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 definite integral, very important.
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In the last lesson, we used the following definition to find the area under the curve.
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Let me see, what should I do today?
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In the last lesson, we use the following definition to find the area under a curve, f(x).
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f(x) is the function, the curve of the function but we just say curve f(x), that is fine, on the interval ab.
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We said that the area was equal to the limit as n goes to infinity.
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N is the number of approximating rectangles, we take the rectangles thinner and thinner,
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of the f(x₁) Δx, + f(x₂) to Δx, so on and so forth, f(x sub n) Δx.
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If we have 15 rectangles, we have 15 terms in the sum.
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Δx was the width of the particular rectangle, we just added them up.
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I will go ahead and express this as a = the limit as n goes to infinity.
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We have the summation notation, the shorthand for this.
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When we picked some index i, which runs from 1 to n, whatever that happens to be.
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F(x sub i) × Δx, here the Δx is equal to the b – a/ n.
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It is the right endpoint of our domain, the left endpoint of our domain of our interval divided by the number of rectangles that we wanted.
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x sub i is actually in the little sub interval x sub i - 1 x sub i.
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All this means is that when we break up the interval, let us say we have x₃, and we have x₄,
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something like that, and we have a over here, b over here.
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In this interval, our x sub i is somewhere in here.
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Sometimes, we pick the left endpoint, sometimes we pick the right endpoint.
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Sometimes, we pick the midpoint, and that it can also be any other point in between.
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That is all that that means, this part right here.
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Let me go ahead and actually draw a little picture of that.
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We have our axis, we had a curve, we had a, we had b.
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A is the one that we called x0.
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Let us say we had 1, 2, 3, 4, this would be our x₁, this would be our x₂, x₃, x₄,
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and our x₅ is going to be our b.
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This just means somewhere in there.
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Our x₄ is going to be somewhere in that little sub interval, that is all that means.
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We are going to define the definite integral.
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We now define the definite integral.
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I will go ahead and put this in red.
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You know what, I think I want to use red.
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I think to prefer to just go back to black, how is that?
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We will say let f(x) be defined on the closed interval ab, divide ab into n sub intervals.
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Everything is exactly as we did before, each of length Δx which is equal to b – a/ n.
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We will let x₀ = a, then we have x1, x2, and so forth.
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We will let x sub n = b.
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The x₀ is the left endpoint, the x sub n is the right endpoint.
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Then, the integral a to b, f(x) dx, the definite integral of the function f from a to b is equal to the limit that we had.
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It is equal to the limit as n goes to infinity of the sum i = 1 to n of f(x sub i) Δx.
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Again, where x sub i is in the sub interval, x sub i – 1, or the x sub i is somewhere in the interval x sub i - 1 x sub i.
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In other words, x₃ is going to be somewhere between x₂ and x₃.
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That is all that means.
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If this limit exists, if the limit on the right exists, in other words,
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we go through the process of finding the Δx which is the b – a/ n, we find the f(x sub i) based on the x sub i.
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We form this thing, we form the sum of this thing, and then what we get is going to be some function of n, and then we take n to infinity.
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If we end up getting a finite limit, if the limit exists, we say that f(x) is integrable.
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This, the limit that we get is that definite integral of f(x).
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If not, if the limit does not exist, then not.
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If not, then not integrable.
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This is the definition of the definite integral.
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We started up with an area, we use this limit to find an area.
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Now we are actually using this definition of area.
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No longer is area, we are actually defining it as something that has to do with the function itself.
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Something called the integral of that function.
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This thing, a to b of f(x) dx is a symbol for the entire process that we go through,
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entire process of summation + the limit as n goes to infinity.
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This is a symbol for the entire process of forming the function, taking the sum of the function, then taking the limit.
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Let us label a couple of things here, say what they are.
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This thing, the f(x), this is called the integrand.
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This right here is called the lower limit of the integral.
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This right here, analogously, is called the upper limit of the integral.
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When we read this, we always say the integral of f(x) from a to b, not from b to a, from bottom to top.
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This thing, this reminds us of the independent variable.
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If the independent variable of the function is x, this has to be dx.
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If the independent variable of the function is r, this has to be dr.
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Reminds us of the independent variable and is the differential version of the Δx.
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This symbol, it is in the shape of an elongated s to remind us
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that integration is just a long summation problem.
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As it turns out in mathematics where the only thing that you can never ever do is add two numbers,
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or 3 numbers, or in the case of calculus an infinite number of numbers.
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That is what this says, this is reminding us that all we are doing,
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when we are taking the definite integral of something is we are forming a big sum, a hundred numbers, 2000 numbers.
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Or when we pass to the integral from the limit, when we take the limit and we passed to something else, in this case, the integral an infinite sum.
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That is what is there to do, it is their to remind us that all we are doing is we are adding a lot of numbers together.
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The definite integral is a number, if it exists.
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If f(x), if the function that we happen to be dealing with is greater than 0 for all x in the particular interval ab,
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in other words, greater than 0, it means above the x axis.
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Then, this number just happens, that is the important thing, it just happens to be the area under a curve for f(x).
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We started off by dealing with curves and we use this limit that we defined, the limits of the sum of such and such.
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It turns out this integral is the deeper concept, the integral is a number.
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It is a number that is associated with a function over a certain integral.
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We define that number by going through this process of doing the limit.
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It so happens that, if the function is greater than 0, the area under the curve happens to equal the integral.
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We are not defining the integral in terms of area.
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What we are actually doing is we introduced area first
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to sort of give some physical conceptual motivation, for this thing that we are trying to do.
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As opposed to just introducing that this is the integral.
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Here, you form the sum, you take this limit and that is that.
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That is very abstract, we start with the area.
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We are defining this, we use that limit to define the integral.
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It is the integral that is the deeper concept.
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It just so happens that it happens to match the area under the curve.
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It is the integral that is the deeper concept.
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It can stand alone, it does not need to be associated with area.
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As a mathematical tool that happens to give us the area, that is what is happening.
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Let us talk about that.
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We just said that if f(x) is bigger than 0 happens to be equal to the area.
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But if we have something like this, let us go ahead and go this way.
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Something like that, let us say this is a and let us say this is b.
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Let us just call this a’ and let us call this point b’.
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For regions from a to a’, the graph of the function is above the axis.
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If I take the integral from here to here, I’m going to get an area.
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The same thing here, from b’ to b, the function is above the x axis, it is positive.
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Therefore, if I take the integral, if I take that limit, I’m going to get a and area.
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What happens though if the function is actually below?
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Here, the area is positive, here, the area is positive.
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The approximating rectangles are just, you are taking some Δx and you are essentially taking some f(x).
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F(x) is negative, what you are going to end up with is a negative area.
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For the regions below the x axis, the area, I will go ahead and put it in quotes, is negative.
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Because this f(x sub i) dx, because these f(x sub i), these are negative.
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They are below the x axis.
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F(x) here is down here, -16, -14, -307, whatever it is.
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These are negative.
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When you take the integral from a to b f(x) dx from here, all the way to here, which involves this and this, this is the definite integral.
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But it does not give you the area under the curve, it is a number.
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What it does is it gives you sort of a net area.
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It is a net area, if you will.
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Let us say this area here was 10 and let us say this area under here was 30.
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Let us say this area under here was -50.
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The integral from a to b of this function is going to be 10 - 50 + 30.
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I mean, technically, we can go ahead and take the absolute value, if we were asked what is the area under the curve?
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We could take 10 + the absolute value is -50 which is 50, 60, and 70, 80, 90.
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Yes, the total area, in terms of actual physical area is 90 but the integral is not 90.
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The integral is 10 - 50 which is -40 + 30 – 10.
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The integral is a number, it can be associated with an area if the function as above the x axis.
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But it does not have to be associated with an area.
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It just so happens to be the same sort of tool that we can use to do so.
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Integral, separate mathematical object altogether, has nothing to do with area.
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That is the important part.
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The definite integral is a deeper concept, I should say it is the deeper concept.
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It is a number which we can associate with a given function on a specified interval.
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In other words, if I gave you some f(x) and if I gave you some interval ab,
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with that function on that interval, I can associate some number.
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In other words, I can map a function on an interval to a number.
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That is what this is, mathematically, that is what is happening here.
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Given a function and the interval, I can do something to it.
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In other words, I can integrate it.
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What I get back, what I spit out, after I turn my crank and do whatever it is that I'm doing to this function, I get some number.
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For functions greater than or equal to 0, we can then associate area with this number, the integral.
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The integral comes first, the number comes first, the area afterward.
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We just introduced it in a reverse way.
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Note the definite integral, if integral of a function, I guess the real morale that I’m trying to push here,
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the definite integral of a function does not have to have a physical association.
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The definite integral exists as a mathematical object.
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It exists as a mathematical object, independent of physical reality.
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The mathematics that we used to describe the physical world, it is not there to describe the physical world.
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The mathematics exists, it is a tool.
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Whether we discovered that we can associate this math,
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we can use it to describe the physical world but it does not come from the physical world.
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In other words, the existence of the physical world does not give birth to mathematics.
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Mathematics exist independently as mathematical objects.
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It just so happens that physical reality happens to fall in line with mathematical description.
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That is what is going.
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The integral exists, it happens to be associated with an area, if we needed to.
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As we will see later on in future lessons, it is going to be associated with more than just area.
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It is going to be associated with volumes and all kinds of things, which is why calculus is so incredibly powerful.
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Let us say a little bit more.
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We chose Δx = b – a/ n, in other words, I should that is, we made Δx the same width for every integral.
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Δx the same for every approximating rectangle.
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They do not have to be equal, they do not have to be the same.
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Always approximating rectangles, we made them that way, of uniform width,
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simply to make our lives easier so that we can actually do some work, do some math.
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They do not have to be the same, each one can be different.
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You actually end up getting the same answer.
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They do not have to be the same, it simply makes things easier to handle.
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Yes, that is the only reason.
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It simply makes things easier to handle.
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Also, we said that x sub i is in x sub i -1 to x sub i.
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By choosing right endpoints, in other words by choosing x sub i as the right endpoint of that particular sub interval, that is x sub i = x sub i,
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again, we just make things easier and we make things more uniform, more consistent.
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It is not so haphazard, not without just taking random points.
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We are being nice and systematic.
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In any sub interval, we take the right endpoint.
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That is it, nice and systematic, keep things orderly.
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Again, we just make things easier.
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Or I should say, not every function is integrable.
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Just because a function exists, it does not mean that we can actually integrate it.
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In other words, when we run through the process of finding it, taking the summation, taking the limit, the limit may not exist.
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Not every function is integrable.
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Our theorem, if f(x) is continuous on the closed interval ab, if it is continuous,
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or if f(x) contains at most a finite number of discontinuities,
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the operative word being finite, then f(x) is integrable.
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We have a criteria now to decide whether a function is integrable or not.
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If the function is continuous on its domain, on its domain of definition, on its interval that we have chosen.
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If it is continuous, it is integrable, I can integrate it.
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I can find the number associated with it.
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Or if that function on its sub interval, the ab has a finite number of discontinuities, I can integrate it, it is integrable.
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For now, we will evaluate the integrals, the definite integrals.
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We will evaluate the definite integral using the definition.
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In other words, we will use the summation limit.
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We will use the definition to evaluate these things.
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It is going to be a little tedious, but that is fine.
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We want to get some sense of what is actually involved here, using the definition.
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Later of course, the same way that we did with differentiation.
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Remember, we have the differentiation, we formed the quotient f(x) + h – f(x) divided by h.
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We simplified that, then we took the limit, that gave us the derivative.
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That is the definition of the derivative.
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Then, we came up with my simple quick ways of finding the derivative.
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We will do the same here.
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We are going to use the definition first and get our feet wet with that, just to get a sense of what is going on.
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And then, we are going to come up with quick ways of doing the integration.
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You have actually already done some of them, antiderivatives.
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An antiderivative is the integral, that is what is going to be happening.
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For now, we will evaluate the definite integral using the definition.
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As such, we may need the following to help us.
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We may need the following properties to help us.
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If I have a sum i goes from 1 to n of i, there is a closed form expression for that.
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It is going to be n × n + 1/ 2.
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In other words, what this is saying is that, if I take, let us say n is 15.
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If I say, out of the numbers from 1 to 15, 1 + 2 + 3 + 4 + 5 + 6 . . . + 13 + 14 + 15,
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if I add all of those up, I can actually just take 15 × 16 and divide by 2.
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This gives me a closed form expression for the sum, that is all this is.
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Just some formulas that are going to help us along.
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I have another one, the sum as i goes from 1 to n of i², that also gives us a closed form expression.
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+ 1 to 1 + 1/ 6.
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I happen to have a closed form expression for i = 1 to n of i³.
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If I take 1³ + 2³ + 3³ + 4³ + 5³, however high I want to go, that is going to be n × n + 1/ 2².
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Let us see b as a constant, if c is a constant and if I have the summation as i goes from 1 to n,
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I have just plain old c, no i anywhere, that is just equal to n × c.
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It is just telling me, i = 1, let us say n is 10, that means add c 10 times.
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C + c + c + c and another 5×, that is 10 × c, n × c.
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This is just a formula for, when I see this, I can substitute this.
00:33:24.700 --> 00:33:26.900
That is all we are doing here.
00:33:26.900 --> 00:33:35.300
It is just writing down some formulas that we can use, when we actually start evaluating these integrals using the definition.
00:33:35.300 --> 00:33:49.300
Another thing that we can do is 1 to n c × ai, if I have some expression that involves a c, I can pull the constant out in front of it.
00:33:49.300 --> 00:33:52.000
The same that I do, when I distribute.
00:33:52.000 --> 00:33:55.600
When I factor things out, I can factor out this c.
00:33:55.600 --> 00:34:00.700
Because this is just ca₁, ca₂ + ca₃.
00:34:00.700 --> 00:34:05.200
They all involve c, just pull the c out.
00:34:05.200 --> 00:34:12.100
It equal c × the sum i = 1 to n of a sub i.
00:34:12.100 --> 00:34:49.300
What this means is that you can pull a constant out from under the summation symbol.
00:34:49.300 --> 00:34:54.700
Let us do the sum as i go from 1 to n.
00:34:54.700 --> 00:35:01.700
If we have a sub i + b sub i, let us say + or -, I can separate these out.
00:35:01.700 --> 00:35:08.500
I can write this as, the sum i go from 1 to n of a sub i, + or -, depending on whether it is + or -.
00:35:08.500 --> 00:35:12.800
In other words, the summation distributes over both.
00:35:12.800 --> 00:35:20.300
You can think of it that way, b sub i 1 to n.
00:35:20.300 --> 00:35:23.600
Properties of the definite integral.
00:35:23.600 --> 00:35:35.700
Let us go to properties of the definite integral.
00:35:35.700 --> 00:35:42.300
These are all going to be very familiar because integration is just a fancy form of summation.
00:35:42.300 --> 00:35:54.600
The summation symbols, everything that we do for summation, we can do for the integral.
00:35:54.600 --> 00:36:06.000
The integral from b to a of f(x) dx is equal to the negative of the integral from a to b of f(x) dx.
00:36:06.000 --> 00:36:17.400
In other words, if I switch the order of integration, if I do here, my Δx is b – a/ n.
00:36:17.400 --> 00:36:24.000
Here it is b – a/ n, here it is going to be a – b/ n.
00:36:24.000 --> 00:36:28.200
If I integrate from b to a, instead of from a to b, all I do is switch the sign of the integral.
00:36:28.200 --> 00:36:33.000
That is it, very simple, that is all.
00:36:33.000 --> 00:36:38.800
In calculus, we are adding from left to right.
00:36:38.800 --> 00:36:42.400
We have some function and we have broken up into a bunch of rectangles.
00:36:42.400 --> 00:36:46.600
If I add in this direction from a to b, I get a number.
00:36:46.600 --> 00:36:50.200
If I decide to add in this direction, I get the negative of that number.
00:36:50.200 --> 00:36:58.900
That is all this is saying, switch the upper and lower limit, change the sign.
00:36:58.900 --> 00:37:06.700
The integral from a to a of f(x) dx that is equal to 0.
00:37:06.700 --> 00:37:19.000
The integral from a to b of cdx, c is any constant, that is just equal to c × b – a.
00:37:19.000 --> 00:37:31.300
All this says is, if I have a constant function from a to b, the integral, it is just c,
00:37:31.300 --> 00:37:37.000
the value of c which is this height × this distance, the area.
00:37:37.000 --> 00:37:40.300
Or it could be negative because it is the integral, integral.
00:37:40.300 --> 00:37:42.800
Now we are thinking, areas can now be negative.
00:37:42.800 --> 00:37:49.500
If we introduce this notion of a negative area, it is not a problem.
00:37:49.500 --> 00:38:07.800
Some further properties, the integral from a to b of c × f(x) dx, you can pull the constant out, c × the integral from a to b of f(x) dx.
00:38:07.800 --> 00:38:20.700
The integral from a to b of f(x) + g(x) dx, the integral of a sum is equal to the sum of the individual integrals.
00:38:20.700 --> 00:38:30.900
In other words, the integral sign distributes over both.
00:38:30.900 --> 00:38:46.200
I will just write up the whole thing, of f(x) dx + the integral from a to b of g(x) dx.
00:38:46.200 --> 00:39:01.500
The integral from a to c of f(x) dx = the integral from a to b of f(x) dx + the integral from b to c of f(x) dx.
00:39:01.500 --> 00:39:05.400
This last one says I can actually break this up.
00:39:05.400 --> 00:39:12.900
If I have a here and if I have c here, if i have b here, and if that is my function,
00:39:12.900 --> 00:39:18.300
the integral from a to c is just the integral from a to b + the integral from b to c.
00:39:18.300 --> 00:39:30.000
That is all I’m doing, just adding them up.
00:39:30.000 --> 00:39:51.000
If f(x) is greater than or equal to 0 on the interval ab, then the integral of f(x) dx is greater than or equal to 0.
00:39:51.000 --> 00:39:55.800
Very simple, nothing happening here.
00:39:55.800 --> 00:40:02.700
Integration is something called an operator.
00:40:02.700 --> 00:40:07.800
An operator is just a fancy word for do something to this function, in other words, operate on it.
00:40:07.800 --> 00:40:12.300
If I give you a function f(x) and I say integrate it, integrate it means do something to it.
00:40:12.300 --> 00:40:17.800
Take this function through a series of steps and spit out a number at the end.
00:40:17.800 --> 00:40:26.500
When you see f(x) is greater than or equal to 0, you already know from years and years, elementary, junior high school math,
00:40:26.500 --> 00:40:36.200
that whenever you have an quality or an inequality, as long as you do the same thing to both sides of the quality or the inequality,
00:40:36.200 --> 00:40:38.900
you retain that equality or inequality.
00:40:38.900 --> 00:40:41.300
Here, f(x) is greater than or equal to 0.
00:40:41.300 --> 00:40:49.200
If I integrate the left side, integrate the right side, I get the left side of the integral of f(x) dx.
00:40:49.200 --> 00:40:53.400
The integral that of 0 is just a bunch of 0 added together.
00:40:53.400 --> 00:40:56.600
It is 0, it stays.
00:40:56.600 --> 00:40:58.500
Think about it that way.
00:40:58.500 --> 00:41:02.700
When you see an equality or inequality, you can integrate both sides.
00:41:02.700 --> 00:41:06.900
It retains the equality or inequality, it retains the relationship.
00:41:06.900 --> 00:41:12.600
Just like if you take the logarithm of both sides, if you exponentiate both sides, if you multiply both sides by 2,
00:41:12.600 --> 00:41:17.100
if you divide both sides by 5, as long as you do it to both sides, you are fine.
00:41:17.100 --> 00:41:24.300
Treat the integral as some operator, as some thing that you do.
00:41:24.300 --> 00:41:39.100
If f(x) is greater than or equal to g(x) on ab, then exactly what you think.
00:41:39.100 --> 00:41:48.500
The interval from a to b of f(x) is greater than or equal to the integral from a to b of g(x).
00:41:48.500 --> 00:41:56.300
This is greater than that so the integral is greater than integral, very simple.
00:41:56.300 --> 00:42:12.100
If m is less than or equal to f(x), less than or equal to M on ab,
00:42:12.100 --> 00:42:28.000
then m × b - a less than or equal to the integral a to b, f(x) dx less than or equal to m × b – a.
00:42:28.000 --> 00:42:35.000
If some function on this interval happens to have a lower bound and an upper bound,
00:42:35.000 --> 00:42:42.900
then the integral of that function is going to be greater than the lower bound × the length of the interval.
00:42:42.900 --> 00:42:46.800
It is going to be less than the upper bound × the length of the interval.
00:42:46.800 --> 00:42:50.700
All we have done here is take the integral of this, the integral of that, the integral of that.
00:42:50.700 --> 00:42:54.000
The integral of that, that is the symbol.
00:42:54.000 --> 00:42:59.400
The integral of a constant m dx, we said it is equal to m × b – a.
00:42:59.400 --> 00:43:06.700
The integral of M is M × b – a.
00:43:06.700 --> 00:43:16.000
Just integrate, integrate, solve for the things you can solve and leave those that you cannot, as the symbol.
00:43:16.000 --> 00:43:18.700
Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.