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

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

### AP Practice Exam: Section II, Part A Calculator Allowed

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
- Problem #1: Part A 1:14
- Problem #1: Part B 4:46
- Problem #1: Part C 8:00
- Problem #2: Part A 12:24
- Problem #2: Part B 16:51
- Problem #2: Part C 17:17
- Problem #3: Part A 18:16
- Problem #3: Part B 19:54
- Problem #3: Part C 21:44
- Problem #3: Part D 22:57

### AP Calculus AB Online Prep Course

### Transcription: AP Practice Exam: Section II, Part A Calculator Allowed

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

*Today, we are going to continue with our AP practice exam.*0004

*We are going to start the free response section.*0007

*Again, this is going to be the same thing.*0010

*For part of it, you are going to be allowed a calculator.*0012

*For part of it, you are not going to be allowed a calculator.*0015

*Let us jump right on in.*0021

*In case you need to pull this up, there is a link down below or let me write it over here.*0026

*It is going to be www.online.math.uh.edu/apcalculus/exams.*0032

*It would be nice if I actually wrote it, /exams.*0054

*You are going to be clicking on where it says section 2, written.*0058

*We are going to be doing version 2.*0064

*Version one is the same thing, it is just different numbers, different arrangement, different order.*0068

*But it is essentially the same exam.*0072

*Let us get started.*0074

*Question number 1, we have a particular region that is going to be bounded by a couple of functions.*0076

*In this particular case, y = -x² and y = 4x², and the y axis.*0086

*The first thing we want to do is go ahead and graph this.*0092

*We see what we are looking at.*0095

*We got y = 4x², it is going to be some parabola, something like that.*0105

*This is our y = 4x².*0111

*Then we have our y = -x² which is going to be a bump function.*0116

*I will go ahead and just mark that there.*0120

*It is going to be something like that.*0123

*This is our function y = e ⁻x², they are all going to look like that.*0127

*These little bump functions.*0134

*It is bounded by the y axis.*0135

*The particular region that we are interested in is that region right there, that little triangular looking thing.*0138

*Part A, find the area of s.*0147

*This is are our region s.*0151

*Find the area of s.*0157

*The first thing we want to do, what we are going to do is we are just going to take the difference in the two graphs.*0166

*We are going to take a little vertical strip and we are going to integrate from 0 to whatever this x value is.*0170

*We need to find where the two graphs meet, so that we have an upper limit for our integral.*0176

*Let us go ahead and do that.*0183

*We set 4x² equal to e ⁻x².*0184

*We get 4x² - e ⁻x² = 0.*0189

*What you are going to do is, again, calculator is allowed here, you are just going to use this to find the roots of this.*0195

*Find the root of this equation, where does it hit the x axis.*0201

*It is going to be this and this, those two.*0209

*We end up with the first root is equal to 0.4515.*0215

*The second root is going to be symmetric.*0221

*On the other side, it is going to be -0.4515.*0223

*Therefore, the area of this region is going to equal the integral.*0228

*S, this region right here.*0232

*That is just going to be from 0 to 0.4515 and this is going to be the length of this strip.*0236

*The length of this strip is going to be, this is some x value,*0245

*it is going to be the height of the top function - the height of the bottom function.*0248

*Height of the top function - the bottom function gives me this height.*0253

*I’m going to add all of those little strips together.*0258

*The top function is e ⁻x².*0261

*The bottom function is 4x² dx.*0264

*I get an answer of 0.300, that is it.*0268

*Nice and straightforward application of an integral to find the area between two curves.*0273

*Let us now move on to part B.*0285

*Part B wants us to find the volume of the solid that is generated when this region is rotated around the x axis.*0288

*Let us go ahead and redraw real quickly, just a small portion of it.*0297

*We have that and we have the bump function.*0301

*This was our region s.*0307

*What we are going to do is we are going to rotate that around the x axis.*0310

*We are going to end up with this region roughly.*0314

*We are taking this region, we are rotating it about the x axis.*0325

*We want to find the volume of that solid.*0329

*I’m going to go ahead and use washers because again, I want to integrate with respect to the x axis.*0334

*Let us recall what our functions were.*0340

*We had y = our outside function, this was our y = e ⁻x².*0342

*Our lower function there which is y = 4x².*0353

*When I take a little strip like a washer, I’m going to take that, I’m going to turn it this way so that I’m looking at it.*0359

*I’m going to be looking at, essentially that.*0369

*I want to take this area and I want to add up all the little volume elements.*0374

*I think I will call this radius 2, I think I will call this outer radius 1.*0385

*The volume is going to equal the integral from a to b, it is going to be from here to here.*0391

*It is going to be the 0.4515 of the area of this circle - the area of this circle which is going to give you this area.*0398

*It is going to give me the area of that × dx.*0409

*It is going to be π × r1² - π × r2² dx.*0415

*r1 = e ⁻x², r2 is equal to 4x².*0426

*r1 is that radius, r2 is that radius.*0432

*Therefore, what we have is the volume equals the integral.*0441

*I pull out the π, 0 to 0.4515 of e ⁻x²² - 4x²² dx.*0446

*When I solve that, I get 1.059, very straightforward.*0468

*Let us go ahead and go on to part C.*0476

*Now what they want us to do is, now what they say is, this region is the base of a solid,*0484

*where the cross section that is perpendicular to the x axis is an equilateral triangle.*0491

*Find the volume of the solid.*0499

*Let us go ahead and draw it again.*0500

*I have got a 4x² part, I have the x² part.*0506

*Basically, what they are telling me is this.*0513

*This is our region s that we have.*0514

*This function right here, this function, this was our y = e ⁻x².*0517

*This function right here was y = our 4x².*0526

*They are telling me that this is the base of a solid.*0531

*If I take a vertical slice perpendicular to the x axis, the cross section, if I were to take this and flip it,*0534

*in other words look at it from that direction, it is going to be an equilateral triangle.*0545

*What they are saying is this.*0550

*I’m going to take this and I’m going to turn it this way.*0552

*What I’m looking at is an equilateral triangle.*0556

*If I’m looking at it from that direction, I see a bunch of equilateral triangles getting bigger.*0566

*That is what is happening.*0571

*They want to know what the volume of the solid is.*0573

*This region is the base of the solid.*0575

*The cross section is going to be an equilateral triangle.*0577

*Let me draw my triangle a little bit better here.*0582

*In fact, I can draw my triangle over here.*0587

*I’m going to move over here, this, that, and that.*0589

*This right here is this right there.*0602

*That height is equal to e ⁻x² - 4x².*0608

*I’m going to draw out a perpendicular and I’m going to call that h.*0615

*H because this is an equilateral triangle, all of the angles are 60.*0619

*When I draw out the perpendicular, I get a 30, 60, 90.*0623

*This is a perpendicular, therefore h is equal to half of this thing × √3.*0626

*It is equal to half of e ⁻x² - 4x, √3.*0634

*The area of that triangle is equal to ½ the base × the height that = ½ the base is that thing,*0647

*e ⁻x² – 4x² × the height × √3/2 × e ⁻x² - 4x.*0661

*Volume = the area of this thing integrated from here to 0.4515.*0678

*I’m going to add up now all of these triangles.*0687

*I’m going to add up all the triangles.*0690

*The volume = the integral from a to b of the area × dx,*0692

*the integral from 0 to 0.4515 of the area which is now √3/4 × e ⁻x² - 4x²² dx.*0701

*I get a volume equal to 0.1037.*0728

*I hope that makes sense.*0735

*Let us go ahead and go on to problem number 2.*0740

*We are given, we have f(x) is equal to the integral from 0 to x of cos².*0748

*A function that is expressed in terms of an integral cos².*0760

*I’m changing my variables dt, on the interval from 0 to 3π/ 2.*0769

*What do we want here?*0782

*Number 2, they want us to approximate f(π) using a trapezoidal rule with n equal to 4.*0784

*Let us see what we have got.*0798

*f(π) = the integral from 0 to π cos² t dt.*0801

*The trapezoidal rule where n is equal to 4, that is going to equal Δ x/ 2 × f(x0) + 2 × f(x1) + 2 × f(x2).*0811

*You definitely want to write this out on your exam so they know what you are using.*0832

*+ 2 × f(x3) + f(x4), you see that you have five terms.*0835

*It is equal to 4, you have 1, 2, 3, 4, 5 terms, 0 to 4.*0843

*Δ x here, Δ x is equal to b - a/ 4.*0850

*It is going to be π - 0/4 = π/ 4.*0857

*Therefore your x0, x1, x2, x3, x4, are going to be 0, π/4, 2π/ 4, 3π/ 4, 1.*0863

*That is what is happening.*0872

*At x = 0, x(0) = 0, we have f(0) is equal to,*0876

*f is this, small f, this is small f.*0890

*We are finding F(π) which is an integral.*0899

*We are evaluating and approximating that integral with trapezoids.*0904

*F is that.*0909

*f(0) = 1² = 1, x1 = π/4, f(π/4) = 1/ √2² because the cos of π/4 is 1/ √2.*0910

*1/ √2² is ½, x2 = π/2, f(π/2) = 0² = 0, x3 = 3π/ 4, f(3π/4) = -1/ √2² which is ½.*0927

*x4 is equal to π, f(π) = -1, -1² is equal to 1.*0959

*Therefore, our approximation t4 is equal to π/4/ 2 × 1 + twice this, twice this, twice this, + that.*0970

*1 + 2 × ½ + 2 × 0 + 2 × ½ + 1 + π/8 × 4 = π/2.*0985

*Just a nice application of the trapezoidal rule, which I actually personally dislike very much.*1008

*I do not know why.*1012

*Part B, find f’(x).*1016

*f(x) = the integral from 0 to x of cos² t dt.*1021

*f’(x) is nothing more than cos² (x).*1031

*Straight application of the fundamental theorem of calculus.*1035

*C, average value of f’ on the interval.*1040

*They want the average value of f’ on the interval.*1048

*The average value = 1/ b – a, integral from a to b, f’ is cos² x dx 1/,*1055

*What was the particular interval, over the interval of 0 to π.*1075

*It is going to be π – 0, the integral from 0 to π, cos² x dx and at 0.5.*1079

*That is it, straightforward, nothing strange happening here.*1088

*Let us do question number 3 and we will done with the first half, very nice.*1095

*We have a particle that is moving along the x axis and its acceleration is given by, our acceleration function is 24t – 12.*1101

*Let me write this a little bit better.*1114

*It equals 24t – 12.*1117

*They are telling me that the velocity at 1 is equal to 8.*1121

*They are telling me that the position at 1 is equal to 10.*1125

*Part A wants us to find an expression for the velocity of this.*1132

*We know what the velocity is, the velocity of t is nothing more than the integral of the acceleration*1135

*which is the integral of 24t - 12 dt, which is going to give us 12t² - 12t + c.*1144

*They gave us a value of v1.*1160

*v1 which is going to equal 12 × 1², just plug in 1 for t - 12 + c.*1161

*They tell us that this equals 8.*1169

*When we solve this, we get c = 8.*1172

*12 – 12, we get c = 8.*1178

*Therefore, we plug this back into there.*1180

*Our velocity function which is what we wanted is going to be 12t² - 12t + 8.*1183

*There we go, that is part A.*1191

*Part B, at what values of t does the particle change direction?*1197

*The particle changes direction when the velocity goes from positive to negative.*1206

*The particle changes direction when v(t) changes sign, positive to negative or negative to positive,*1220

*where it crosses the x axis completely, not touches it.*1240

*Find the root of v(t).*1252

*When I plug this in my calculator and I check the roots, there are no real roots.*1259

*Does not change direction, the particle does not change direction.*1272

*It might slow down, then speed up again, but it does not change direction.*1291

*Now part C, now they want us to find an expression for the position of the particle.*1300

*We know what that is.*1307

*We know that the position of the particle is nothing more that the integral of the velocity function.*1309

*It is going to be the integral of 12t² - 12t + 8 dt.*1315

*We are going to end up with 4t³ – 6t² + 8t, again + another constant.*1325

*They told us what x sub 1 is, what the position at t = 1 is.*1337

*It is going to be 4 × 1³ - 6 × 1² + 8 × 1 + c.*1342

*They tell me that that = 10.*1351

*When I solve this, I get c is equal to 4.*1353

*I plug that back in, therefore my position function = 4t³ – 6t² + 8t + 4.*1356

*That is what I want.*1374

*D, the total distance traveled from t = 1 to t = 4.*1377

*The particle does not change direction.*1389

*I do not know what is the best way to do this?*1399

*Total distance traveled.*1403

*It does not change direction.*1413

*I did it one way but now that I'm sort of sitting here and thinking what is the best way to do this.*1419

*The total distance traveled by the particle.*1424

*The velocity × time gives me total distance.*1433

*That is fine, I can just go ahead and take,*1436

*Our total distance traveled, because it did not change direction, I do not have to split up the,*1439

*From t = 1 to t = 4 because it does not change direction,*1455

*I do not have to break it up into moving to the left, moving to the right, things like that.*1460

*I can just go ahead and I can find the distance at t = 4.*1465

*I can find the distance at = 1 which is equal to 196, which is equal to 10,*1470

*and I can just take the difference which is going to be essentially the same as the total distance traveled.*1478

*If you want, you could have done it this way, the total distance.*1489

*Distance = rate × time, rate is velocity, velocity × time.*1493

*Because velocity is not constant, the total distance = the integral from 1 of 4 of the velocity function dt.*1499

*You can go ahead and put the velocity function which is this thing into here and integrate that,*1507

*which is the same essentially as this.*1513

*Because when you integrate the velocity function, you are going to get the distance function.*1515

*When you evaluate the distance function at 4, this is function at 1, you are going to subtract.*1518

*It is the same thing, however you want to look at it.*1523

*That is it for this part, where the calculator was allowed.*1528

*The next part is going to be another three questions where no calculator is allowed.*1531

*Thank you so much for joining us here at www.educator.com.*1536

*We will see you next time, bye.*1538

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