.0049

Evaluate the following definite integral.0061

We are going to go ahead and separate this out because we have a 2 + a sum on the top and a single entity on the bottom.0064

We can go ahead and write this as the integral from 0 to π/3 of 2/ cos² θ0072

+ the integral from 0 to π/3 of 1, which is just cos² θ/ cos² θ.0087

Sorry, I forgot my dθ, dθ there.0097

This is just the integral from 0 to π/3 of 2 × sec² θ + the integral from 0 to π/3 of dθ.0103

The integral of 2 sec² θ is 2 × tan(θ).0119

We are going to evaluate that from 0 to π/3.0125

This is just going to be θ evaluated from 0 to π/3.0129

This is going to be 2 × tan(π/3), we have got √3.0136

The tan(0) is 0 + π/3 – 0.0144

Our final answer is 2 √3 + π/3.0155

That is all, straight application of the fundamental theorem of calculus.0163

The second fundamental theorem, find the antiderivative, evaluate the lower limit - the upper limit.0166

If water is draining from a tank at f(t) m³/min.0181

We have got m³/min.0191

When we are dealing with physical applications, units are going to be very important.0194

In fact, units will often tell you in which direction you want to move, what the answer is,0198

what you are going to be multiplying by what.0206

Watch the units very carefully.0208

If water is draining from a tank at f(t) m³/min, then what does the following integral represent?0210

The integral from 10 to 30 of f(t) dt.0217

Let us take a look at this.0222

This f(t) is our function and it is m³/min.0223

This dt is just the differential time element.0227

Its unit is just time, minute.0230

We have f(t) m³/min and t is in minutes.0239

dt is just a differential time element, it just means it is really tiny time element.0250

Just a small, I will put that in quotes, very small time element.0260

That is it, it is just minute.0266

f(t) dt, the integrand, that is equivalent to m³/min × minute.0273

The minute cancels, the integrand alone gives me something which is volume, m³.0286

What it gives me is a differential element of volume.0299

In other words, f(t) dt is just equal to some small volume element.0304

If f(t) m³/min is draining from a tank, if I multiply it by that really tiny minute, that multiplication, the integrand alone,0318

without having done the integration gives me the tiny little bit amount of volume that is drained out.0328

Integration is just the sum, if I add up all of these little volumes,0334

I'm going to get the total volume that is drained out between 10 min and 30 min.0339

That is what is happening here.0344

Remember, integration just means sum, just means add everything up.0351

The integral from 10 to 30 of f(t) dt is the volume of water0364

that drains out of the tank between 10 min and 30 min, after you have opened the valve.0380

What we have done is this, we said that a differential volume element is equal to the function × this differential time element.0419

Because this gives us cubic meters, that is a differential volume.0428

When I integrate both sides, the integral of dv is just v.0433

The integral of f(t) dt from 10 to 30 gives me my total volume for that time period, from 10 to 30.0438

I add up all the little time elements, that is all I'm doing with integration.0452

That is all integration is, it is just a long sum of a bunch of tiny numbers.0458

That is all hope, I hope that make sense.0462

For the following velocity function in meters per second,0471

find the net displacement over the given time interval and the total distance traveled by the particle.0474

We are just talking about one dimensional motion.0481

It is a particle to start someplace and it starts going this way, that it might go this way, that it might go this way.0483

We do not know how many times it goes this way, that way.0492

At some point, it is just going to stop at certain time interval.0494

Wherever it is, from where it started, that is the net displacement.0499

The total distance is how far it actually traveled all together, in order to get to its net displacement.0504

That is all that means.0512

If I start at my room and I go to my kitchen, and I come back to my room,0513

my net displacement is 0, because I started at my room, I ended at my room, I have not moved.0518

My total distance is the distance to the kitchen and the distance from the kitchen back, two different things.0524

Net displacement, where you start and where you end, nothing in between.0531

Total distance is the distance you traveled, in order to where you are trying to get to.0535

This is a velocity function that they gave us.0544

Our s(t) is our displacement function.0547

You remember, when you have a displacement function s(t), s’(t) is equal to the velocity function.0565

s” which is equal to v' that is equal to the acceleration function, that is the relationship.0575

They gave us a velocity function, let us integrate this to find what the displacement function is.0584

Tell me where the particle is at a given time t.0590

s(t), our displacement function is equal to the integral of our velocity function, 2t² - 3t - 6 dt.0598

This is going to equal s(t) is equal to 2t³/ 3 - 3t²/ 2 - 6t + c.0612

This is an indefinite integral.0627

We are going to be starting at some point.0631

Before we move at time 0, we are staring some place.0635

Let us set that place as the origin, we will just call that our 0 point.0638

s = 0, we set s(0) when t = 0, we set it equal to 0 where we are on the x axis.0645

Then, s(0), I put 0 in for all of these, I get 0 - 0 - 0 + c.0660

We said that s sub 0 is 0, that implies that our constant is equal to 0.0673

Therefore, we can finally write our final function, s(t).0681

We found our constant is equal to 2t³/ 3 – 3t²/ 2 – 6t, that is our displacement function.0685

Net displacement means where are we when t is equal to 4, the end of our time interval.0700

Let us go ahead and do that, let us find out where exactly we are.0725

We have got s(4), that is going to equal 2 × 4³/ 3 - 3 × 4²/ 2 - 6 × 4.0730

We are going to get -5.33, I’m going to make this a little clear, it looks like a 6.0748

-5.33 that is our net displacement.0758

What that is telling me is, here is my 0, I have got 1, 2, 3, 4, 5, 6.0763

I’m about right there, -5.33.0774

After 4 seconds has past, I can be located right there.0778

We have done it, in my net displacement I’m at -5.33.0786

I’m 5.33 units away from where I started which was our starting point, our origin.0789

Part B, the total distance traveled.0796

Here is where I am at t = 4 seconds.0801

I will do this in blue.0809

That is fine, I will do it over here.0814

The particle could have gone like this.0815

I could have started at 0 and I could have gone this way, and then come over here to end up at -5.33.0826

I could have started at my origin, I could have gone this way really far, and come back and ended up at -5.33.0834

Or I could have started and gone a bunch of times and ended up at -5.33.0844

Net displacement just tells me where I am in a certain time.0853

It does not tell me how I got there, what path did I follow, what was the total distance?0857

Now we need to do that.0861

In order to do that, we need to find out, when velocity is positive, I'm moving in the positive direction.0865

When velocity is negative, I'm moving in the negative direction.0872

Between my time interval 0 to 4 seconds, from 0 to 4 seconds,0875

I need to know when I was traveling to the right, positive velocity.0882

When I was traveling to the left, negative velocity.0886

Let us do that.0891

Let us find where the velocity is greater than 0 and where the velocity is less than 0.0897

In other words, when is it moving to the right, when is it moving to the left?0926

The velocity function that is equal to 2t² - 3t - 6 is equal to 0.0931

We are going to find the values of t where it equal 0.0942

We are going to find it where it is less than, where it is greater than.0947

This is not factorable.0950

Because it is not factorable, we can either use Newton's method to find where it equal 0,0957

we can use the quadratic formula to find where it equal 0.0961

Or we can just graph it using a graphing utility, your graphing calculator or some application on the internet,0965

some mathematical software, to let us know where this graph actually hits 0.0971

Now when I solve it this way, I use a graphing utility.0977

I just graph it and see where it hits.0980

v(t) is equal to 0, when I get 2 points, when t is equal to -1.137 which I'm not going to use.0984

The reason I'm not is because our time interval is from 0 to 4.0995

We are not interested in negative time values, when t is equal to 2.637.0999

Our velocity function = 0, when t is equal to 2.637.1010

When I check values from 0 to 2.637, my velocity is actually going to be less than 0.1017

The particle is moving to the left, from 0, 1 second, 2 seconds, to 2.637 seconds, the velocity,1032

that particle is actually moving to the left from our origin.1039

This is moving left.1047

And then from 2.637 all the way to 4, the end of our interval, the velocity is going to be greater than 0.1052

I check this, this is the whole point.1064

Once I know where it hits 0, I check to the left of that point and I check to the right of that point,1066

to see which values of t to c, I put them in here.1070

2.637, let us say put in a 2 in here.1074

For when I put 2 in the original velocity function, it is going to give me a negative number.1077

Which means from 0 to 2.637, I'm negative.1081

When I check a number like 3, 4, 5, anything bigger than 2.637,1084

when I put it into here, the velocity ends up being greater than 0.1089

That is how I know this.1092

That means the particle is moving to the right.1095

I already know what the displacement is at 4, it was -5.33.1103

I know where I am when time = 0, I’m at the origin.1110

I need to find out where I am at 2.637 seconds.1114

s(2.637 seconds), when I put into my s function which is 2 × 2.6377³/ 3 - 3 × 2.6377²/ 2 - 6 × 2.6377.1121

That was my s function, it is -10.55.1151

At 2.637 seconds, I’m at -10.55.1157

This is what happened, here is 0.1163

I have ended up going -10.55 in those 2.637 seconds.1170

I already know where I am at 4 seconds.1180

At 4 seconds, I’m over here at -5.33 which means from 2.627 seconds to 4, I turn around and I come back this way.1184

I already know that, that I’m going to the right because from 2.637 seconds,1194

that is when the velocity equal 0, that means it stopped.1199

The velocity becomes positive after that, it means I’m moving to the right.1202

The total distance that I traveled is 10.55, this distance.1208

This distance + that distance, from here to here, and from here to here.1223

This distance is 10.55 + 10.55 - 5.33.1229

10.55 - 5.33 that gives me that distance right there.1240

My total distance is 15.77, I hope that make sense.1247

Let us do another one.1256

The following is the linear density function in g/m of a 5m rod, find the total mass of the rod.1260

Let us see what we have got.1270

I have this rod, let us make it a cylindrical rod, 5m long.1272

They are telling me that the linear density is this.1286

Linear density means, as x gets bigger and bigger, the density changes where it is.1292

In other words, the density here is different than the density here,1301

it is different than the density here, because it is a function of x.1303

This is the 0, this is the 5 mark, I will go ahead and set this as the origin.1311

The density is in grams per meter, I have got g/m.1318

If I multiply by meter, meters cancel, I'm going to be left with grams which is what I’m looking for.1329

I’m looking for the total mass.1337

Therefore, my differential mass element DM, my differential mass element is equal to my density function 14 + 3 × x¹/31340

which is in g/m × the differential length element dx.1364

That is going to give me the mass of that little element.1375

When I add all of the little masses together, in other words, when I integrate, I get the total mass.1379

I integrate both sides.1387

Integrate both side which means nothing more than add up all of the differential mass elements to find the total mass.1397

The integral of dm is just m.1429

The total mass is equal to the integral from 0 to 5 of this 14 + 3x¹/3 dx = 14x +,1439

that is fine, I will just go ahead and write it out.1465

3 × x⁴/3 / 4/3 from 0 to 5 = 14x + 9x⁴/3 / 4 from 0 to 5.1468

When I do that, I get a final answer of, it is going to be, I put the 5 in here, I put the 0 in here.1487

I get 89.23 - 0 = 89.23 grams.1495

That is all, I hope that make sense.1504

An acceleration function and an initial velocity over a particular time interval are given below.1515

Find the velocity function at time t, find the total distance traveled during the time interval.1524

We have our acceleration function 3 sin 2t – 5.1532

We have an initial velocity of 2 and we are looking at a time interval from 0 to 6.1535

We already now that our displacement function is s(t).1542

When we take the first derivative of s(t), we get our velocity function.1550

When we take the second derivative of our displacement function1555

which is the first derivative of the velocity function, we get our acceleration function.1558

Here they gave us the acceleration function and they want us to find the velocity function.1564

They want us to find this.1569

We go backwards, we integrate up.1570

Therefore, our velocity function v(t) is equal to the integral of the acceleration function dt,1574

which is equal to the integral of 3 × sin(2t) – 5.1584

That is going to equal -3/2 × cos(2t) - 5t + c.1594

v(0) which means we put 0 in for here, v(0) is equal to 2.1613

v(0) = -3/2 × cos(2) × 0 - 5 × 0 + c.1622

They set that that = 2.1634

I’m going to get, is equal to -3/2 - 0 + c is equal to 2, that is going to give me c = 7/2.1639

I plug that back in to here.1659

Therefore, I get a velocity function.1669

My velocity function is equal to -3/2 × cos(2t) - 5t + 7/2.1673

This was the answer that I was looking for.1685

My velocity function at time t.1688

I integrate the acceleration function.1691

If you get a velocity function and they want to find the displacement function, you integrate the velocity function.1693

If you get an acceleration function and you want to go to displacement function, you integrate twice.1699

I found the velocity function, now I integrate this one up one more time.1704

Now part B, what am I going to do for part B?1709

The total distance traveled during the time interval.1715

Part B, once again, we want the total distance travelled during the time interval.1720

The total distance, we need to know when it is moving to the right, when it is moving to the left.1725

We need to know where the velocity function equal 0.1731

Because before that it is going to be positive or negative, after that it is going to be positive or negative.1734

At that point, it is going to be 0.1738

In other words, it is going to turn and go in the other direction.1740

We set the velocity function equal to 0 to find out where it is bigger than 0, where it is less than 0.1746

Let us go ahead and do that.1754

I did that graphically.1756

I went ahead and I graphed the function -3/2 cos(2x) 2t - 5x + 7/2.1759

I got this little bit of a graph, closed in on it.1765

I found out from 0 to 0.58, the velocity function is above the x axis, it is positive.1769

From 0.58 all the way to 6, the velocity function is negative.1778

Let us go ahead and put this over here.1785

From 0 to 0.58 seconds, our velocity function is going to be greater than 0, which means the particle is moving to the right.1788

And then from 0.58, which is our 6 seconds, I believe it was, our velocity function is less than 0.1799

I can see it right off the graph.1811

We need to go ahead and find our s(t).1814

Our s(t), our displacement function which is equal to the integral of the velocity function,1822

which is equal to the integral of this, our velocity function.1829

-3/2 cos(2t) -5t + 7/2 dt.1836

I get an s(t) is equal to -3/4 × sin(2t) – 5t²/ 2 +7/2 t + c.1860

If we take our starting point as the origin, in other words, if s at time equal 0 = 0,1878

we are going to end up with c = 0.1893

That is fine, I will just go ahead and do it over here.1899

We end up with, s(0) = -3/4, sin(0) is 0.1901

0 is 0 + 0 + c is equal to 0.1913

Therefore, c is equal to 0, therefore, it just drops out.1919

We are left with our displacement function is equal to -3/4 × sin(2t) – 5t²/ 2 + 7/2 t.1923

From 0 to 0.58, s(0.58) is equal to 0.501 which means that starting at the origin, 0.5 seconds later,1955

we had gone to the point 0.501.1974

From 0.58 all the way to 6, we need to know where s(6) is.1984

At 6 seconds, where are, displacement wise.1993

s(6) is equal to -68.59, way out here some place.1998

I will just go ahead and put it like right there.2009

Basically, I have turned around and I end up over here - 68.59.2012

My total distance, the 68.59 is from this mark.2022

My total distance is, I went this far 0.501 +, and I went back 0.501, and then I went further 68.59.2030

That is all, thank you so much for joining us here at

We will see you next time, bye.2061