AP Calculus AB The Fundamental Theorem of Calculus

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

Today, we are going to talk about the fundamental theorem of calculus, profoundly important.0004

This central theorem of calculus, probably one of the most important theorems in all human intellectual history.0008

Absolutely fantastic, let us jump right on in.0015

Let us start by evaluating an integral.0020

Let me go to black here.0024

Let us start by evaluating the integral from a, some constant, to x, which I'm going to leave variable.0030

The upper limit can be anything of the function t dt.0046

Here is what we are doing.0058

Let me draw this out real quickly.0060

This function f(t) is t, that is just this function right here, that is just a line.0068

We are saying pick some a, whatever that is, it does not matter where it is, here or there.0088

We are saying integrate to some point x.0093

We are leaving x open ended.0101

This can be, let us say 1 to 5, 6, 7, 8, we are just leaving it open ended.0104

t is the variable of integration here.0111

t is the variable of integration, in other words, it is the independent variable of this function.0115

This is the t axis, this is the f(t) axis.0121

It is the variable of integration, in other words, the independent variable of the function.0126

We simply leave x unspecified.0152

In other words, it can be anything.0166

Again, oftentimes, especially when you are just learning something, it is best to work formally.0170

In other words, do not worry if you have not wrapped your mind around what everything means, just work symbolically.0178

x does not mean anything, the interval symbolism does not mean anything.0185

a does not mean anything, t does not mean anything, d does not mean anything.0188

But you know how to manipulate these symbols, that is what is important.0191

Let us go ahead and evaluate this integral.0195

We know that the integral from a to x of f(t) dt is the limit as n goes to infinity of the sum0199

as i goes from 1 to n of f(t) sub i × Δt.0227

We know that, we are going to do the same thing.0235

This thing, our f(t) is t.0238

We are going to find Δt, we are going to find f(Δt).0245

We are going to multiply them, we are going to evaluate the sum.0249

And then, we are going to take the limit and that is going to give us the integral.0252

Let us start off with Δt here.0258

Δt, that is equal to x - a/ n.0265

I know that the t sub 0 is equal to a.0276

I know that t sub 1 is equal to a + Δx which is x - a/ n.0280

I know that t sub 2 = a + x – a/ Δn + Δx which is another x – a Δn.0290

It is going to be +2 × x - a/ n.0298

Therefore, I know that t sub i is going to equal a + i × x - a/ n.0305

I found my t sub i, I have my Δt.0315

I’m going to plug my t sub i a + i × x - a/ n into my f which is that.0321

Here, the f(t sub i) is equal to t sub i, because f(t) is equal to t.0331

Therefore, f(t sub i) × Δt is equal to a + i × x - a/ n × x - a/ n.0343

This is my f(t sub i), this is my t sub i.0367

When I do all of my multiplication, I get myself a × x - a/ n + i × x - a/ n × x - a/ n.0371

It is going to equal ax/ n - a²/ n + i × x² - 2xa + a²/ n².0390

It is going to equal ax/ n - a²/ n + ix²/ n² - 2i xa/ n² + i a²/ n².0413

This is my f(t) i Δt.0439

I’m going to evaluate the sum.0454

I’m going to evaluate this sum.0458

It is going to equal the sum of this, the summation symbol distributes over each of these.0462

I have 1, 2, 3, 4, 5 summations, equals the sum as i goes from 1 to n ax/ n - the sum as i goes from 1 to n of a²/ n +0468

the sum as i goes from 1 to n of ix²/ n² - the sum as i goes from 1 to n of 2ix a/ n² +0490

the sum as i goes from 1 to n of i a²/ n².0512

We pull out the things that we can pull out.0521

I’m going to get ax/ n × the sum, I’m just going to leave off the i = 1 to n of 1.0524

We know that we are going from 1 to n - a²/ n × the sum of 1 + x² n².0535

I have to leave the i underneath, + x²/ n² × the sum of i - 2x a/ n²0547

× the sum of i + a²/ n² × the sum of i.0564

= ax/ n × n - a²/ n × n + x²/ n² × n × n + 1/ 2.0582

Because the summation from 1 to n of i is equal to this closed form expression, -2xa/ n × n × n + 1/ 2.0604

I’m just going to write it down here.0619

+ a²/ n² × n × n + 1/ 2 = ax - a² + x²/ n² +,0620

What is going on here, + a² + x²/ 2.0662

I’m going to multiply all this out, multiply all of these out.0676

+ x²/ 2n -, this is n².0680

I have to make sure that I got everything right here.0699

n, n, n², n², n², -x/ a - xa/ n + a²/ 2 + a²/ 2n.0701

We have our final summation that we have evaluated.0736

Now we take the limit as n goes to infinity of this expression.0740

I’m just going to go ahead and do it as is, instead of rewriting it.0752

First of all, let us go ahead cancel a few things.0757

ax and xa cancel, as n goes to infinity, this term goes to 0.0761

This term goes to 0, this term goes to 0.0770

I'm left with -a² + x²/ 2 + a²/ 2.0774

This and this, I'm left with x²/ 2 - a²/ 2.0790

There we go, the integral from a to x of t dt, I picked a specific function t dt is equal to x²/ 2 - a²/ 2.0798

No matter what x happens to be.0817

Notice, this integral gave us a function of x because we said x can be anything, the upper limit.0821

What I end up with is sum function of x.0836

Now I do this, now take the derivative of the thing that you just got.0845

Now take ddx of your x²/ 2 - a²/ 2.0852

The derivative of x²/ 2 is x and the derivative of this is 0.0862

There you go, here is what we did.0870

We have a function t, we integrate it to get this as a function of x.0874

We differentiated that and we ended up actually getting our x.0880

In some sense, the t and the x are the same.0885

You will see that in just a second.0889

I just wanted to throw that out there.0891

We found the integral, I just happen to take the derivative.0893

I will tell you why in just a second.0897

What we have is this, what we have is this, ddx of the integral from a to x of t dt is equal to x.0903

In general, the derivative with respect to x of the definite integral from a to x0926

of whatever function happens to be f(t) dt will always give you f(x).0940

In other words, this t is just a variable of integration.0948

If x is the upper limit of that integral, if I integrate the function0954

and then take the derivative of what I integrated, I end up just getting my x f(x) back.0960

In other words, this ddx operator and this integral operator, they cancel each other out leaving you just f(t),0966

but this upper limit goes into that t leaving you a function of x.0974

Differentiation and integration are inverse processes.0979

The logarithm of the exponential cancel, they cancel each other out.0986

The exponential of the logarithm, they cancel each other out.0991

The derivative of the integral, they cancel, leaving you just f(t).0995

The integral of the derivative, they cancel, leaving you just f(t).0999

They are inverse processes.1004

If the upper limit happens to be just the variable x, what you end up with is just your function of x.1005

In other words, this symbol and this integral symbol go away, all you have to do is put this into here.1011

Whatever f happens to be, just write it as a function of x, that is what is actually happening.1017

Now if I define a function as a definite integral with the upper limit being x,1027

if I define a function by means of an integral which I can do, it is not a problem.1033

By integral such as, if I set the g(x) is equal to the integral from a to x of f(t) dt,1048

it is a strange way of actually defining the function.1063

Not strange in the sense that you have never seen it before.1065

But all this is saying that, if I’m given some random function f(t), if I actually form the integral of it,1068

leaving the upper limit of integration as a variable x, what I end up with is a function of x g(x).1075

I'm saying that g(x) is this.1082

Instead of giving you the f(x) explicitly, I'm giving it to you in terms of another function f(t).1085

I'm saying f take f(t), integrate that function, evaluate that sum, leave x variable that gives me a function of x.1092

As you just saw, the integral of t dt was x²/ 2 – a²/ 2, that is a function of x.1100

It is strange in a sense that you have never seen it, but it is perfectly valid to define a function this way.1109

If I define a function by means of an integral such as this, if you ever see this,1115

then, g’(x), the derivative of this g is nothing more than f(x).1124

In other words, just get rid off the symbol and just put x in for t.1132

Differentiation and integration are inverse processes, very profound, very deep.1142

There is no reason in the world to believe that this should be true.1159

The two branches of calculus actually developed independently.1163

Without any sort of relationship between the two, we ended up discovering that there is a relationship between the two.1169

That is a really profound moment in intellectual history.1176

Two people working on separate things, or the same person working on two separate things, and somehow they come together.1183

The relationship is, these things are inverse processes of each other.1188

If I have a function, I could integrate it.1192

If I differentiate that, I get my function back.1194

If I have a function, I can differentiate it.1198

If I integrate what I get, I get my function back, that is amazing.1200

Let us go ahead and write the first fundamental theorem of calculus.1206

The first fundamental theorem of calculus, abbreviated as FTC.1213

If f(x) is continuous, if f is continuous on the closed interval ab,1232

and we define g as g(x) equals the definite integral from a to x of f(t) dt,1253

then g(x) itself, 1, is continuous, 2, it is continuous on ab.1273

It is differentiable on ab, and 3, g’(x) is nothing more than f(x) in essence.1294

Start with f, integrate to get some other function.1318

Take that function, if you differentiate, it brings you back to f, very simple.1333

The second fundamental theorem of calculus which will allow us to evaluate definite integrals directly.1343

It is actually one theorem, you can call it first part, second part.1357

I just tend to think of it as first fundamental theorem, second fundamental theorem.1359

They are the same thing really.1364

Second fundamental theorem of calculus.1365

If f(x) is continuous on the closed interval ab, then the integral from a to b of f(x) dx1375

= f(b) – f(a), where f is any antiderivative of f.1396

F’ is equal to f.1423

If I have any integral from a to b of f(x) dx, I find the antiderivative.1426

The antiderivative, I stick in b, and then I stick in a, and I subtract the f(a) from the f(b).1434

That actually gives me the value of that, that is the second fundamental theorem of calculus.1443

Thank you so much for joining us here at www.educator.com.1450

We will see you next time for example problems on the fundamental theorem of calculus.1452

Take care, bye.1457