WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to talk about the fundamental theorem of calculus, profoundly important.
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This central theorem of calculus, probably one of the most important theorems in all human intellectual history.
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Absolutely fantastic, let us jump right on in.
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Let us start by evaluating an integral.
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Let me go to black here.
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Let us start by evaluating the integral from a, some constant, to x, which I'm going to leave variable.
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The upper limit can be anything of the function t dt.
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Here is what we are doing.
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Let me draw this out real quickly.
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This function f(t) is t, that is just this function right here, that is just a line.
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We are saying pick some a, whatever that is, it does not matter where it is, here or there.
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We are saying integrate to some point x.
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We are leaving x open ended.
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This can be, let us say 1 to 5, 6, 7, 8, we are just leaving it open ended.
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t is the variable of integration here.
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t is the variable of integration, in other words, it is the independent variable of this function.
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This is the t axis, this is the f(t) axis.
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It is the variable of integration, in other words, the independent variable of the function.
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We simply leave x unspecified.
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In other words, it can be anything.
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Again, oftentimes, especially when you are just learning something, it is best to work formally.
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In other words, do not worry if you have not wrapped your mind around what everything means, just work symbolically.
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x does not mean anything, the interval symbolism does not mean anything.
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a does not mean anything, t does not mean anything, d does not mean anything.
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But you know how to manipulate these symbols, that is what is important.
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Let us go ahead and evaluate this integral.
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We know that the integral from a to x of f(t) dt is the limit as n goes to infinity of the sum
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as i goes from 1 to n of f(t) sub i × Δt.
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We know that, we are going to do the same thing.
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This thing, our f(t) is t.
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We are going to find Δt, we are going to find f(Δt).
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We are going to multiply them, we are going to evaluate the sum.
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And then, we are going to take the limit and that is going to give us the integral.
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Let us start off with Δt here.
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Δt, that is equal to x - a/ n.
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I know that the t₀ is equal to a.
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I know that t₁ is equal to a + Δx which is x - a/ n.
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I know that t₂ = a + x – a/ Δn + Δx which is another x – a Δn.
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It is going to be +2 × x - a/ n.
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Therefore, I know that t sub i is going to equal a + i × x - a/ n.
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I found my t sub i, I have my Δt.
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I’m going to plug my t sub i a + i × x - a/ n into my f which is that.
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Here, the f(t sub i) is equal to t sub i, because f(t) is equal to t.
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Therefore, f(t sub i) × Δt is equal to a + i × x - a/ n × x - a/ n.
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This is my f(t sub i), this is my t sub i.
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When I do all of my multiplication, I get myself a × x - a/ n + i × x - a/ n × x - a/ n.
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It is going to equal ax/ n - a²/ n + i × x² - 2xa + a²/ n².
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It is going to equal ax/ n - a²/ n + ix²/ n² - 2i xa/ n² + i a²/ n².
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This is my f(t) i Δt.
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I’m going to evaluate the sum.
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I’m going to evaluate this sum.
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It is going to equal the sum of this, the summation symbol distributes over each of these.
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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 +
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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 +
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the sum as i goes from 1 to n of i a²/ n².
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We pull out the things that we can pull out.
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I’m going to get ax/ n × the sum, I’m just going to leave off the i = 1 to n of 1.
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We know that we are going from 1 to n - a²/ n × the sum of 1 + x² n².
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I have to leave the i underneath, + x²/ n² × the sum of i - 2x a/ n²
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× the sum of i + a²/ n² × the sum of i.
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= ax/ n × n - a²/ n × n + x²/ n² × n × n + 1/ 2.
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Because the summation from 1 to n of i is equal to this closed form expression, -2xa/ n × n × n + 1/ 2.
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I’m just going to write it down here.
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+ a²/ n² × n × n + 1/ 2 = ax - a² + x²/ n² +,
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What is going on here, + a² + x²/ 2.
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I’m going to multiply all this out, multiply all of these out.
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+ x²/ 2n -, this is n².
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I have to make sure that I got everything right here.
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n, n, n², n², n², -x/ a - xa/ n + a²/ 2 + a²/ 2n.
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We have our final summation that we have evaluated.
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Now we take the limit as n goes to infinity of this expression.
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I’m just going to go ahead and do it as is, instead of rewriting it.
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First of all, let us go ahead cancel a few things.
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ax and xa cancel, as n goes to infinity, this term goes to 0.
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This term goes to 0, this term goes to 0.
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I'm left with -a² + x²/ 2 + a²/ 2.
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This and this, I'm left with x²/ 2 - a²/ 2.
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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.
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No matter what x happens to be.
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Notice, this integral gave us a function of x because we said x can be anything, the upper limit.
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What I end up with is sum function of x.
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Now I do this, now take the derivative of the thing that you just got.
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Now take ddx of your x²/ 2 - a²/ 2.
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The derivative of x²/ 2 is x and the derivative of this is 0.
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There you go, here is what we did.
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We have a function t, we integrate it to get this as a function of x.
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We differentiated that and we ended up actually getting our x.
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In some sense, the t and the x are the same.
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You will see that in just a second.
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I just wanted to throw that out there.
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We found the integral, I just happen to take the derivative.
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I will tell you why in just a second.
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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.
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In general, the derivative with respect to x of the definite integral from a to x
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of whatever function happens to be f(t) dt will always give you f(x).
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In other words, this t is just a variable of integration.
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If x is the upper limit of that integral, if I integrate the function
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and then take the derivative of what I integrated, I end up just getting my x f(x) back.
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In other words, this ddx operator and this integral operator, they cancel each other out leaving you just f(t),
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but this upper limit goes into that t leaving you a function of x.
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Differentiation and integration are inverse processes.
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The logarithm of the exponential cancel, they cancel each other out.
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The exponential of the logarithm, they cancel each other out.
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The derivative of the integral, they cancel, leaving you just f(t).
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The integral of the derivative, they cancel, leaving you just f(t).
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They are inverse processes.
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If the upper limit happens to be just the variable x, what you end up with is just your function of x.
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In other words, this symbol and this integral symbol go away, all you have to do is put this into here.
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Whatever f happens to be, just write it as a function of x, that is what is actually happening.
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Now if I define a function as a definite integral with the upper limit being x,
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if I define a function by means of an integral which I can do, it is not a problem.
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By integral such as, if I set the g(x) is equal to the integral from a to x of f(t) dt,
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it is a strange way of actually defining the function.
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Not strange in the sense that you have never seen it before.
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But all this is saying that, if I’m given some random function f(t), if I actually form the integral of it,
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leaving the upper limit of integration as a variable x, what I end up with is a function of x g(x).
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I'm saying that g(x) is this.
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Instead of giving you the f(x) explicitly, I'm giving it to you in terms of another function f(t).
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I'm saying f take f(t), integrate that function, evaluate that sum, leave x variable that gives me a function of x.
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As you just saw, the integral of t dt was x²/ 2 – a²/ 2, that is a function of x.
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It is strange in a sense that you have never seen it, but it is perfectly valid to define a function this way.
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If I define a function by means of an integral such as this, if you ever see this,
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then, g’(x), the derivative of this g is nothing more than f(x).
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In other words, just get rid off the symbol and just put x in for t.
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Differentiation and integration are inverse processes, very profound, very deep.
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There is no reason in the world to believe that this should be true.
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The two branches of calculus actually developed independently.
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Without any sort of relationship between the two, we ended up discovering that there is a relationship between the two.
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That is a really profound moment in intellectual history.
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Two people working on separate things, or the same person working on two separate things, and somehow they come together.
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The relationship is, these things are inverse processes of each other.
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If I have a function, I could integrate it.
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If I differentiate that, I get my function back.
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If I have a function, I can differentiate it.
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If I integrate what I get, I get my function back, that is amazing.
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Let us go ahead and write the first fundamental theorem of calculus.
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The first fundamental theorem of calculus, abbreviated as FTC.
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If f(x) is continuous, if f is continuous on the closed interval ab,
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and we define g as g(x) equals the definite integral from a to x of f(t) dt,
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then g(x) itself, 1, is continuous, 2, it is continuous on ab.
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It is differentiable on ab, and 3, g’(x) is nothing more than f(x) in essence.
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Start with f, integrate to get some other function.
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Take that function, if you differentiate, it brings you back to f, very simple.
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The second fundamental theorem of calculus which will allow us to evaluate definite integrals directly.
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It is actually one theorem, you can call it first part, second part.
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I just tend to think of it as first fundamental theorem, second fundamental theorem.
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They are the same thing really.
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Second fundamental theorem of calculus.
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If f(x) is continuous on the closed interval ab, then the integral from a to b of f(x) dx
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= f(b) – f(a), where f is any antiderivative of f.
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F’ is equal to f.
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If I have any integral from a to b of f(x) dx, I find the antiderivative.
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The antiderivative, I stick in b, and then I stick in a, and I subtract the f(a) from the f(b).
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That actually gives me the value of that, that is the second fundamental theorem of calculus.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time for example problems on the fundamental theorem of calculus.
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Take care, bye.