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

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

### 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

### AP Calculus AB Online Prep Course

### Transcription: More Example Problems, Including Net Change Applications

*Hello, and welcome back to www.educator.com, 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 θ *0072

*+ 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 water *0364

*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 × x¹/3*1340

*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 function*1555

*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 www.educator.com.*2056

*We will see you next time, bye. *2061

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