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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

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

Hello, welcome back to, 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 integral0173

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 20822

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

We will see you next time, bye.2514