AP Calculus AB Example Problems for The Definite Integral

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

Today, we are going to do some example problems for the different integral that we introduced in the last lesson.0004

Let us get started.0010

Approximate the following definite integral using mid points and the given number of sub intervals.0014

This is a lot like the problems that we had before, when we are dealing with area.0021

We want to just go ahead and start with approximation using midpoints.0025

I think I will work in blue here.0031

We have the integral from 1 to 6.0040

Δx is equal to, we want 5 sub intervals.0043

We have 6 - 1/5 which is equal to 1.0050

Our Δx is equal to 1.0056

We are going from 1, 2, 3, 4, 5, 6.0058

This is 1, this is 6, we are going to be looking at these endpoints.0069

These are going to be our x sub i.0074

This is going to be x sub 1, this is going to be x sub 2, x sub 3, x sub 4, and x sub 5.0077

1.5, 2.5, 3.5, 4.5, and 5.5, because we are approximating with midpoints.0086

Let us go ahead and set up a little table here.0099

Let me do it over here, actually.0104

I have got my x sub i and I have my f(x sub i).0106

I have got 1.5, 2.5, 3.5, 4.5, and 5.5.0115

When I put these numbers into this function to find out what the y value is, I get 0.753, 1.283, 1.295, 1.012, and 0.6799.0124

Therefore, the integral from 1 to 6 of x³ e ⁻x dx is going to be approximately equal to the sum as i goes from 1 to 5.0152

5 points of the f(x sub i) Δx.0174

That is just equal to Δx which is 1 × the sum,0181

I’m sorry, I will write all of this out.0193

This thing is nothing more than Δx × f(x sub 1) f(x sub 2) + f(x sub 3) + f(x sub 4) + f(x sub 5).0196

Let us go ahead and be as explicit as possible.0218

That is going to equal 1 × 0.753 + 1.283 + 1.295 + 1.012 + 0.6799.0220

When I do all of that, I end up with 5.0229, that is my approximate integral using midpoints.0238

We could have used right endpoints, we could have used left endpoints, 1, 2, 3, 4, 5.0253

You are going to get numbers that are sort of similar.0260

We decided to use midpoints for this particular problem.0262

Just to show you what it actually looks like.0269

Here is my list of x values, 1.5, 2.5, 3.5, 4.5.0272

These are the actual values of f(x) carried out to a greater degree of precision.0277

Our function happens to look like this.0283

We were going from 1 to 6 midpoints, numbers.0285

Midpoints give me all the numbers.0296

What we found was the definite integral.0297

In this particular case, between 1 and 6, the function happens to be positive, it is above the x axis.0299

Therefore, this number that we got, the 5.02 something, it happens to be the area under the curve.0304

It happens to be, it is the integral from 1 to 6 of this function using midpoints.0313

An approximation of the integral, it just happens to correspond with the area under the curve.0320

Express the following limit as a definite integral.0330

Little practice using the actual symbol, very simple.0333

The limit is an goes to infinity 1 to n, here is our function.0337

This is going to be our integrand, this we are going to turn into dx instead of Δx.0343

This is going to be our lower limit integration, this is going to be our upper limit of integration.0350

We write this as the integral from π/3 to π cos² x/ x² dx.0354

We are done, this is the definition, the limit of the sum.0369

The summation symbol becomes the integral symbol.0375

That is what is happening here.0379

For the real problem, evaluate the following definite integral using the definition.0390

This is going to be a bit of a process here.0395

One of the reasons why we want to come up with quick ways of doing the integral was because0398

we do not want to take this process that we are about to go through, for this problem in the next, every time we take an integral.0401

This is going to equal the limit as n goes to infinity of the sum from 1 to n of the x sub i Δx.0409

First, we find this, in other words we find an expression for that given the fact we have f(x) and we have -2 and 4.0436

First, we find that.0443

Second, once we find that, we form the sum.0445

Second, we are going to evaluate the sum.0453

When we evaluate the sum, you will end up with the function of n.0467

The function of n, there is going to be n in it.0478

The last thing, we evaluate the limit.0484

Third, evaluate the limit, in other words, take n in that function that we get.0488

This part, take n to infinity and see what you get.0497

Evaluate the limit as n goes to infinity.0503

That is what we are going to do, first, second, third, we will get our answer.0506

We have -2 to 4, that is where we are integrating in.0513

A is equal to -2.0517

We know that b is equal to 4.0520

We know that Δx = b - a/ n is equal to 4 - -2/ n is equal to 6/n.0523

That is our Δx.0536

That takes care of our Δx part, that is going to be the 6/n.0541

Let us see if we can come up with something for x sub i first, some expression for x sub i,0545

that we can put in to f(x), in order to get f(x sub i).0555

We are going to take that, multiply it by the 6/ n to get this thing.0559

That is all we are doing here.0565

We have taken care of the Δx, let us see if we can find x sub i.0568

Let us see if we can elucidate a pattern.0572

We have x sub 0, that is equal to -2.0575

x sub 1 is equal to -2 + Δx which is 6/n.0580

x sub 2 = -2 + 6/ Δ n + another Δx which is + 12/n.0587

x sub 3 = -2 + 12/n + another Δx which is 6/n, which is 18/n.0598

Notice, this number is 1 × 6, 1, 1.0609

This number is 2 × 6, 2 is here, 2 is here.0618

This number is 3 × 6, 3 is here, 3 is here.0622

We have our pattern, our x sub i is equal to -2 + i × 6/n.0629

Which I’m going to go ahead and write it as 6i/ n.0640

Great, now we found our x sub i.0645

Perfect, we found our x sub i, now we are going to stick our x sub i.0647

This -2 + 6i/ n into this function, to find f(x sub i).0652

F(x sub i) is equal to 2 + 2 × x sub i which is equal to 2 + 2 × -2 + 6i/ n,0670

that equals 2 - 4 + 12i/ n = -2 + 12i/ n, that is our f(x sub i).0690

F(x sub i) × Δx is going to be -2 + 12 sub i/ n, that is my f(x sub i).0712

My Δx was 6/n.0723

Therefore, this is going to equal -12/n + 72/ n².0728

This is my f(x sub i) Δx.0739

I’m going to evaluate the sum and I'm going to take the sum as i goes from 1 to n.0746

We have taken care of the first part, we are going to evaluate the sum of this.0754

The sum of the f(x sub i), -12/n + 72i/ n².0761

The summation symbol distributes over that.0776

I get the sum i from -1 to n -12/ n + the sum of i1/ n 72i/ n².0780

i is the index, i has to stay under the summation symbol.0798

Everything else can be taken out as a constant.0803

In other words, there is no i here.0805

Therefore, I can pull out the -12n.0808

Here I have to leave the i but I can pull out the 72/ n².0811

Therefore, this is going to equal -12/n × the sum of 1 to n of 1 + 72/ n² × the sum i = 1 to n of i.0815

Let us rewrite that so we have it on the page.0843

-12/n × the sum from 1 to n of 1 + 72/ n² × the sum i from 1 to n of i.0848

The sum from 1 to n of 1, we just add 1 n ×, it is just n + 72/ n².0864

We have a closed form expression for this, remember.0878

It is equal to n × n + 1/ 2.0880

It equals -12 + 72 divided by 236, 36 × n²/ n².0887

This is going to be n² + n.0898

72n²/ 2n², gives me 36n² + 36n/ n².0904

I just went ahead and divided the 2 into it , and then multiplied.0921

72 divided by 2 is 36, 36n² 36n/ n² - 12.0925

36n²/ n² is 36.0936

36n/ n² is 36/n which is equal to 24 + 36/n.0941

The sum of i = 1 to n of f(x sub i) dx is equal to 24 + 36/n.0957

Now we evaluated the sum, we have our function of n.0970

Now we take the limit as n goes to infinity.0975

The limit as n goes to infinity of 24 + 36/n, this one goes to 0.0980

As n goes to infinity, we are left with 24, that is our answer.0991

The integral from 1 to 6 of that function, going through the entire process.0996

Finding the Δx, finding the x sub i, forming f(x sub i).1002

Multiplying f(x sub i) × Δx, evaluating the sum, and then taking the limit, gives us a final answer of 24.1009

That is it, run through the process.1017

Tedious but reasonably straightforward, as long as you have the formulas that you need.1022

Evaluate the following integral using the definition, same thing.1029

We know that this is going to equal, let us go ahead and write the definition,1034

so that we know what we are dealing with here.1039

The definition, the integral from a to b of f(x) dx equals the limit as n goes to infinity.1041

I think it is a good idea to write down the definition of the equation over and over again, that way you remember it.1050

The limit to infinity, the sum as i go from 1 to n of f(x sub i) Δx.1058

We are going to run through the same process.1067

We are going to find Δx, we are going to find an expression for x sub i.1069

We are going to put x sub i into f.1073

We are going to form f(x sub i).1077

We are going to multiply by Δx.1078

We are going to evaluate the sum of that thing.1080

We are going to get a function of n and then we are going to take n into infinity to get our final answer.1082

That is what we do.1086

Let us go ahead and do a is equal to 1, b is equal to 6.1090

Therefore, the Δx = b - a/ n which is 6 - 1/ n which is 5/n.1097

That takes care of the Δx.1113

x sub 0 that is equal to 1, x sub 1 = 1 + Δx which is 5/n.1117

x sub 2 = 1 + 10/ n, x sub 3 = 1 + 15/n.1131

We see the pattern, x sub i = 1 + 5i/ n.1140

Therefore, our f(x sub i) is equal to, you put this into here, into there, f(x sub i).1152

Therefore, it is 1 + 5i/ n² + 4 × 1 + 5i/ n – 7.1168

It is going to equal 1 + 10i/ n + 25i²/ n² + 4 + 20i/ n - 71185

which is going to equal -2 + 30i/ n + 25i²/ n².1204

This is just our f(x sub i), you need to now multiply that by our Δx which is 5/n.1219

Our f(x sub i) × Δx is equal to -2 + 30i/ n + 25i²/ n² × 5/n.1228

This is going to equal -10/n + 150i/ n² + 125i²/ n³.1249

We have our expression, now we evaluate the sum.1272

The sum from 1 to n of the f(x sub i) × Δx = the sum of 1 to n of this expression.1277

-10/n + 150i/ n² + 125i²/ n³.1293

Distribute, separate, sum as i goes from 1 to n of -10/ n + the sum i from 1 to n of 150i/ n²1306

+ sum i from 1 to n of 125i²/ n³.1327

There is no i here, pull it all out.1339

I pull out the 150/ n², I pull out the 125/ n³ = -10/n × the sum i goes from 1 to n of 1 + 150/ n² ×1341

the sum as i goes from 1 to n of i + 125/ n³ × the sum as i goes from 1 to n of i².1359

This is going to equal -10/n × n + 150/ n² × n × n + 1/2, because that is this.1375

We also have an expression for this one.1393

This is going to be + 125/ n³ × n × n + 1 × 2n + 1/ 6.1395

It is just arithmetic, that is all it is.1410

We get -10 + 75n² + 75n/ n² + 125/ n³1415

× 2n³ + 3n² + n/ 6 = -10 + 75 + 75/n + 250/6.1435

I multiply everything, cancel everything.1457

+ 375/ 6n + 125/ 6n² = 640/6 + 825/ 6n + 125/ 6n².1462

This is our function of n, now we evaluate the limit, we take n to infinity.1489

The limit as n goes to infinity of this thing, 640/6 + 825/ 6n + 125/ 6n².1498

As n goes to infinity, this goes to 0, this goes to 0.1514

We are left with 640/6 or 106.67.1519

One painful process, we definitely need a quicker way to do this.1530

We have a quicker way to do this.1534

We actually done it with antiderivatives, we will do some more.1535

Evaluate the following definite integral by using areas.1543

6 to 30, 1/3x – 4, this is a line.1548

Let us go ahead and draw this out.1553

1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7,1569

8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.1588

Actually, I meant this to be 20 not 30.1596

Therefore, I’m going to go ahead and change this to 20.1602

1/3x – 4, 4 up 1/3, up 1/3, that takes us to 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.1607

We are going to get that line.1629

We are integrating to 20.1636

We are looking for the integral of this function.1638

They want us to do it in terms of the areas.1641

Therefore, I’m going to find area number 1 and I’m going to find area 2.1643

I'm going to add them.1649

Area number 1 is going to be negative because it is below the x axis.1650

Area number 2 is going to be positive.1653

When I add them up, I’m going to get some net area that is equal to the integral.1655

Great, nice and simple.1660

Therefore, this integral, the integral from 6 to 20,1664

What is going on here?1677

I’m not integrating from 6, I got all my numbers mixed up here.1693

That is okay, very easy to fix, I’m sorry.1702

The integral here is actually not from 6 to 30, it is from 0 to 20.1704

I do not know where the 6 to 30 came from.1709

Fortunately, everything still stands, let us recap.1713

We have a function, we graph that function.1717

We are integrating from 0 to 20.1720

We graphed it from 0 to 20.1722

Yes, we are finding this area which is going to be negative.1723

We are going to be adding it to this area.1727

Everything else is just fine.1729

This is going to be 1/3x – 4 dx is going to be area number 1 + area number 2.1734

This is 12, from 0 to 12, f(x) is below the x axis.1744

Therefore, area 1 is going to be negative.1761

Area 1 is going to be negative, base × height/ 2, it is just a triangle.1764

Base is 12, height is 4/2.1771

That gives me -24.1779

Area 2 is going to be from 12 to 20, that is just going to be positive.1784

It is going to be base × height/ 2.1790

The base of this triangle is 8, the height is 20.1792

I just put it into here and I end up with a height of 2.667/2.1801

I get 10.668.1808

Therefore, I just add these two together.1813

a1 + a2, I get a -13.32.1818

The integral of the function is a negative number, the net area.1825

More area here than there is here.1830

If the integral from 7 to 14 of f(x) = 27 and if the interval from 10 to 14 = 9, find the integral from 7 to 10.1839

We have a nice property that takes care of this for us.1849

We want the integral from 7 to 10 of f(x) d(x).1853

That equals the integral from 7 to 14 of f(x) d(x) + the integral, as long as this and this are the same, the interval from 14 to 10.1860

This and this are that and that.1876

If these are the same, I can just add these integrals to give me this final one.1880

Here we are fine, the integral from 7 to 14 of f(x) dx.1889

Here, this is the integral from 14 to 10, what they gave us is 10 to 14, lower to upper 10 to 14.1896

Now it is 14 to 10, it is switched.1905

What I have to do is, when I switch the limits, I change the sign of the integral.1907

It is - this is equal to negative of the 10 to 14 of the f(x) dx.1912

Therefore, it is just equal to 27 - 9 = 18.1922

Thank you so much for joining us here at www.educator.com.1931

We will see you next time, bye.1933