WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, and welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to talk about the problem of finding the area under a curve from one point to another point.
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Let us jump right on in.
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Let us see, I want to find the area under the curve f(x) = x² from x = 1 to x = 5.
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That is my task, here is what I want to do.
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I’m going to draw my x² curve, something like that.
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I will say that this is my 1.
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I will say that maybe 5 is over here.
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I want to find the area under the curve from the x axis up to the curve, that is what I'm looking for.
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How can I do that?
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What I’m basically going to do what is I’m going to approximate it first with rectangles
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because I have a formula for finding the area of a rectangle.
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It is based × the height.
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I'm going to draw a bunch of rectangles, approximating it that way.
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Let us start and see what happens.
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The first thing I only to do is to choose how many rectangles I need to start with.
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In this case, I’m just going to start by approximating it with 4 rectangles.
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I’m going to break up my domain into 4 sections.
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This is my halfway mark, that is my halfway mark, that is my halfway mark.
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I have from 1, 2, 3, and I have 4.
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Now I’m going to draw my rectangles.
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Let us say I start with an approximation using 4 rectangles.
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The number 4 is arbitrary, I could have used 3, I could have used 7, I could have used 10, I could have used 114.
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It is just 4 gives us a chance to actually, it is not too many and
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allows us to see what is going on theoretically without burdening us with too much drawing.
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Here are the rectangles that I’m going to draw.
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It is going to be one rectangle, two rectangles.
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This is my first one and I want that area.
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This is my 2nd, this is my 3rd, and this is my 4th.
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I’m going to find the areas by multiplying the base × the height, base × the height, base × the height, base × the height.
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That is going to give me an approximation to the area under the curve.
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These parts right here, obviously they are not going to be included.
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It is going to be an underestimate, we will deal with that in just a second.
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My first area, area 1, that is just going to be 1.
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This is from 1 to 5, this distance is 4.
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If I break it up into 4 rectangles, each distance, each base length of a rectangle is 1.
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The first area is 1 × f(1) because this height right here, it is just f of whatever 1 is.
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It is up to the graph, this is my graph.
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F which is equal to x², it is just 1, = 1 × 1 = 1.
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My area 2, my area 2 = 1 which is the base × f(2).
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This is 3, this is 4, this is 5.
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This is going to equal 1 × f(2).
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The f(2) is 4, it is going to be 1 × 4 which is 4.
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I take my third area, my third year is going to be this one right here.
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It is going to be 1 × f(3), because the height of the rectangle is f(3).
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1 × f(3) which is equal to 1 × 9 which is equal to 9.
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My 4th area which is going to be 1, which is the base of the rectangle.
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F(4) is going to be the height of the rectangle.
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1 × f(4) which is equal to 1 × 16 which is 16.
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I add these together to get an area l = 3.
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I will put a little 4 here.
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This symbolism that I’m using the area l4, 4 stands for the number of rectangles that I actually chose,
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the number of divisions that I make to the domain.
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L stands for left the endpoint.
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I chose left endpoints, the rectangle, the left endpoint, to go up to my graph.
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Left endpoint for this rectangle, left endpoint for this rectangle, left the endpoint for that rectangle.
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That is what the symbolism means.
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My first approximation using left endpoints as my height.
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The f of the left endpoint to make the height of the rectangle.
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That gives me an approximation of 30.
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Clearly, this is an underestimate, not a problem.
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Let us go over here, let me redraw real quickly.
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I have got this, I have this one, I have this one, I have this one, and I have this one.
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This is 1, 2, 3, 4, 5.
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This area was area 1, this was area 2, this was area 3, and this was area 4.
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Notice that this al4 = 30 is an underestimate.
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It is an underestimate because the approximating rectangles are below the graph.
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The approximating rectangles are below the graph.
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Our arbitrarily shows 4 rectangles.
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The left endpoints of each rectangle, they are the ones that correspond to x = 1, x = 2, x = 3, and x = 4.
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Left, left, left, left of the points that I can choose in my domain.
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I chose left endpoint starting from the left.
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The heights of the rectangles corresponded to f(1), f(2), f(3), f(4).
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It was the distance from 1 to 2 × the height at the left endpoint.
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This base × that, this base × that, this base × this, this base × that.
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Let us do the same thing, same graph, I pick x², same number of rectangles 4.
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But let us use 4 different rectangles, but all over the same domain.
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Let us do this again, let us go back to blue.
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Let us do this again but choose 4.
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Let us choose a different 4 rectangles.
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A different 4 rectangles but still 4 rectangles.
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I have got my drawing again, I got my x² graph.
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I have 1, I have 5, midpoint, midpoint, midpoint.
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Now I have got 1, 2, 3, 4.
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This time, I’m going to choose these rectangles.
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I have area 1, area 2, area 3, and area 4.
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My area 1 is equal to, the base of the rectangle is still just 1.
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I will go ahead and do this in red.
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Is still 1, it is 1 ×, but this time I’m going to take, notice the height of the rectangle is this height right here.
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That is f(2), it is going to be 1 × 4 = 4.
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Area 2 is 1 × this time, it is f(3), that is my rectangle right there.
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That is the height of my rectangle.
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It is going to be 1 × f(3) which is going to be 1 × 9 which is 9.
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My 3rd area is going to be 1 × f(4) which is 1 × 16 is 16.
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My 4th rectangle, the height of the rectangle is going to be 1 × f(5) which is going to be 1 × 25, which is 25.
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Notice this time, I have taken right endpoints, 4 for each rectangle.
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From the right endpoint, I went up to the graph.
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And from there, I drew my rectangle to the next x value, drew my rectangle to the next x value.
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I add these together to get an area using right endpoints, 4 rectangles, I end up with 54.
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I will draw a quick representation of it again.
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This time I have got this and I have got that, 1, 2, 3, 4, 5.
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This time my rectangles are 1, 2, 3, 4.
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It is this area, this area, this area, and this area.
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A right 4 is an overestimate, you can see that it actually goes, here is the graph, the rectangles go above the graph.
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It is an overestimate, that is clear to see.
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Again, we have 4 rectangles.
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The right endpoints, it corresponds to 5, 4, 3, and 2.
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I will write it in order this way, but I will right it backwards.
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X = 5, x = 4, x = 3, x = 2, to demonstrate that I actually started from the right working left.
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The heights of the rectangles ,they correspond to f(5), f(4), f(3), and f(2).
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The true area is something between the overestimate and the underestimate.
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It is going to be more than 30 and less than the 54, whatever it is that we got.
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The true area which we will call a is between these two.
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In other words, the true area is less than or equal to a right.
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It is greater than or equal to a left, 4 and 4.
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This one was 30, I believe, this one was 54.
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That much we know, we know we have an upper limit and we know we have a lower limit.
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We know that the area is going to be somewhere in between them.
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In general, if we want to approximate with rectangles, the area under some f(x),
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from a point a to a point b, we can do so in two ways.
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The first way is we can form rectangles whose height is based on a left endpoint.
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Or we can form our rectangles whose heights are based on right endpoints.
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Let us do a left endpoint, general scheme.
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This is all just theory that we are throwing out here.
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Just want to make sure that this is really solidly understood, before we actually talk about anything specific.
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Left endpoint, our scheme is going to be as follows.
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We have some graph, it does not matter what the graph actually looks like.
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We have an a and we have a b.
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In this particular case, I’m just going to choose 5.
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Once I choose the number of rectangles, n is the number of rectangles.
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In this case, for the sake of argument here, I’m just going to choose n = 5.
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My Δx which is going to be the length of my base, that is just equal to the right endpoint - the left endpoint.
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In other words, the length of the domain divided by 5.
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That is it, I’m just breaking this up into 5 things.
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I have got 1, 2, 3, 4, there we go.
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I'm going to call a, x₁, I’m going to call this x₂, x₃, x₄, x₅.
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B, I’m going to call x₆.
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The length is just one endpoint + the Δx, whatever this is + the Δ.
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The Δx is actually the same for all of them.
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I take the difference between this right endpoint and this left endpoint.
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I divide by the number of rectangles that I’m going to have under this graph, that gives me the length of the bases.
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We are just doing left most endpoints and here is how it works.
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Here is the procedure for doing left most endpoints.
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You take the left most endpoint which is our a, which is we are going to call x1.
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And then, you go up to the graph.
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We start here, we go up to the graph.
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The next thing we do, you go right to the next x value which is x₂.
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From here, you go up to the graph, and then, you go to the right, to the next x value, that is your first rectangle.
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Now from x2, starting at x2, you go up to the graph, that puts us here.
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Then, you go right to the next x value which is x3.
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You go right which is x3.
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You keep going this way, from x3 you go up to the graph, you go to the right to the next x value.
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That is your 3rd rectangle.
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X4, you go up to the graph, that gives me my next rectangle.
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It is not going to be that if you use left endpoints, you are going to be an underestimate, right endpoints overestimate.
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That worked out that way because of the graph that we chose, x², it was that way.
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The graph itself does not matter.
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When you use left endpoints, you start to the left endpoint, you go up to the graph, that is your height.
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You go up to the next x value, that is your rectangle.
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Sometimes it is going to be above, sometimes it is going to be below, depending on the shape of the graph.
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Now last one, from x⁵, we go up to the graph.
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It is where we hit the graph and then we go to the right, to the next x value.
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That is our area 1, that is our area 2, area 3, area 4, area 5.
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We have to have this many rectangles as that.
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We have the left endpoints that we choose because we have that many rectangles.
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That is how many endpoints we have.
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We have 5, in this case, x1, x2, x3, x4, x5.
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Left endpoints, we started with a.
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The procedures are just the same.
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Whatever point that you choose, go up to the graph, and then go to the right until you hit the next x value.
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That is your height, where you hit the graph.
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That is all we are doing.
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Now our area is just equal to a1 + a2 + a3 + a4 + a5.
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This is equal to f(x1) × Δx, f(x1) is this, this is our Δx + f(x2) × Δ.
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This is f(x2), this is our Δx + f(x3) × Δx + f(x4) × Δx + f(x5) × Δx.
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I’m going to factor out the Δx, area = Δx × f(x1) + f(x2) + f(x3) + f(x4) + f(x5).
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That is it, the Δx is the same.
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It is the right endpoint of the domain - the left endpoint of the domain.
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The length divided by the number of rectangles I choose.
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Δx is the same in all cases, I can just factor it out.
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It just becomes f(x1) + f(x2) + f(x3) + f(x4) + f(x5).
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That gives me my 5 rectangles.
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I have to have as many terms as I have n, that is how I keep track.
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This is left endpoints.
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Let us do the case with right endpoints.
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I have a graph, I have a, I have b.
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1, 2, 3, 4, I have broken it up into 1, 2, 3, 4, 5 sections.
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N is equal to 5, my Δx which is the length of one section, that is b - a/5, in this case.
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I call my ax1, this is my second point, my 3rd point, my 4th point, my 5th point.
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B, I call my 6th point.
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Now using right endpoints, this is right endpoints.
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That procedure, you go to the right most endpoint which is b.
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Nice standard procedure, you go to the rightmost endpoint which is b which is x = 6.
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You go up to the graph, and then you go left to the next x value below it.
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The left to the x value below it, which in this case is x₅.
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That right endpoint, you go up to the graph, you go to the left until you hit this x value.
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That is your first rectangle.
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Now from x5, starting from x5, you go up to the graph then go left to the next x value.
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Starting at x5, you go up to the graph, you hit here.
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You go to the left to the next x value, that is your second area, and so on.
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I will do one more, this one is going to be x4.
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Notice we stop at x4.
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From x4, you go up to where it hits the graph.
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And then, you go to the left, that is your 3rd rectangle.
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From x3, you go up to where it hits the graph, and then you go to the left to where your next x value.
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That is your third rectangle.
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From x3, you go up to where it hits the graph.
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You go over to the next x value.
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This last x value is a, you stop there.
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I’m going to call this a1, this a2, a3, a4, and a5.
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I started from the right using right endpoints.
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For a particular interval, the last one I chose, left endpoints, now I’m choosing right endpoints.
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The last one we do the left, left, left, left, left.
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Now it is right, right, right, right, right, starting from the right.
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Again, my area is equal to a1 + a2 + a3 + a4 + a5.
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My area = Δx f(x6), Δx(x6) that gives me this area, +Δx f(x5).
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F(x5) that gives me this area, and so on.
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+ Δx f(x4) + Δx f(x3) + Δx f(x2).
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Where do I stop? 1, 2, 3, 4, 5 terms, that is where I stop.
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My area = Δx × f(x6) f(x5) f(x4), this is all just notation here for something that is clear geometrically.
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I'm just taking a height × width.
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F(x4) + f(x3), this is all just algebra because we need to turn geometry into algebra.
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F(x2), that is that.
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The rectangles that I got, if I’m choosing right endpoints and going up to the graph,
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are not the same as the rectangles that I choose from the left going to the left endpoints.
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We want to see that, let us superimpose them.
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Let us go back to blue.
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These left endpoint rectangles are not the same as the right endpoint rectangles.
00:31:28.000 --> 00:31:33.400
We want to make sure you see that.
00:31:33.400 --> 00:31:42.700
Here is how, nice and big.
00:31:42.700 --> 00:31:48.400
We have our function, we have a, we have b, breaking it up into 5.
00:31:48.400 --> 00:31:55.600
1, 2, 3, 4, this is 1, 2, 3, 4, 5 left endpoints.
00:31:55.600 --> 00:31:58.000
I will do those in blue.
00:31:58.000 --> 00:31:59.700
Starting from the far most left endpoint.
00:31:59.700 --> 00:32:06.600
I go up to the graph, I go across, that is my first rectangle.
00:32:06.600 --> 00:32:14.100
This is x1, this is x2, x3, x4, x5.
00:32:14.100 --> 00:32:16.800
B, of course is x6.
00:32:16.800 --> 00:32:24.200
From x2, I go up to the graph and I go across to the right, that is my second rectangle.
00:32:24.200 --> 00:32:29.600
From x3, I go up to the graph, I go to the right, I come down.
00:32:29.600 --> 00:32:35.500
At x4 I go up to the graph, go there, and I come down.
00:32:35.500 --> 00:32:38.300
Go up the graph there and I come down.
00:32:38.300 --> 00:32:44.000
Those are my left endpoint rectangles, 1, 2, 3, 4, 5.
00:32:44.000 --> 00:32:47.300
Now I begin with the rightmost endpoint, I’m going to do this in red.
00:32:47.300 --> 00:33:00.500
From the right most endpoint, from here, I go up to the graph, it is here, and I go to the left, that is my first rectangle.
00:33:00.500 --> 00:33:06.800
I go up to hit the graph, it is my second rectangle.
00:33:06.800 --> 00:33:12.500
I go up to hit the graph, my 3rd rectangle.
00:33:12.500 --> 00:33:17.300
Go up to hit the graph, it is my 4th rectangle.
00:33:17.300 --> 00:33:22.400
I go up to hit the graph, it is my 5th rectangle.
00:33:22.400 --> 00:33:37.400
The ones in red, those are the right endpoints.
00:33:37.400 --> 00:33:41.000
The ones in blue, those are my left endpoints.
00:33:41.000 --> 00:33:47.100
They are different, they are going to give me different numbers.
00:33:47.100 --> 00:33:50.100
I wanted to make sure that you knew that they were different rectangles.
00:33:50.100 --> 00:33:54.900
One of them starting with left endpoints going up to the graph going across.
00:33:54.900 --> 00:34:04.200
The other for the right endpoints, I’m starting from the right going up and cutting across.
00:34:04.200 --> 00:34:08.400
Let us give the general form, let us go back to blue.
00:34:08.400 --> 00:34:19.200
In general, for any n, I’m not going to specify what n is.
00:34:19.200 --> 00:34:33.900
For any n there are n + 1 x values.
00:34:33.900 --> 00:34:39.200
N is the number of rectangles.
00:34:39.200 --> 00:34:46.800
If you decide that you are going to do 5 rectangles, you are going to have six points total.
00:34:46.800 --> 00:34:57.000
x1, x2, x3, x4, x5, x6, that is going to be your a x2, x3, x4, x5, and then b.
00:34:57.000 --> 00:35:05.700
Let us draw this out.
00:35:05.700 --> 00:35:12.300
We have our general graph of any function, this is our f(x), no matter what it is.
00:35:12.300 --> 00:35:17.100
We have our a that is going to be our x1.
00:35:17.100 --> 00:35:23.100
We go all the way to our b, this is going to be our x n + 1.
00:35:23.100 --> 00:35:30.900
In other words, if I choose n rectangles, my b is actually my x n + 1 terms.
00:35:30.900 --> 00:35:36.600
If I choose 10 rectangles, a is my x1, b is going to be my x sub 11.
00:35:36.600 --> 00:35:49.300
If I choose 30 rectangles, b is going to be my x31.
00:35:49.300 --> 00:35:52.300
However, I break it up.
00:35:52.300 --> 00:36:00.400
Δx is equal to b - a divided by n.
00:36:00.400 --> 00:36:03.900
Final point of the domain, initial point of the domain.
00:36:03.900 --> 00:36:05.700
I subtract them, that gives me the distance.
00:36:05.700 --> 00:36:14.400
I divide by the number of rectangles that I choose, that gives me my Δx which is the length of the base of one of my rectangles.
00:36:14.400 --> 00:36:34.300
Left endpoint, area is equal to Δx × f(x1) f(x2), all the way to f(x sub n).
00:36:34.300 --> 00:36:56.200
My right endpoint, area is going to be Δx which is the same but this time it is going to be f(x2) + f(x3) f(x) n + 1.
00:36:56.200 --> 00:37:02.200
If I start from the left, it is going to be this, this, this, this, this, this, this.
00:37:02.200 --> 00:37:07.400
If I start from the right, it is going to be this, this, this, this, this, this, this.
00:37:07.400 --> 00:37:10.100
It does not include the x1.
00:37:10.100 --> 00:37:13.400
If I go from left, it does not include the x n + 1.
00:37:13.400 --> 00:37:15.800
That is why it ends at x sub n.
00:37:15.800 --> 00:37:17.900
It ends here from going to the left.
00:37:17.900 --> 00:37:23.600
Going from the right, it includes everything except it ends here.
00:37:23.600 --> 00:37:27.200
It does not include that one, that is what is going on.
00:37:27.200 --> 00:37:30.200
This is the definition of the left endpoint area.
00:37:30.200 --> 00:37:36.800
This is the definition of the right endpoint area.
00:37:36.800 --> 00:37:53.600
As you take more rectangles, that is n higher and higher, 10, 20, 30, 40, 50,
00:37:53.600 --> 00:37:58.500
clearly your Δx is going to get smaller because you are divided b - a/ Δ n.
00:37:58.500 --> 00:38:15.400
N higher and higher, you get a better approximation of the true area.
00:38:15.400 --> 00:38:31.900
That make sense, you are taking thinner rectangles so there is not as much a gap up near the graph.
00:38:31.900 --> 00:38:36.700
Let us start again here, let us erase this.
00:38:36.700 --> 00:38:55.900
As you take more and more rectangles, n higher and higher, you get better approximations to the true area.
00:38:55.900 --> 00:39:18.500
We define the true area as the limit as we take n, the number of rectangles, to infinity.
00:39:18.500 --> 00:39:23.200
That is what we do in calculus, we always take something to infinity.
00:39:23.200 --> 00:39:32.200
N to infinity of the left endpoint area or the right endpoint area.
00:39:32.200 --> 00:39:38.200
Once you form that thing, the left endpoint area, right endpoint area, with Δx and f(x),
00:39:38.200 --> 00:39:44.000
once you have a function, that is going to be some function of n, you take n to infinity.
00:39:44.000 --> 00:40:01.600
Our definition is this, the true area = the limit as n goes to infinity of the right,
00:40:01.600 --> 00:40:10.000
a true = the limit as n goes to infinity of the left.
00:40:10.000 --> 00:40:12.800
It can be shown, it can be demonstrated, which you will do if you are a math major and
00:40:12.800 --> 00:40:16.400
you go on to take some course called analysis,
00:40:16.400 --> 00:40:20.000
you have to go to demonstrate that these two limits end up being the same.
00:40:20.000 --> 00:40:26.100
It make sense, remember when we first started this lesson, we had a lower sum of 30.
00:40:26.100 --> 00:40:29.200
And then, we have an upper sum of 54.
00:40:29.200 --> 00:40:35.200
If I went a little higher, let us say to 10 rectangles, I will be a little bit more than 30, I will be a little bit less than 54.
00:40:35.200 --> 00:40:37.800
There is going to be some number that they are going to converge to,
00:40:37.800 --> 00:40:40.500
as the number of rectangles become smaller and smaller.
00:40:40.500 --> 00:40:42.900
That is the true area, that is what we have done.
00:40:42.900 --> 00:40:46.800
We found an upper sum, we found a lower sum.
00:40:46.800 --> 00:40:52.500
If we take n smaller and smaller rectangles, the lower sum is going to rise, the upper sum is going to drop.
00:40:52.500 --> 00:40:56.400
There is going to be some point where they are going to meet.
00:40:56.400 --> 00:41:24.600
It can be shown that these two limits are equal and the limit is the true area.
00:41:24.600 --> 00:41:26.500
It gets even more interesting than that.
00:41:26.500 --> 00:41:44.700
In fact, for your x₁, x₂, x₃, and so on,
00:41:44.700 --> 00:42:02.300
you can actually choose any point between an x sub n and x sub n + 1.
00:42:02.300 --> 00:42:07.400
In other words, for any rectangle, you do not have to pick a left endpoint or a right endpoint,
00:42:07.400 --> 00:42:11.800
that is just very systematic because it gives us a systematic procedure for doing things.
00:42:11.800 --> 00:42:15.500
You can actually choose any point you want in there, it absolutely does not matter.
00:42:15.500 --> 00:42:19.400
For every different rectangle, you can choose a different point.
00:42:19.400 --> 00:42:26.000
It is better if you just choose the same point consistently, either left or right, or midpoint is often a good one that we use.
00:42:26.000 --> 00:42:40.700
The truth is that you can choose any point you want.
00:42:40.700 --> 00:43:10.800
It does not have to be left or right endpoints, it can be midpoints which is used often.
00:43:10.800 --> 00:43:14.400
Often case is when you take the midpoint, you actually get a better approximation.
00:43:14.400 --> 00:43:18.900
You get an approximation, and then you take the limit as n goes to infinity.
00:43:18.900 --> 00:43:23.700
You get your particular area.
00:43:23.700 --> 00:43:30.300
Let us finish this off with a discussion of something called the sigma notation.
00:43:30.300 --> 00:43:31.900
Let me go back to blue here.
00:43:31.900 --> 00:43:40.300
Sigma notation is a shorthand notation, you have seen it before back in algebra, pre calculus, things like that.
00:43:40.300 --> 00:43:43.600
It is a shorthand notation for long sums.
00:43:43.600 --> 00:43:51.100
We do not wand to write out, f(x1) Δx, we need a shorthand notation for that.
00:43:51.100 --> 00:43:52.500
Let us go ahead.
00:43:52.500 --> 00:44:09.300
The left area of n, we said was equal to Δx × f(x1) f(x2) +…+ f(x sub n).
00:44:09.300 --> 00:44:12.900
In sigma notation, it looks like this.
00:44:12.900 --> 00:44:17.700
The sum I to n, i is the index.
00:44:17.700 --> 00:44:26.700
1, 2, 3, 4, 1, 2, 3, 4, 5, 6, it goes up to however many you want to add.
00:44:26.700 --> 00:44:31.200
It is going to be Δx f(x sub i).
00:44:31.200 --> 00:44:37.100
That is it, you put i1 in for here, that is the 1st term.
00:44:37.100 --> 00:44:40.200
And then, you go 2, that is the 2nd term.
00:44:40.200 --> 00:44:42.000
The 3, that is the 3rd term.
00:44:42.000 --> 00:44:48.000
All the way to n, that is the last term.
00:44:48.000 --> 00:45:08.800
This is where the index starts and this is where the index stops.
00:45:08.800 --> 00:45:29.500
If right, n is equal to, we said that the right one, the right endpoint was f(x2) + f(x3) + f(x) n + 1.
00:45:29.500 --> 00:45:36.100
Let us actually do it so you can read it.
00:45:36.100 --> 00:45:54.400
That sigma notation, the sum i goes from 1 to n, Δx (x) i + 1.
00:45:54.400 --> 00:46:07.900
We said that the area was the limit as n goes to infinity of ar n.
00:46:07.900 --> 00:46:13.900
Which was the same as the limit as n goes to infinity of the al n.
00:46:13.900 --> 00:46:21.100
Area = the limit as n goes to infinity.
00:46:21.100 --> 00:46:35.800
How is this for a pretty intimidating looking thing, Δx f(x sub i).
00:46:35.800 --> 00:46:40.000
Let me put the left one first and let me put the right one next.
00:46:40.000 --> 00:46:47.600
Left are x sub i = the limit as n goes to infinity.
00:46:47.600 --> 00:46:54.800
I just put this in for this here.
00:46:54.800 --> 00:47:08.000
1 to n, Δx f(x sub i) + 1.
00:47:08.000 --> 00:47:10.400
This is how we do it.
00:47:10.400 --> 00:47:19.200
You form this thing, you sum it up, you get some expression in n, and then, you take n into infinity, that gives you your area.
00:47:19.200 --> 00:47:20.400
That is what we are ultimately going to do.
00:47:20.400 --> 00:47:22.200
Again, this is all the theoretical stuff.
00:47:22.200 --> 00:47:26.700
What is important, what I want you to take from this lesson is the idea of a rectangle.
00:47:26.700 --> 00:47:30.600
Choose your left endpoints, go up to the graph, go over.
00:47:30.600 --> 00:47:33.300
Building your rectangles, that is what is important.
00:47:33.300 --> 00:47:36.700
Or right endpoints, or midpoints, whatever it is.
00:47:36.700 --> 00:47:42.400
For whatever point that you choose, you are going to go up to the graph and you are going to build your rectangle from there.
00:47:42.400 --> 00:47:46.000
That is what is really important here.
00:47:46.000 --> 00:47:50.200
As long as you can understand where this came from, that this is the base of the rectangle and
00:47:50.200 --> 00:47:53.900
these are the heights of the rectangles, that is all that is necessary.
00:47:53.900 --> 00:47:57.200
The rest it just tedious notation.
00:47:57.200 --> 00:48:23.600
Let us finish it off by saying, when you have to explicitly solve something like,
00:48:23.600 --> 00:48:36.200
the limit as n goes to infinity of the sum from 1 to n of Δx f(x sub n),
00:48:36.200 --> 00:48:51.200
the first thing you are going to do is work from the inside out, like everything else in math.
00:48:51.200 --> 00:48:57.200
I know it is notation but notation is just a shorthand for telling you what to do, it is just an algorithm.
00:48:57.200 --> 00:49:03.900
This says, the first thing you want to do is first find an expression for this.
00:49:03.900 --> 00:49:08.400
First, find an expression for this, for whatever it is.
00:49:08.400 --> 00:49:10.700
If they give you an f, you find a Δx.
00:49:10.700 --> 00:49:17.300
Remember, Δx = b - a/ n.
00:49:17.300 --> 00:49:54.600
Second, if there is a closed form expression for this sum, which are often will be for our purposes,
00:49:54.600 --> 00:50:00.300
you calculate the sum, calculate this expression.
00:50:00.300 --> 00:50:02.900
When you calculate it, it is going to make the sum go away.
00:50:02.900 --> 00:50:08.100
In other words, there is a way of finding what the sum of something is often.
00:50:08.100 --> 00:50:09.100
Not always, but often.
00:50:09.100 --> 00:50:17.200
The last thing you do, whatever expression you get, you take the limit as n goes to infinity.
00:50:17.200 --> 00:50:23.800
You know how to do limits.
00:50:23.800 --> 00:50:42.400
You take the limit as n goes to infinity of the expression you just got.
00:50:42.400 --> 00:50:46.600
Do not worry, we are going to be doing example problems.
00:50:46.600 --> 00:50:52.000
You are left with the answer which is your area.
00:50:52.000 --> 00:50:55.000
I will go to go ahead end this lesson here.
00:50:55.000 --> 00:50:59.300
The next lesson is going to be the example problems for this particular discussion.
00:50:59.300 --> 00:51:01.400
Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.