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

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

## Discussion

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## Table of Contents

## Transcription

### AP Practice Exam: Section 1, Part A No Calculator, cont.

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 #15 0:22
- Problem #16 3:10
- Problem #17 5:30
- Problem #18 8:03
- Problem #19 9:53
- Problem #20 14:51
- Problem #21 17:30
- Problem #22 22:12
- Problem #23 25:48
- Problem #24 29:57
- Problem #25 33:35
- Problem #26 35:57
- Problem #27 37:57
- Problem #28 40:04

### AP Calculus AB Online Prep Course

### Transcription: AP Practice Exam: Section 1, Part A No Calculator, cont.

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

*Today's lesson is going to be a continuation of the Section 1 Part A of the AP practice exam.*0004

*The one with no calculator allowed.*0010

*Let us continue on.*0012

*We finished off with question number 14, in the last lesson.*0014

*Now we are going to start with question number 15.*0017

*Let me go ahead and go to blue.*0020

*Question number 15 is asking us to find the average value of a given function over a particular interval.*0024

*The function at our disposal is 2x + 3³.*0031

*The particular interval over which we want to take the average value is -3 to -1.*0043

*The average value is really easy to calculate.*0050

*It is just, basically, the integral of the function divided by the length of the interval.*0054

*It is usually written as 1/ b – a.*0061

*A is the first number, b is the second number, × the integral from a to b of gx dx.*0063

*It is a simple evaluation of integral.*0073

*This is going to be 1/ 1 – a - 3 × the integral from -3 to -1 of 2x + 3² dx.*0076

*I think it was squared, was not it cubed?*0092

*Yes, squared not cubed.*0095

*I’m just going to go ahead and multiply this out.*0099

*= ½ × the integral from -3 to 1 of, this is going to be 4x² + 12x + 9 dx.*0102

*That is going to equal ½ 4x³/ 3 + 12x²/ 2.*0118

*That is going to be + 6x² + 9x all evaluated from -3 to -1.*0132

*When we do the evaluation, we get ½.*0141

*This is going to end up being.*0145

*I would not go through all the stuffs, I will just do what it is that we have got.*0151

*= ½ - 13/3 + 27/3.*0159

*You are going to end up with a value of 7/3 which is going to be c.*0168

*That is it, just evaluate the definite integral*0173

*Here there is no particular technique to use, just straight application of antiderivatives.*0177

*Number 16, what does number 16 asks us?*0184

*Excuse me, let me move that over.*0189

*Number 16 is asking us to evaluate a limit.*0195

*In this particular case, it is the limit as t goes to 0 of the quotient of tan(π/4) + t – tan(t)/ t.*0198

*The important thing here is instead of actually evaluating limit, you are not going to evaluate the limit.*0226

*You are going to recognize that this limit is the definition of derivative for a particular function.*0231

*This limit is going to be the derivative evaluated at π/4.*0238

*In this particular case, it is going to be y =, this is the derivative of the tangent function evaluated at π/4.*0250

*You just want to recognize that is what it is, instead of actually evaluating this limit.*0260

*Let us write that down.*0266

*Recognize this as the definition of the derivative.*0269

*That is all this is, definition of the derivative of the tan(x) at x = π/4.*0274

*We have y is equal to tan(x).*0295

*We have y’ of tan(x) which is sec² (x).*0298

*y’(π/4) = sec² of π/4 which is equal to √2² which is equal to 2.*0304

*That is choice e.*0319

*Just recognize that that is the definition of derivative.*0322

*Let us see, that is number 16, let us go ahead and go to number 17.*0328

*Let us see what number 17 is asking us.*0335

*17 is giving us a particular function, a function of t.*0337

*They want us to find the instantaneous rate of change at t = 0.*0342

*Nice and straightforward, instantaneous rate of change.*0349

*We know that an instantaneous rate of change is the derivative.*0351

*Our derivative is the slope, it is the rate of change.*0355

*The slope of the secant line is equal to the average rate of change.*0362

*The slope of the tangent line which is the derivative is the instantaneous rate of change.*0365

*Let me write number 17, our function of t is equal to 2t³ - 2t + 4 × √t² + 2t + 4.*0372

*We want to evaluate f’ at 0.*0393

*Let us go ahead and f’.*0401

*f’(t), it is going to be product rule.*0404

*It is going to be a little tedious and long, but that is not too big of a deal.*0409

*We have got 2t³, this × the derivative of that + that × the derivative of this.*0413

*-2t + 4 × ½ t² + 2t + 4⁻¹/2 × 2t + 2 + t² + 2t + 4.*0420

*Instead of running the radical sign, I will go ahead and write it that way.*0445

*× the derivative of what is inside here which is going to be 6t – 2.*0447

*Of course, we just plug in t, we do not have to simplify this.*0453

*We can just plug in our particular 0.*0457

*When you plug 0 in to t for all of this and evaluate that, you are going to end up with 2 - 4 = -2.*0461

*That is our answer and that is choice b.*0471

*Nice and straightforward.*0476

*Let us try number 18.*0482

*Let us see, what is number 18 asking us to do?*0487

*It wants us to calculate, it wants us to do ddx of 11 ⁺cos(x).*0491

*We know that the derivative of some constant e ⁺u is equal to e ⁺u × ln(a) × du.*0507

*If we have 11 ⁺cos(x) and we subject it to the differential operator, in other words, take the derivative of it.*0527

*It is a fancy term for that.*0535

*We subject it to differentiation.*0537

*We get the derivative of 11 ⁺cos(x) is equal to 11 ⁺cos(x) × natlog of 11 × the derivative of the u.*0539

*Because u is a function of x, it is not just x.*0552

*The derivative of cos(x) is - sin(x).*0556

*That is it, basically, you just end up with -sin x 11 cos(x) ln(11) which is d.*0561

*Again, if you want to separate them out, so you can actually not get confused as to which is which with the symbols,*0576

*just put some parentheses around it, not a problem.*0581

*Great, nice and straightforward.*0584

*Those are the ones that we love.*0588

*Number 19, let us see what we have got here.*0591

*Number 19, we have a solid which is generated by rotating a particular region that is enclosed by these graphs and these lines.*0596

*We are rotating about the x axis and they want to know which one of these integrals*0610

*actually gives you the volume of the particular solid that is generated.*0615

*Let us see what we got.*0621

*We have y = √x and we have x = 1, x = 2, and y = 1.*0622

*Let us go ahead and draw this out.*0632

*Something like this, we know what the √x function looks like.*0637

*It looks something like that.*0643

*X = 1, let us just put it here.*0646

*x = 2, let us go ahead and put it here.*0650

*y = 1, that is this line right here.*0654

*This is our origin, y = 1.*0663

*I will just go ahead and put it right there.*0665

*It is the region that is bounded by all of these.*0671

*Bounded by x = 1, x = 2, the function y = √x, and the line y is 1.*0673

*This is the region that we are looking for.*0688

*That is the region that we are going to rotate around the x axis.*0689

*Basically, we want to come down here.*0694

*Now we have that one, we are going to have this.*0697

*This is the cell that is going to be generated.*0705

*We are going to have this solid, we want to find the volume of that solid or the integral representing that volume.*0708

*I think what I’m going to do here is I’m going to go ahead and use washers.*0717

*I'm going to integrate from 1 to 2.*0721

*I’m going to integrate in the horizontal direction.*0724

*I’m definitely going to be using dx.*0726

*My upper limit and lower limits of integration, my lower to upper is actually I’m going to go from 1 to 2.*0730

*Essentially, what I'm doing is I’m taking a little bit of a washer here.*0739

*It looks like this.*0744

*It is going to be something like this.*0754

*This is the region.*0756

*We are going to take the area of that washer and we are going to add up all the areas along the x axis.*0759

*Let us go ahead and call this radius the outer radius.*0774

*We will call this one the inner radius.*0778

*The volume is going to be the integral from a to b of an area element × dx.*0781

*We said we are going to integrate from 1 to 2, that is 1 to 2.*0792

*The area of this region is going to be π × the outer radius² - π × inner radius².*0797

*In other words, the big circle - the inner circle.*0805

*That is just π × the outer radius² - the inner radius² × dx.*0808

*Now we just need to know what the outer and inner radiuses are, as functions of x.*0818

*This is going to equal, I’m going to pull the π out of the integral sign, it is 1 to 2*0822

*The outer radius is, if this is my x value, the outer radius is √x.*0830

*The inner radius is 1.*0840

*It is going to be the outer radius, it is going to be √x²,*0842

*that is the outer radius² - the inner radius which is from here to here, from here to here, which is 1² dx.*0853

*We have π × the integral 1 to 2 of x - 1 dx.*0864

*That is our answer, it did not ask us to evaluate it.*0876

*This is going to be choice d.*0879

*In this particular case, we decided to use washers.*0883

*I’m going to stick with red, I guess it is kind of nice.*0888

*This is going to be question number 20.*0891

*There is no calculator involved in this particular section, the first section.*0895

*The first section is no calculator, the second section is there are going to be calculator.*0898

*When you do the free response questions, one of the sections is going to be no calculator,*0903

*the other one is going to be with calculator.*0908

*Let us see what 20 is asking us.*0912

*It is asking is to calculate a limit.*0915

*Let us see what we have got.*0918

*They want us to evaluate the limit as x goes to 0 of 4x/ sin(3x) + x/ cos(3x).*0919

*The limit of the sum is the sum of the limits.*0939

*Let me actually separate this out.*0942

*This is equal to the limit as x approaches 0 of 4x/ sin(3x) + the limit as x approaches 0 of x/ cos(3x).*0944

*This one right here, when I put 0 in, I get 0 sin(3x), sin(3) × 0, sin(0) is 0.*0962

*What I ended up with is 0/0.*0968

*This is an indeterminate form.*0971

*We have something great for this, L’Hospitals rule.*0973

*I just take the derivative of that or the derivative of that, and I take the limit again.*0978

*This limit is actually equal to the limit as x approaches 0.*0983

*The derivative of 4x is equal to 4.*0989

*The derivative of sin of 3x is 3 cos of 3x.*0992

*Now when I take x to 0, the cos(3) × 0 is cos(0).*1000

*The cos(0) is 1.*1004

*The limit of 4/3, this is actually equal to 4/3.*1006

*This first limit right here, that is equal to 4/3.*1013

*This one, when I put 0 in, I end up with 0/1 which is 0.*1017

*My final answer is 4/ 3 + 0 = 4/3.*1025

*My answer is b.*1032

*Again, just a straight application of l’Hospitals rule, any of the indeterminate forms.*1034

*If you have 0/0, if you have infinity/ infinity, if you have infinity × 0, things like that.*1038

*Let us try number 21.*1049

*What is 21 asking us?*1053

*For 21, it says that y is greater than 0 and it is telling us that dy dx is equal to 3x² + 4x/ y.*1058

*They are telling us that the point 1√10 is on the graph.*1080

*What is 0y?*1090

*In other words, find the y value.*1096

*They are telling us the y is greater than 0.*1102

*They are telling us that the derivative is equal to 3x² + 4x/ y.*1104

*They are telling us that the graph itself contains the point 1 √10.*1109

*They we want to know what the y value is, when x = 0 of the function.*1112

*Let us what we have got.*1121

*We know y’, that is dy dx.*1124

*y’ is equal to 3x² + 4x/ y.*1126

*I'm going to go backwards.*1137

*This is essentially an implicit differentiation.*1139

*Somebody who has done implicit differentiation and they come up with this.*1143

*Just go backwards.*1146

*What you end up with is yy’, just multiply through by y = 3x² + 4x.*1148

*What you will end up, when you rearrange this, you get -3x² - 4x + y, y’ is equal to 0.*1159

*Essentially, now you are just going to integrate this function one at a time.*1173

*The integral of -3x² is x³.*1177

*We are just going backwards to what y is equal.*1181

*This is -2x², the integral of -4x is -2x² because the derivative of -2x² is -4x.*1184

*Here, this derivative is ½ y², the antiderivative.*1193

*Because if I take ½ y², take the derivative of it, implicitly, it is going to be 2 × ½ y × y’.*1200

*The 2 and ½ cancel, you are left with y and y’.*1210

*That is equal to some constant c.*1214

*We do not know what c is.*1219

*Let us go ahead and work it out.*1223

*We figured out what our original function is, in implicit form.*1225

*That is this thing.*1229

*We also know that we have an x and y value.*1230

*We know that 1√10 is on this graph.*1234

*This is –x³, sorry.*1240

*-1³ - 2 × 1² + 1/2 y which is √10² is equal to c.*1245

*Here what we end up with is c is equal to 2.*1263

*If c is equal to 2, let us plug that back in.*1270

*Now we know that –x³ - 2x² + ½ y² is equal to 2.*1274

*We want to know what y is, when x is 0, just plug this in.*1286

*I get 0 - 0 + ½ y² is equal to 2.*1291

*y² is equal to 4.*1301

*y is equal to + or -2.*1304

*But they said early on, y is greater than 0 so that means that y = 2.*1307

*I took the implicit derivative formula and I worked my way backwards integrating to the original function.*1316

*I hope that made sense, this choice was a.*1322

*Let us see, that was number 21.*1329

*We are almost done with this section.*1332

*We are actually moving along pretty well.*1334

*This is 22, what is 22 asking us to do?*1335

*It looks like it is asking us to evaluate,*1340

*Excuse me, let me turn the page here.*1343

*I have difficulty turning these things.*1345

*It wants us to evaluate an integral.*1348

*Let us evaluate the integral from 1 to 2 of 1/ 4 - t², all under the radical.*1350

*Again, this is just a question of recognizing antiderivatives and doing a little manipulation.*1362

*I’m going to rewrite this as.*1368

*I think I want to go back to blue, I like it better.*1370

*The integral from 1 to 2, I’m going to factor out a √4 in the bottom, 1 - t²/ 4.*1373

*Let me write this a little bit better.*1388

*It is going to be 1/ √4 × √1 - t²/ 4.*1393

*I hope that makes sense that this is this.*1402

*I factored out a √4 under the radical sign.*1404

*I just pulled it out of the radical sign, because a radical × a radical is equal to a radical.*1408

*All I’m doing is a little bit of mathematical manipulation.*1414

*This 1/ √4 comes out as ½ × the integral 1 to 2, 1/ 1 -, t²/ 4 is the same as t/2² under the radical dt.*1416

*Now I do a u substitution, u = t/2, du = ½ dt, dt = 2 du.*1439

*Therefore, this integral is equal to ½ × the integral from 1 to 2 of 1/ 1 - u² × 2 du which is equal to the integral from 1 to 2.*1456

*This is 2/2 which turns into 1, 1/ 1 - u² du.*1485

*I recognize this as the inv sin(u), from 1 to 2.*1494

*I plug u back in as the inv sin of t/2, from 1 to 2 = the inv sin of, when I put 2 in there,*1503

*it is going to be the inv sin of 1 - the inverse sin of ½.*1516

*The inverse sin of 1 is π/2, the inverse sign of 1/2 is π/6.*1523

*I'm left with π/3 which is choice a.*1532

*That is it, just a little bit of manipulation to turn it into an antiderivative that we recognize.*1538

*In this case, the inverse sin function.*1543

*That is 22, let us try number 23.*1548

*23 is asking us to evaluate an integral.*1558

*Again, this time it looks like an indefinite integral.*1562

*We have the integral e ⁺2x × √e ⁺x + 1 dx.*1566

*This is going to be a little strange.*1580

*I’m going to start with u = e ⁺x + 1.*1582

*Du = e ⁺x dx.*1590

*I’m also going to write dx is equal to du/ e ⁺x.*1596

*What this gives me is, when I plug these into that, I get the integral of e ⁺2x × √e ⁺x + 1, that is just u and dx.*1604

*dx is equal to du/ e ⁺x.*1621

*This is equal to e ⁺2x divided by e ⁺x.*1628

*I get the integral of e ⁺x × that u du.*1634

*I have the u, I have the du, I just need to take care what this e ⁺x is.*1642

*Let me write that down.*1652

*I need to get everything in terms of u, one variable.*1655

*I have taken care of that, I just need to take care of e ⁺x.*1659

*We said that u is equal to e ⁺x + 1.*1664

*U - 1 = e ⁺x which means x is equal to natlog of u – 1.*1670

*Therefore, this integral now becomes the integral,*1684

*I do not really need to do that.*1694

*Should I leave it as u⁻¹?*1697

*I do not need to do this part, that is not even necessary.*1703

*Again, I have got e ⁺x, now it is u⁻¹.*1710

*Now I have the integral of u⁻¹ × √u × du.*1713

*Perfect, this is great, this is equal to the integral of, this distribute.*1720

*This says u³/2 - u ^½.*1725

*du = 2/5 u⁵/2 - 2/3 u³/2 + c.*1732

*When we substitute back in what u is, we get that this integral is equal to 2/5 × e ⁺x + 1⁵/2 - 2/3*1754

*× e ⁺x + 1, because u was e ⁺x + 1³/2 + c.*1772

*This is choice e.*1779

*That is it, nice application.*1780

*We set the mess around with that e ⁺x.*1783

*We can work one function at a time.*1785

*Whenever you are doing the u substitution, everything has to be changed into u.*1787

*You may not be able do it in one step, you may have to take that extra step.*1791

*Number 24, let us see what does number 24 asks us.*1797

*Let us come over here.*1806

*Number 24, we are given an acceleration and we are given an initial position of velocity.*1809

*What is the position of the particle at another time?*1818

*I’m given acceleration a ⁺t is equal to 12t + 4.*1824

*They are telling me that my initial position x(0) is equal to 2.*1831

*They are telling me that my velocity at 1 is equal to 5.*1835

*They want to know what is my position at time t = 2.*1841

*We know that if you are given that position graph, the first derivative is velocity, the second derivative is acceleration.*1850

*If you are given an acceleration, integrate once, you get velocity.*1855

*Integrate twice, you get position.*1859

*v(t) or velocity of t = the integral of a(t) dt which is equal to the integral of 12t + 4 dt, which is equal to 6t² + 4t + c.*1862

*That gives us our velocity, we have to find c.*1891

*We have an initial value.*1893

*V(1) which is equal to putting 1 in here.*1899

*You get 6 + 4 + c.*1902

*They tell me that v(1) is actually equal to 5.*1906

*I get c is equal to -5.*1909

*My velocity function v(t), I just put this -5 back into here.*1912

*I get 6t² + 4t – 5.*1918

*For f(t), f(t) is equal to the integral of v(t) which is equal to the integral of 6t² + 4t - 5 dt is equal to 2t³, not²,*1924

*2t³ + 2t² - 5t, + this time I will just call it d.*1945

*They are telling me that x(0) is equal to 2.*1956

*X(0) which is equal to 0 + 0 - 0 + d is equal to 2.*1958

*I get d is equal to 2.*1968

*Therefore, my function, my x(t) is equal to 2t³ + 2t² - 5t + 2.*1976

*Therefore, my function at t = 2, just plug the 2 in, is equal to 2 × 2³ + 2 × 2² - 5 × 2 + 2.*1987

*You end up with 16, e is our choice.*2003

*Very nice and straightforward.*2008

*What is number 25 asking us?*2017

*They ask us to determine an integral, combination of integrals here.*2020

*We have the integral from 0 to π/2 of the sin(3x) dx + the integral from 0 to π/6.*2026

*Is that correct, yes, π/6 of the cos(3x) dx.*2040

*Just straight evaluation of an integral.*2048

*Sin(3x) dx, this is going to be a u substitution.*2052

*Hopefully, you have done enough of these to recognize that,*2055

*when you are taking the trig of some constant × x, that constant comes out as fraction.*2058

*This is going to equal, the integral of sin x - cos x.*2064

*It is going to be – 1/3 of cos(3x), evaluated from 0 to π/2 + the integral of cos.*2068

*It is going to be + 1/3 × sin(3x), evaluated from 0 to π/6.*2078

*You are going to end up this, plug in these, this – that, this – that.*2088

*You are going to end up with 0 - a - 1/3 + 1/3 – 0.*2092

*You are going to end up with 2/3.*2112

*This is choice c.*2115

*That is it, straight application.*2117

*All you have to worry about is that.*2119

*The integral of sin of a(x) is equal to -1/ a cos of a(x).*2121

*The integral of cos of a(x) is equal to 1/a × sin of a(x).*2133

*When we do a u substitution, it will work out.*2146

*It will make itself clear.*2149

*Let us see number 25.*2155

*Let us see number 26, we are almost there.*2157

*Number 26, it is asking us to determine the derivative of this particular function at π/2.*2161

*We got f(x) is equal to cos(2x) - 2².*2167

*They want us to find f’ at π/2.*2179

*F’(x), this is just an application, a very careful application of the chain rule.*2185

*Easy, just you got to be careful, that is all.*2189

*We are going to get 3 × cos(2x) - 2² × the derivative of what is inside which is - sin(2x) – 2*2193

*× the derivative of what is inside, × 2 = -6 × cos(2x) - 2² × sin(2x) – 2.*2212

*Now you just basically just plug π/2 in.*2236

*F’(π/2), here is -6 × cos(2) × π/2 - 2² × sin(2) × π/2 – 2.*2238

*I think if I'm not mistaken, the choice is c.*2271

*Number 27, we have come to the end of this first part.*2278

*What is number 27 asks you to do?*2283

*We are going to compute the derivative of something that is given to us, in terms of an integral.*2290

*f(x) is defined as the integral from 0 to x² of the natlog of t² + 1 dt.*2296

*We have learned from the fundamental theorem of calculus that whenever f(x) is defined as an integral from some constant to x,*2312

*really all you do is you are taking the derivative of an integral.*2319

*Since, the derivative of the integrals are inverse processes, all you are doing is getting rid of the integral sign.*2323

*Your answer would be have been ln(t)² + 1 or ln(x)² + 1.*2327

*However, this upper limit is not x, it is x².*2332

*Really, all you have to do is, remember, f’(x), everything is exactly the same.*2337

*It is going to be the ln of,*2345

*You are going to put this into here.*2354

*It is going to be x⁴ + 1.*2357

*You just have to remember, because this is an x, you have to multiply by the derivative of this thing.*2361

*The derivative of x is just 1. It would be just 1, normally.*2368

*But the derivative of x² is 2x.*2371

*It is literally that simple.*2375

*This is choice e, the only difference is that they put the x and the 2x in front.*2380

*Once again, here, just plug in this into here, that will give you the function.*2386

*But remember that, if this is not x, then you need to just multiply by the derivative of this thing.*2394

*It looks like we have one more, I apologize, there are 28 questions not 27.*2403

*Number 28, here it wants us to evaluate the derivative with respect to x of the ln of 2 - cos x.*2412

*That is one parenthesis, that is two parentheses, and a bracket.*2429

*This is just an application of chain rule, it is not a not a big deal.*2432

*f’(x) which is ddx is equal to the derivative of natlog is 1/ the argument.*2436

*This is the argument, it is going to be 1/ derivative of ln(u), it is 1/ u du dx.*2449

*1/ ln(2) - cos x × the derivative of the argument ×, the derivative of ln(2) - cos x is 1/ 2 - cos x*2457

*× the derivative of the argument which is the derivative of -cos x is sin x.*2484

*That is it, it is choice c.*2493

*Just have to watch the differentiation.*2495

*That takes care of Section 1, Part A, no calculator allowed.*2499

*Next lesson, we will start with section 1 part B, where the calculator is allowed on the multiple choice sections.*2505

*Thank you so much for joining us here at www.educator.com.*2512

*We will see you next time, bye.*2514

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