WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to do some example problems for the area under a curve.
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Let us get started.
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Use the graphs on the following pages to find left endpoint,
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right endpoint approximations for the area under the curve from 0 to 12.
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First use n = 6, we are going to use 6 rectangles then do the same for n = 12.
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Draw all the appropriate rectangles for each case.
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There are 4 copies of the graph, one for each case, do the work, write on the graph, find areas, things like that.
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Let us get started.
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For here, we have n = 6 and we are going to do left endpoints.
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We know the a is equal to 0, we know that b is equal to 12.
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We have our 0 here and we have our 12 over here.
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We know that Δx, that is going to be b - a/n, which is 12 - 0/ 6, which is equal to 2.
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Our Δx is going to equal 2.
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Left endpoints, left endpoint, go up to the graph, draw your rectangle.
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This is the endpoint, go up to the graph, draw your rectangle.
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Go up to the graph, draw your rectangle.
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Go up to the graph, go to the right to the next x value, draw your rectangle.
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Go up to the graph to your next x point, draw your rectangle.
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Go up to the graph, rectangle.
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There you go, 1, 2, 3, 4, 5, 6, Δx is 2.
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Therefore, our area, left, 6 rectangles, is equal to 2 which is the Δx.
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It is going to be f of this + f of that + f of this + f of that + f of this + f of that.
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1, 2, 3, 4, 5, 6, read it right off the graph.
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This x value is going to be 9.
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I read it right off the graph, it is 8.8.
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It is going to be 8.4, the next one is 7.6 + 6.4.
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My last one is going to be 5, there we go.
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I get a value of 90.4 using these.
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My left endpoint approximation using 6 rectangles gives me 90.4.
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We are going to do again n = 6.
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We are going to do right endpoints.
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Once again, a is equal to 0, b is equal to 12.
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It is the area I’m looking for, which means my Δx is going to be 12 - 0/6.
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Again, we are looking at 2 right endpoints.
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I start from the right, go up to the graph, and I go to the left, until I hit the next point.
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It is that one, 12, this is 2.
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That is what this means, the base of my rectangles have a length of 2.
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From here, I go up to the graph, and I go to the left to the next x endpoint.
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That is my second rectangle.
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I go up to the graph, go to the left, that is my 3rd rectangle.
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Up the graph, to the left, that is my 4th rectangle.
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Up to the graph, to the left, that is my 5th rectangle.
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Up to the graph, left, that is my 6th rectangle.
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Let us go ahead and do 1, 2, 3, 4, 5, 6.
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You can number them anyway you want.
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You can go from the left to the right, it does not really matter.
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My area with right endpoints using 6 rectangles is 2 Δx ×, we are starting over here, or I can start over here.
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It just depends on how you want to do it.
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We are using right endpoints, I just going to go ahead and start over here.
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When I do that, it is going to be 3.2 + 5 + 6.4 + 7.6 + 8.4 + 8.8.
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3.2, 5, 6.4, 7.6, 8.4, 8.8.
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When I do that, I get 78.8.
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Now we are going to do 12 rectangles.
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This time, n = 12.
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Once again, Δx is equal to 12 - 0/12.
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Now the Δx is 1, I have 12 rectangles.
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We are going to do the left endpoints.
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From the left endpoint 0, right endpoint is 12.
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Up, go across, that is one rectangle.
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Up, go across, 2nd rectangle.
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Up, go across, 3rd rectangle.
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Up to the graph, go across, 4th rectangle.
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Up, go across, up to the graph, go across.
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Up to the graph, go across, up to the graph, go across.
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Up to the graph, go across, up to the graph, go across.
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Up to the graph, go across, up to the graph, go across.
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There you go, the points 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
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Those are the points that I'm going to be looking at the area.
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Using left endpoints 12, it is going to equal, Δx is 1.
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That height, that height, that height, 9.
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I have got 9 + 8.9 + 8.8 + 8.6 + 8.4 + 8 + 7.6 + 7 + 6.4 + 5.7 + 5 + 4.2.
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I get a value of 87.6.
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These are my f values.
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Those numbers that I have read off the graph are these right here.
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Left endpoints, right endpoints, n = 12, Δx = 12 - 0/12 = 1.
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Starting from the right, starting from here, go up to the graph, go across, that is my first rectangle.
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Go up, go across, 2nd rectangle.
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Go up, go across, 3rd rectangle.
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Go up, go across, 4th rectangle.
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Go up to the graph, go across, 5th rectangle.
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Go up to the graph, go across, 6th rectangle, 7th rectangle, 8th rectangle, 9th rectangle, 10th rectangle, 11th rectangle, 12th rectangle.
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4, 5, 6, 7, 8, 9, 10, 11, 12.
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
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The heights are going to be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
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The area that you get is area using right endpoints.
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12 rectangles is 1 × we are going to have 3.2 + 4.2 + 5 + 5.7
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+ 6.4 + 7 + 7.6 + 8 + 8.4 + 8.6 + 8.8 + 8.9.
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We are going to get a value of 81. 8.
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Here is what we have, a left 6 was 90.4.
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A left 12 was 87.6, a right 6 was 78.8.
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A right 12 was 81.8, this is dropping as I went from 6 to 12.
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As I made more rectangles, smaller rectangles, this one went up.
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There is going to be some value that they are going to hit, some a true value.
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The true value actually happens to be 84.96.
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You will find out how to get that, in a couple of lessons from now.
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This drops, this rises, lower sum comes up, upper sum comes down.
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There is some point where they meet, that point is the true area.
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It turns out to be 84.96.
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Using 5 rectangles, approximate the area under the graph of f(x) = x² + 2, from x = -2 to 3 using midpoints.
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That is use the point midway between the left and right endpoints of a given approximating rectangle.
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We have got n = 5, we are using midpoints Δx = b - a/ n which is 3, - and - 2/ 5 which is 1.
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The base length of the rectangle is equal to 1.
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We are using midpoints.
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The points I’m interested are, this is one rectangle right here, midpoint between the x values.
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This is going to be a rectangle but I’m using the midpoint of the rectangle, not the left endpoint, not the right endpoint.
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I'm using midpoints, that is another point, that is another point, that is another point, that is another point.
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These x values, that is going to be my x.
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x₁,x₂, x₄, x₅.
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Now from those points, I go up to the graph and I go to the left endpoint, to the right, the right endpoint, that is my rectangle.
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I’m going to go ahead and draw this as a dotted line.
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From whatever point that I choose, that is what I go up to the graph.
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That is going to be my f value.
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From this midpoint, I go up to the graph, that is going to be my f value.
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From there, I go to the right endpoint, to the left endpoint which is 1.
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1, that is the width of my rectangle, I just happen to be taking points in between.
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From the midpoint, I go up to the graph, I go to the right, to one x value, to the left x value.
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That is my other rectangle, 1 rectangle, 2 rectangle, 3 rectangle, midpoint.
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I go up to that point, I go to the right.
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To one x, I go to the left to one x, that is my rectangle.
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Midpoint, I go up there, I go to the right, to one, to left, to one, that is my other rectangle.
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1, 2, 3, 4, 5, and these are my rectangles.
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My area, midpoints using 5 rectangles is going to be 1 ×, it is going to be f(-1.5) + f(-0.5)
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+ f(0.5) + f(1.5) + f(2.5).
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These are the points that I'm going to take the f value of, that I’m going to put into my function x² + 2 to get the y value.
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The area of the midpoints 5 is equal to 1 ×, it is going to be 4.25 + 2.25 + 2.25 + 4.25 + 7.0625.
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The area is going to equal 20.0625, the approximate area.
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I hope that made sense.
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I have my points but I’m using midpoints.
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From my rectangle, from that midpoint, that is where I go up to the graph and
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I go left and right to the x value below it, the x value above it.
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This is the height of my rectangle now.
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Not left endpoint, up to the graph, or right endpoint, up to the graph, but a midpoint up to the graph, and go left and right.
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Let us see, given the function f(x) = 3√x from 0 to 27, write an expression for the true area under the graph using limits in sigma notation.
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Let us find Δx first, let me go to blue.
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I have got Δx = b - a/ n, that is going to be 27 – 0/ n, which is going to be 27/n, that is my Δx.
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My area is the limit as n goes to infinity of the sum i = 1 to n of, we said Δx.
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The equation was the limit as n goes to infinity of the sum of Δx × f(x) sub i.
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Δx is 27/ n.
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F is this, so it is going to be 3√x sub i.
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We need to find what x sub i is, in terms of n.
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We know what the x sub i are going to be, we have a beginning point, we have an ending point.
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We actually can find some expression for x sub I, to put into this.
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Let us list them, we have x₁, that is just 0, that is the left endpoint.
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x₂, that is 0 + Δx, that is going to be 27/n.
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x₃, that is just x₂ which is 27/n + another Δx, that is going to be 54/n.
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x₄ equal, it is going to be 54/n + another Δx, that is going to be 81/n.
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x sub i is going to be some, we are going to be looking for some function of i/n.
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Let me not write this just yet, I'm looking for some sort of a pattern I can find here.
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Some relationship between, I see that there are n’s in the denominator and I see a 27, 54, and 81.
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I need to find a relationship between 27, 2, 54, 3, 81, and 4.
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In other words, is there a relationship between the I and some f(i) which will give me these numbers.
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I'm looking for an expression that I can substitute.
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I’m looking for an x sub i that I can substitute into here, that I can substitute for x sub i.
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I need some relation between i and n.
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In other words, I'm looking for some x sub i which is going to be some function of i/n.
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Based on this pattern that I elucidated from the first x₁, the 2nd point, the 3rd point, the 4th point.
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I saw this pattern, I’m looking for something here.
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I have got it, this is 2 and this is 27.
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This is 3 and this is 54.
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This is 4 and this is 81.
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27 is 1 ×, if i is 2, then this is 2 - 1 × 27.
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If i is 3, then this is 3 - 1 which is 2 × 27.
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If i is 4, 81 is equal to 3 × 27.
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Therefore, my x sub i is going to equal i - 1 × 27/n.
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This relationship, as i runs 1, 2, 3, 4, 5, 6, 7, gives me these numbers.
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Now I can put this into here and what I end up with is the following.
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A = the limit as n goes to infinity, the sum as i runs from 1 to n of 27/n × 3√27 × i - 1/n.
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Those are perfectly acceptable, I can absolutely leave it like this.
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However, there is another way to simplify this thing, to simplify the function.
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I can adjust the index as follows.
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Now I had x₁ = 0, x₂ = 27/n, I had x₃ = 54/n, I had x₄ = 81/n.
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Therefore, I had x sub i is equal to i - 1 × the 27/n.
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I can do this, we can do, we can set a equal to 0.
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And then, we can call x₁ 27/n, x₂ 54/n, x₃ is 81/n.
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Therefore, x sub i is just 27i/ n.
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It does not have that i - 1 quality.
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This is just 1, 27, 2, 2 × 27 is 54, 3, 3 × 27 is 81.
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i × 27, it makes it a little bit easier.
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All I have done is instead of calling x1 0, I just shifted the index down.
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I just called a = 0, I have called x1 the 27/n, I shifted the index.
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I end up with a is equal to the limit as n goes to infinity, the sum from 1 to n of 27/n × 3√27 i/ n.
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It is a lot simpler and I can actually pull out the 3√27 here.
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I can write a = the limit as n goes to infinity of the sum i = 1 to n.
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3√27 is 3, 3 × 27 is 81.
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We get 81/n × 3√i/n.
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When I actually evaluate the sum which I will do in the next problem, a version of that,
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I just pull out all the n, all that I'm left with under the summation sign is an i.
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Everything else comes out as a constant.
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It is really nice.
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That is our final answer, or this one, either one is fine.
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Given the function f(x) = 4x³ from 0 to 2, find the true area by actually evaluating the limit expression that you obtain.
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Use the following fact about the sum of cubes in the first n integers.
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When I end up with a sum of cubes, 1³ + 2³ + 3³, all the way to whatever number n³,
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I have a closed form expression for that, that is this.
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If I have a sum that I can express as this, I can express in sigma notation, I can replace that sigma notation with this.
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That is what this is saying.
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Let us go ahead and do this problem.
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Let us start with our Δx, I’m going to start over here.
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Δx = b - a/ n, it is just going to be 2/n.
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I know, therefore, that my area is going to equal the limit as n goes to infinity of the sum
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as i goes from 1 to n of Δx, which is my 2/n × 4 × x sub i³.
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That is my definition.
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My question is, what can we put for this x sub i in there?
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Let us see what we have got.
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Let us come over here.
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I think I’m going to do, I’m going to do what I did last time.
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I’m going to set a equal to 0, and I’m going to set x₁ equal to, it is 0 + Δx which is 2/n.
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x₂ = 2/n + 2/n which is 4/n.
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x₃ = 4/n + 2/n which is 6/n.
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I got my relationship 1, 2, 2, 4, 3, 6, x sub i is equal to 2 sub i/n.
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A is equal to the limit as n goes to infinity, i goes from 1 to n of 2/n × 4 x sub i³.
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x sub i is this, 4 × 2i/ n³.
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This is going to equal the limit as n goes to infinity of the sum i = 1 to n.
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2 × 4 is 8, this is 8/n × 8 i³/ n³.
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We are just doing everything that we need to do.
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This is a function of i, the index is what the variable is under a summation sign.
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8 × 8 is 64, n × n³ = n⁴.
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I’m just going to pull all of that and take it outside of the summation sign, which I can actually do.
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This = the limit as n goes to infinity, 8 × 8 is 64, n × n³ is n⁴ × the summation of i goes from 1 to n of i³.
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The index that I choose, that is the variable.
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Everything else, I can pull out is a constant.
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This is this, now I have something already, I know what this is.
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The sum as i goes from 1 to n of i³, that is just equal to 1³ + 2³ + all the way to n³.
00:29:32.100 --> 00:29:36.600
They gave me an expression for that, they said it is this.
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That = n × n + 1/ 2².
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I’m going to put this in for that and I get the following.
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I get a = the limit as n goes to infinity of 64/ n⁴ × n × n + 1/ 2²
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= the limit as n goes to infinity of 64/ n⁴ × n² × n² + 2n + 1/ 4,
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= the limit as n goes to infinity of 64/ n⁴ × n⁴ + 2n³ + n²/ 4.
00:31:04.900 --> 00:31:18.700
64 divided by 4, this is going to equal the limit as n goes to infinity of 16 × n⁴/ n⁴ is 1.
00:31:18.700 --> 00:31:25.600
2n³/ n⁴ is 2/n.
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N²/ n⁴ is n².
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When I take n to infinity, this term goes to 0, this term goes to 0.
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That is it, nice and simple.
00:31:44.500 --> 00:31:57.500
I have my area = the limit as n goes to infinity of the sum i to n of my Δx × my f(x sub i).
00:31:57.500 --> 00:32:01.400
I found an expression for Δx, I put it in.
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I used my x₁, x₂, x₃, x₄, to elucidate some relationship,
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in terms of n and I that I can put in for f(x sub i).
00:32:12.100 --> 00:32:19.000
I put it in, I solved this, pulled out what I needed leaving just i.
00:32:19.000 --> 00:32:22.400
As much as I can, i, underneath the summation symbol.
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I had a closed form expression for that summation symbol.
00:32:25.100 --> 00:32:28.400
I solved that, combined by simplifying.
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And then, my final step, I just took the limit is n goes to infinity.
00:32:31.700 --> 00:32:38.000
This goes to 0, this goes to 0, the limit is 16.
00:32:38.000 --> 00:32:40.400
This is a long way of doing this.
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In a couple of lessons we will find a very quick way of doing,
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but it is really important that we understand where this comes from.
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We did not just pull it out of nowhere.
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Clearly, a lot of intricate detailed work goes into defining some of these higher mathematical concepts,
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like finding area under a curve which we are going to call integration,
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which is going to be essentially anti-differentiation.
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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, bye.