AP Calculus AB Example Problems for the Fundamental Theorem

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

Today, we are going to be doing some example problems for the fundamental theorem of calculus that we just covered.0004

These are going to be extraordinarily straightforward and simple, almost surprisingly.0010

Our first example is find the derivative of the following function.0017

The function is g(x) and it is defined as the integral from 1 to x, whatever x might be.0021

5/ t³ + 9 dt, nice and simple.0027

Since the upper limit is x, we would literally just put x where we have the t.0032

Our final answer is just 5/ x³ + 9, and we are done.0039

That is the beauty of the fundamental theorem.0046

When a function is defined as an integral or the upper integral is a variable,0048

this particular variable right here xx, we just literally put that in there.0054

Let me correct this, a little bit of a notational issue here.0064

g’ because we are looking for the derivative.0071

g’(x), that is a little more clear, I think.0075

We have 5/ x³ + 9.0079

Literally, when we take the derivative of this, we are just getting rid of this integral sign0084

because differentiation and integration or anti-differentiation are inverse processes.0090

That is the whole idea.0095

That is it, nice, straightforward, and simple.0097

Find the derivative of the following function.0102

g(x), upper variable is x, we do the same thing.0105

Therefore, the derivative of this g’(x), all we have to do is we put an x where we see a t.0110

We get x² sin³ x, that is it, very nice.0117

As long as this upper limit of integration is the variable that is inside the function defining it,0128

we can just put it straight into the function itself.0144

Let us try this one, find the derivative of the following function.0150

g(x) = the integral from 1 to x⁴ sin(t) dt.0155

Now this is x, this time, we have not x but we have a function of x.0161

The only thing we do know is we do the same thing.0168

We are just going to be putting in the x here.0171

g’(x) is equal to sin(x).0173

But because this is not x but it is a function of x, by the chain rule,0177

we are just going to multiply by the derivative of this function.0181

Let us write that out.0186

When the upper limit is not just x but a function of x, some more complicated function of x, we do the following.0188

This is the long process, the easy part is just multiply by the derivative of the upper limit, whatever the function happens to be.0220

Essentially, what you are doing is you are letting x⁴, the upper limit integration equals some letter u.0227

Essentially, you are just doing a substitution.0239

And then, du dx is going to equal 4x³.0240

And then, you rewrite this g(x) = the integral from 1 to u of sin(t) dt.0250

Then, when you do this, when you put this into here, in other words, when you do g’(x) is equal to the sin(u),0263

because u is a function of x, by the chain rule, the derivative is sin(u) du dx.0277

You are multiplying by du dx.0285

What you end up getting is g’(x) is equal to sin(x)⁴ × 4x³.0292

This part is exactly the same, you are just putting the upper limit into this variable to get the sin x⁴.0303

And then, you are just multiplying by the derivative of that because now it is a function of x, it is not just x itself.0311

I hope that makes sense.0321

Essentially, what you are doing is you are doing the same thing as you did before.0322

If this were just an x, you would put it in and it would be sin(x).0325

The derivative of x is just 1, you were technically doing du dx.0331

The derivative of x is just 1, which is why it just ends up staying that way.0336

But if it is a function, just multiply by its derivative.0342

That is your final answer, right there, very straightforward.0345

Find the derivative of the following functions, same thing.0356

Now π/3, the upper limit is cos(x).0359

We are just going to put this into the t's and then multiply everything by the derivative of cos(x).0363

g’(x), we would literally just read it off.0372

It is going to be 3 × cos(x) + 7 × cos² (x) × the derivative of that function cos(x) which is –sin(x).0375

We are done, that is all.0391

Just do what the first fundamental theorem of calculus says which is just put the upper limit into those things.0399

Then, multiply by ddx of the upper limit.0415

That is all, moving along nicely.0431

Evaluate the following integral.0435

We want to evaluate a definite integral.0437

We have a lower limit, an upper limit, they are both numbers.0440

We are going to find the antiderivative, the integral.0443

And then, we are going to put in 5, we are going to put in 1,0446

and we are going to subtract the value of what we get from 1, from the value of the upper limit 5.0450

Let us go ahead and do that.0456

The integral of this, let us go ahead and actually multiply this out first because we have x³ - 4x².0457

This is going to be the integral from 1 to 5 of x⁶ – 4x × that is going to be 4x⁴, then another 4x⁴.0468

It is -8x⁴ + 16x², I just multiply this out, dx.0484

And that is going to equal x⁷/7 - 8x⁵/5 + 16x³/3.0495

I'm going to evaluate this from 1 to 5.0512

In other words, I’m going to put 5 in for all of these and get a number, then I'm going to subtract by 1.0514

Putting them into all the x here.0520

When I put 5 into this, I get 78,000, 5⁷ is 78,125/7.0522

When I put 5 into here, multiply by 8 and divide by 5.0535

I get - 2500/5, I hope you are going to be confirming my arithmetic.0539

I’m notorious for arithmetic mistakes.0544

+ 2000/3 that takes care of the 5.0546

I’m going to subtract by putting 1 in.0553

This is going to be 1/7 - 8/5 + 16/3.0556

Find the antiderivative of the integrand, and then evaluate.0565

Put the upper number into this, you are going to get a value.0570

Put the lower number into the x values, you are going to get this and then you subtract.0573

The rest is just arithmetic.0577

I'm just going to go ahead and leave it like this.0579

You can go ahead and do the arithmetic, if you want, to get a single answer.0581

Evaluate the following integral, once again, we are using the second fundamental theorem of calculus.0588

The integrals of f from 2 to 8 is going to be the f(8) - f(2), F(8) – F(2), that is what the second fundamental theorem says.0594

Remember, it is this one, if from a to b of f(x) dx is just equal to f(b) – f(a), where F is the antiderivative of f.0611

This is going to be the integral from 2 to 8, I’m going to separate these out.0629

I’m going to write this as x²/3, it is going to be x/ x²/3 - 3/ x²/3.0638

x/ x²/3 dx - the integral from 2 to 8 of 3/ x²/3 dx.0646

That is going to equal, the integral from 2 to 8 of x¹/3 dx - the integral from 2 to 8 of 3 × x⁻²/3 dx.0662

I can go ahead and integrate these, not a problem.0678

1/3 + 1 is 4/3, I get x⁴/3/ 4/3 - 3 × x – 2/3 + 1 is 1/3/ 1/3.0681

That is going to equal ¾ x⁴/3 - 9x¹/3.0700

I'm going to evaluate that from 2 to 8.0712

When I put 8 in for here, I'm going to get 12.0716

When I put 8 into here, it is going to be -18, that is the first one.0721

-, when I put 2 into here, I end up with 1.89.0727

And then, - when I put 2 into here, I end up with 11.34.0734

There is your answer.0740

Once you have the antiderivative, you put the upper limit into the x value to evaluate, that gives you the first term.0742

You put the lower limit into the x, that gives you the second term.0753

You subtract the second term from the first term, f(b) - f(a).0758

Nice and straightforward.0765

Another good example, the integral from π/6 to 2π/ 3 of cosθ dθ.0772

This is going to equal where the integral of cosθ dθ is going to be sin(θ).0780

We are going to evaluate sin(θ) from π/6 to π/3.0786

Not a problem, that is going to be the sin(π/3) - sin(π/6).0796

We are just putting in upper and lower limit into the actual variable of integration.0809

The sin(π/3) is √3/2 – sin(π/6) which is ½.0815

The integral from 1 to 3 of 5/ x² + 1.0835

You should recognize that the integral of 1/ x² + 1 is equal to the inv tan.0841

Remember, the integral or the antiderivative, whatever you want to call it,0847

of 1 + x² dx that was equal to the inv tan(x).0852

Therefore, this is, that is fine, I will go ahead and separate it out.0860

This is just a constant, it is 5 integral from 1 to 3, 1/ x² + 1 dx = 5 × inv tan(x), evaluated from 1 to 3.0866

That is it, that is all it is.0885

This is just 5 × inv tan(3) – inv tan(1).0887

That is all, you can go ahead and multiply the 5 through.0901

You can do 5 or you can evaluate this from here to here.0904

This - this and then multiply the constant through.0911

The order actually does not matter.0914

I’m just somebody who tends to like to see the constant on the outside, personal choice.0916

g(x) is equal to the integral from 0 to x of f(t) dt.0928

f being the function whose graph is given below.0934

This graph right here, this is a function of f.0937

Answer the following questions.0943

We take a look at this graph, it looks like some sort of a modified sin or cos graph, definitely periodic.0945

We are taking a look real quickly, we notice that we have about a 3. something, then we have a 6.2 something.0952

It turns out that these are actually just π/2, π.0961

I’m going to go ahead and label them that way.0970

I left the labels just as is, but it is nice to recognize the numbers for what it is that they are.0972

Especially, since you are talking about a periodic function, then more than likely the graphs that you are going to be dealing with0980

are going to pass through the major points, π/2, π, π/6, π/3, π/4, things like that.0985

In this case, let me mark these off.0992

3.14, this is π.1003

This over here looks like it is π/2.1010

Over here, this looks like 2π and this looks like 3π/ 2, 4π/ 2.1017

It looks like we have got 5π/ 2.1028

At least that way we have some numbers that we are accustomed to seeing,1030

instead of just referring them as 3.1, 1.5, things like that.1033

Give the values of x for which g(x) achieves its local maxes and mins.1040

You have to be very careful here, this is a graph of f.1047

g is the integral of f from 0 to x.1052

Therefore, g is actually the area under the graph.1059

The graph actually falls below the x axis, the area is going to become negative.1066

Give the values of x for which g(x) achieves its local maxes and mins.1075

As we start moving from 0 x, moving forward g(x), as x gets bigger and bigger,1080

this value is just the area under the graph itself.1091

It is going to hit a high point, when I hit π/2.1095

And then, at π/2, because the graph itself falls below the axis, as I keep going,1101

as x keeps getting bigger and bigger, the integral itself is going to be negative.1108

The area is going to hit a maximum and then the integral is going to start to go negative.1115

The area is going to diminish.1120

At this point, the area is going to go down.1123

At some point, it is going to hit 0.1127

The area is going to rise to a certain number, that is up to here.1130

We are going to start subtracting from the area, at some point it is going to hit 0.1134

Over here, it is going to hit a minimum because now this area under the x axis is a lot bigger than the area above the x axis.1138

The integral from 0 to π is going to be a negative number.1147

It is actually going to hit a maximum.1154

One of the maximums that it is going to hit is right there.1157

And then, it is going to hit a minimum right there.1161

After that, it is going to climb again, the area under the graph.1164

That is what we are doing, g(x) is the area of f, the area under the graph of f because this is the graph of f.1169

The area is going to get bigger again.1178

At this point, it is going to hit another maximum.1180

And then at this point, the area is going to be negative.1185

We are going to hit another minimum, 2π.1187

We are going to increase the area again.1191

It is going to pass this and it is going to hit another maximum.1194

That is what is going to happen.1198

Give the values of x for which g(x) achieves its local maxes and mins.1199

It is going to achieve its local maxes and mins where the area under the graph hits a maximum and a minimum.1205

It is a maximum at π/2, a minimum at π.1215

Maximum at 3π/ 2, a local minimum at 2π, local max and min at 5π/ 2.1218

We have π/2, π, 3π/ 2, 2π, and 5π/ 2.1225

The reason is because g(x) is the net area under the graph of f.1241

This red is the graph of f.1248

Where does g(x) achieves its absolute max, between 0 and 8, this point right here.1255

Let us go ahead and draw this out.1266

Basically, what we are going to have is this.1267

Let me do this in black.1270

g(x) is going to look like this, it is going to rise.1272

It is going to hit a maximum here and it is going to hit a minimum.1277

It is going to start rising again.1282

It is going to hit a maximum and it is going to hit a minimum, and it is going to hit a maximum.1286

It is going to start coming down to there.1296

The area keeps getting bigger and bigger.1300

The absolute value, the area keeps getting bigger and bigger.1304

Our highest point is going to be this point right here, which is going to be our 5π/ 2.1307

Again, because g(x), that is what we want, where does g(x) achieve its absolute max.1316

g(x) is the net area under the graph.1322

Positive, if it is above the x axis, negative if it is below.1326

This is the graph of g.1330

On what intervals is g(x) concave up or concave down?1335

For concave up or concave down, we want to see the sin(g”).1341

g’ is equal to f.1356

By the fundamental theorem of calculus, if I take the derivative of this, I get rid of this.1360

I just have f(x), this is g’ that mean g“ is equal to f’.1364

f’ is just the derivative of the red graph.1373

The derivative is the slope.1379

It is going to be concave up where the slope is positive.1383

It is going to be concave down where the slope is negative.1387

If the slope is positive, that is f’.1392

f’ happens to be g”.1395

Therefore, we have from here to here, is concave up.1400

From here to here, it is concave down.1415

From here to here, concave up.1424

From here to here, concave down.1429

From here to here, concave up.1433

And from here to 8, we got concave down which is exactly what we see.1437

Concave up graph, this is g that we are looking at.1443

Now it is concave down, it starts to be concave up again up to here.1445

From here to here, it is concave down, it matches.1450

I hope that makes sense.1457

You are given a graph f, if g is defined as the integral from 0 to x of f.1460

It is the net area under that particular graph.1467

I hope that made sense.1472

Let us take a look at what it actually looks like.1475

The original function f was this one right here.1478

This one, this is the actual g(x).1484

This was our f(x).1488

It is a maximum here, it is a minimum here.1494

Local, hits a local max here, it is right there.1502

It is a local min here, it is right there.1507

Hits a local max here, that is right there.1511

It is exactly what we said before.1515

Thank you so much for joining us here at www.educator.com.1518

We will see you next time, bye.1520