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The Average Value of a Function

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  • Intro 0:00
  • The Average Value of a Function 0:07
    • Average Value of f(x)
    • What if The Domain of f(x) is Not Finite?
    • Let's Calculate Average Value for f(x) = x² [2,5]
    • Mean Value Theorem for Integrate
  • Example I: Find the Average Value of the Given Function Over the Given Interval 14:06
  • Example II: Find the Average Value of the Given Function Over the Given Interval 18:25
  • Example III: Find the Number A Such that the Average Value of the Function f(x) = -4x² + 8x + 4 Equals 2 Over the Interval [-1,A] 24:04
  • Example IV: Find the Average Density of a Rod 27:47

Transcription: The Average Value of a Function

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

Today, we are going to be talking about the average value of a function.0004

Let us jump right on in, let us work in blue here.0007

Let f(x) be defined by the following table of values.0012

We have x and then we have different values for f(x).0030

Let us take 1, 2, 3, 4, 5, and then, 6, 7, 8, 9, and 10.0036

We have 10 discreet points and our values are going to be 2, 2.7, 3.6, let us say 2.5, 6.8.0050

I have 9.7, 10.2, 10, and 9 is 13.4, we have 11.7.0064

What is the average value of this function?0091

In other words, what is the average value of f(x)?0097

It is very simple, like any average value, you add them all up and you divide by the number of points that there are.0100

What is the average value of f(x)?0109

It is exactly what you think.0123

The average = 1/10 because there are 10 points.0126

The sum as x goes from 1 to 10 of f(x).0130

f(1)+ f(2) + f(3) + f(4), all divided by 10, like any other average.0138

Now the question is what if the domain of f(x) is not finite?0144

It is a finite domain, I have 10 numbers in it.0160

I take an interval, like from 0 to 5.0164

The length of that interval is 5 but the number of points in that interval is infinite,0168

because there is an infinite number of points on the real number line.0175

How do we find the average value of the function then?0180

Such as, if I have the function f(x) is equal to x².0189

If I have x is in the interval from 2 to 5.0195

What is going to be the average value of the function?0200

Between 2 and 5, f(x) is going to take on an infinite number of values.0202

How do I find the average?0208

It is the same thing, we add and we divide by the length, so to speak.0209

Same thing here, except adding when we are dealing with an infinite domain becomes integration.0215

Because we are adding smaller and smaller things, we passed the integration, when we passed the limit.0221

We integrate the function and then we divide by the length which in this case is going to be 5/2.0227

Average value, here the average value = 1/ the length, which is going to be b – a.0233

This is the b, this is the a.0250

The integral from a to b of f(x) dx, that is it, that is the definition of the average value of a function.0251

It is going to be the integral of the function from a to b, the lower limit to the upper limit divided by the length of the interval.0260

It is essentially what we did here.0271

We are adding up all the values of the function and we are dividing by the number there are.0273

Here we are adding up all values of the function, the integral.0277

And we are dividing by the number there are which is essentially going to be the length of the interval.0280

Let us calculate the average value for f(x) = x², on the interval from 2 to 5.0287

Let me go back to blue here.0309

The average value is going to be 1/ 5 - 2 × the integral from 2 to 5 of x² dx.0315

It is just going to equal 1/3 x³/ 3 from 2 to 5 0329

which is going to equal 1/3, 125/3 - 8/3 which is going to equal 13.0339

What does this mean?0355

The average value of f(x)² from 2 to 5 is 13, what is that mean?0357

What does this mean? It means this.0372

I know my x² function look something like that.0376

If I have 2 and if I have 5, from 2 the value of f(x) is going to be 4.0380

At 5, the f(x) value is going to be 25.0388

What this means is that, when x goes from 2 to 5, f(x) takes on values from 4 to 25.0401

From 4 to 25, it takes on all the values from 4 to 25.0424

The average of those values is 13, that is it.0432

The average of those values given the weight of the function is not linear.0439

It is not y = x, the average of those values is 13.0447

Now you have an average, you have a way of finding an average of a finite number of terms.0454

Now you have a way of finding the average of an infinite number of terms.0459

That is all we are doing, we are using integration.0462

That is all, very simple.0467

The hardest issue is going to be integration itself.0470

We can find the average of an infinite sum, if you will, an infinite sum of numbers.0476

We just passed the integration.0497

Again, for f(x) such that x is in the interval from a to b.0503

The average value of f/ that interval is equal to 1/ b - a × the integral from a to b of f(x) dx.0520

Now we know what the integral from a to b of f(x) dx is, it is the area under the curve from a to b.0532

It is the area under the curve from a to b.0548

It turns out, we have something called the mean value theorem for integrals.0562

We have the mean value theorem for derivatives, way back when.0571

The mean value theorem for integrals says if f(x) is continuous on a closed interval ab, 0575

then there exist a number c which is contained in ab.0605

There is a number c between a and b, such that f of that c is actually equal to the average value0619

or another way of saying this, I can move this a/b to the other side.0641

There is a c such that f of that c × the length of the interval actually = the area under the curve.0646

What that means is the following.0664

Something like that.0667

This is a, this is b.0671

Let us go ahead and use our example that we did.0679

This is our x² function, y = x².0683

This is 4, this is 25, this was our 13.0688

What the mean value theorem for integrals is saying is that there is some number c here.0698

There is some number c such that, let us draw this out actually.0715

Let me draw that again.0720

I have got 2, this is not a and b, this was 2 and this was 5.0723

Let us make it specific, 5 and 13.0729

The mean value theorem says that the f(c), there is some c.0744

In this particular case, it is going to be √13 because f (c) is 13.0752

The average value of the function, f(c) which was 13 × 5 - 2 = the integral from 2 to 5 of the x² dx.0758

In other words, f(c) × b - a = the integral from a to b of f(x) dx.0772

What this means geometrically is that the area under the curve, this area right here,0783

there is some c such that f(c) × f(c) which is in this case 13 × the length.0788

This is the 13, that is our height.0799

This is the 5 – 2, that is 3, where the area underneath that rectangle is the area underneath the curve.0801

That is it, average value, that is all that means.0809

Here we have the area under the curve.0814

Here is the area of the interval length × the average value.0825

That is it, mean value theorem for integrals.0840

Let us do some examples.0845

Find the average value of the given function over the given interval y = t³ e⁻²⁴ from 0 to 4, very simple.0849

Let us go ahead and work in blue here.0858

We know that our f average is equal to 1/ b – a, the integral from a to b of f(x) dx.0863

Here that is equal to 1/ 4 – 0, the integral from 0 to 4 of t³ e ⁻t⁴ dt.0876

That is all, that is all we are doing, this is just an integration problem.0888

This is going to equal 1/4 × the integral from 0 to 4 of,0893

Let me do it over here first.0906

I’m going to do a u substitution.0908

I’m going to call u - t⁴, du is going to be -3t³ dt.0912

Therefore, t³ dt is going to equal -1/3 du.0926

The du = -1/4 t³, there we go dt.0941

Take a breath, slow down, there we go.0960

-4t³ dt, there we go.0962

Now we have t³ dt is equal to -du/4.0966

Now I'm going to go ahead and put these in.0979

It is going to be ¼, that is that one, the integral, t³ dt.0982

t³ dt is going to be - du/4 × e ⁺u = -1/16 0⁴ e ⁺u du = -1/16 e ⁺u 0989

= -1/16 e ^- t⁴, from 0 to 4 = -1/16 × 1/ e ⁺256 – e⁰ which is 1.1021

If you want, I can go ahead and just leave it that way.1050

The graph is going to look as follows.1055

This is the function, this is y = t³ e ⁻t⁴.1059

The average value is a particular something.1071

We went from 0 to 4, past here.1073

Clearly, even though the majority of it is you are getting up to like 0.38, 0.39,1080

but we are actually taking all of these average values from about 1.6, 1.7, all the way to 4 is virtually 0, 1089

which is why the average value of this function from here to 4 is actually going to be very small.1096

Find the average value of the given function over the given interval, 1107

then find the number c in the given interval such that f(c) = the f average.1110

Our function is this and our interval is 3,8.1119

Let us see, our f average is equal to 1/ b - a which is 8,3, 3 to 8 of 5 × cos(x) – sin(5x) dx.1124

This is going to equal 1/5, the integral from 3 to 8 of 5 cos x - 1/5 the integral from 3 to 8 of sin 5x and my dx.1149

5 comes out, we are left with just the integral of cos x which is going to be sin x from 3 to 8.1172

This is going to be -1/5 × 1/5.1188

The integral of sin is –cos, this is going to be × a - cos 5x from 3 to 8.1198

Let us see what I have got.1219

When I actually do this, I should have sin x, from 3 to 8 this is going to be, we have +1/25.1223

That is fine, 1/25.1249

You know what, I do not need simplify it, just go ahead and put these in.1258

I have got sin 8 - sin 3 -, when I multiply all these out, putting 8 and 3, keeping this negative sign here, 1265

I left the negative sign in here, I just pulled off the 1/5 to turn it into 1/25.1278

I left it like that.1282

What you are going to end up getting once you expand is going to be cos of 40/25 - cos 15/25.1285

I ended up with the value of 0.852, that is the average value of this function.1296

Here is what it looks like.1304

We are going from 3 to 8, this is the actual graph itself.1306

This is y = 5 cos x – sin 5x.1310

We are going from 3 to 8.1318

Some negative values, some positive values, some negative values.1325

On average, we are going to end up with 0.852.1329

We want to find the c such that f(c) = the average value.1341

f(c) is just 5 × cos(c) - sin 5c = the average value which is 0.852.1360

When I solve this equation for c, I move this over to the left to set it equal to 0.1381

Use my calculator or whatever graphical utility I have, I get c is equal to 4.685 and c is equal to 7.611.1385

The interval was from 3 to 8.1405

Both of those numbers are in the interval from 3 to 8.1410

Therefore, both of these, there are two numbers.1414

It does not have to be just one, you can have more.1418

There are two numbers c, such that f of these c’s is equal to 1/ b - a × the integral from a to b of f(x) dx.1421

Example 3, find the number a such that the average value of the function = 2/ the interval from -1 to a.1446

The average value of the function 1/ b – a, a - -1 × the integral from -1 to a of -4x² + 8x + 4 dx.1462

We want this average value to equal 2.1485

Solve this equation for a.1490

I get 1/ a + 1, the integral here is -4x³/ 3 + 8x²/ 2.1494

This is going to be + 4x² + 4x from -1 to a is equal to 2 – 4a³/ 31515

+ 4a² + 4a - 4/3 + 4 - 4 = 2 × a + 1,1537

multiplied through, solve for that.1557

I end up with -4a/ 3 + 4a² + 4a - 4/3 + 2a + 2.1560

Combine terms, a’s, numbers, I get – 4a³/ 3.1577

I simplify this equation, I bring things together, multiply by 3.1592

My final equation I get is, once I have gotten rid of the fractions, -4a³ + 12a² + 6a - 10 = 0.1596

When I solve this, I get a = -1, I get a = 0.775, and I get a = 3.225.1610

-1 to -1, we can ignore this one.1628

We have -1 to 0.775, two possible values, and -1 to 3.225. 1635

There you go, that function is the graph of the function.1644

We are going from -1, this is one value of a and this is our second value of a.1650

That makes this average value equal 2.1658

That is it, we are just using the definition of average value.1663

The linear density of a rod 10 m long is a function of the distance from one end.1668

The density function is this, 17/ √3x + 5 kg/m, find the average density of the rod.1674

The average density is 1/, what does this rod looks like?1688

We have a rod which is going to be 10 m long.1694

This is 0, this is 10.1698

They are telling me as I move along the x axis, as x increases from 0 to 10, the density changes.1701

It is a variable density, as we move along the rod.1708

The average value is just , the density average = 1/ 10 - 0 1712

the integral from 0 to 10 of the functions 17/ √3x + 5 dx.1728

How are we going to solve that integral?1739

Let us go ahead and do this.1746

Let us do u is equal to 3x + 5¹/2, that is going to make du = ½ 3x + 5 ^½ × 3 dx,1748

which gives me 2 du divided by 3 is equal to dx/ √3x + 5.1781

Perfect, that means our average density is equal to 1/10 from 0 to 10, 1792

2 du/ 3 because now dx/ 3x + 5 is that right there.1815

That is just 2/3 du, then × 17.1825

We are going to get 2 × 17 is 34/30.1830

We are going to get 34/30 × the integral du from 0 to 10 which = 34/30 u from 0 to 10 = 34 ⁺30.1837

u is equal to √3x + 5, from 0 to 10.1862

When we solve this, we end up with 4.17 kg/m.1872

That is it, the average value of the depth, they have given you a function.1877

All this means is that when we move from 0 to 10, this is the density.1884

As you move along the rod, it is more dense here, towards the left end of the rod.1889

As you move to the right, the density starts to decrease.1894

That is it, that is all that is going on here.1897

The average density is going to be the average value of the function.1898

That is all hope, I hope that makes sense.1902

Once again, the average value of the function over the interval from a to b is equal to 1/ b – a, the integral from a to b of f(x) dx.1910

Thank you so much for joining us here at www.educator.com.1930

We will see you next time, bye.1932