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Riemann sums, left endpoints

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Riemann sums, left endpoints

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In this video we're talking about how to use a Raymond Sum and left end points to find the area underneath a curve. And remember that a Riemann sum is just one technique that we can use to approximate area under a curve before we learn how to use an integral or an antiderivative to find exact value underneath the curve. So Raymond Sum is going to let us approximate area and the curve that we've been given in this particular problem is F of X is equal to one divided by x. We've been asked to find the area underneath this curve and above the x axis over the interval x equals one to x equals five. So on this interval right here, we're looking for area underneath this curve and we've been told that n equals four and n is going to be the number of sub intervals that you divide this interval into. Or you can think of it as the number of rectangles you're going to use to approximate area underneath the curve. So whenever you're dealing with a ream in some problem, the first thing you're going to want to do is find Delta X and the formula that we always use to find Delta X think of Delta X is the width of each one of our rectangles. The formula is going to be B minus A divided by N. We'll keep in mind here that the interval we've been given is the interval A to B, so we essentially have a comma B here. So that's where A and B come from. And is this any value we've been given here, the number of rectangles or the number of sub intervals and is equal to four? So if we want to find Delta X, we can just say B minus A or in our case, five minus one divided by N, which we know is four. And then when we simplify, five minus one is four, so we end up with four over four or just one. So Delta X is going to be equal to one. That means that the width of each of our rectangles is going to be one unit. So before we get into the formula that we'll use for the Raymond Sum, let's take a look at the curve. F of X equals one divided by X to figure out what this delta x value really means. So here's what our curve looks like. This is the graph of F of x equals one divided by x, and we've just zoomed in here. We've got the origin and you can see that we have the interval here, x equals one to x equals five. So what we're interested in approximating is the area underneath this curve between these two green lines I just drew. So over this interval here, x equals one to X equals five. Now, remember, we're supposed to divide this interval into four rectangles. We use that information to say that Delta X was equal to one, or that the width of each rectangle was equal to one. So our first rectangle is always going to start at the left edge of our interval, which here is x equals one. If the width of every rectangle is one, then the width of the first rectangle has to go from 1 to 2 and end right here, because the width here between one and two would be one. So this interval from 1 to 2 is going to define the first rectangle, and then the second rectangle would be defined from 2 to 3, and then from 3 to 4 and then from 4 to 5. So the reason that we find Delta X is so that we can start here at our left edge and count up by Delta X so that we can find the right edge of the first rectangle and then the right edge of the next rectangle. And we keep counting up until we get to the end. And what you can see that that results in is four rectangles. So we have the first rectangle here, the second rectangle, the third rectangle and the fourth rectangle. And we should always end up with the number PN that we were given. So we have four rectangles or four sub intervals. Now we've been asked to use left end points with this Raymond Sum to approximate area, which means that the height of each rectangle we already know the width of each rectangle is one, but the height of each rectangle is going to be dictated by the point at which the left edge of each rectangle meets the curve. So this first rectangle, the width, goes here from 1 to 2, so the left edge of that interval is one. If we come up from that point until we meet the curve, we end up right here. So this point right here is going to be the value that dictates the height of this first rectangle. So if we wanted to draw the first rectangle, here's what it would look like. This would be the height, and we would come down and fill up the width of that rectangle. And so we would use this rectangle right here to approximate the area under the curve over this interval. So we're using this rectangle to approximate this area right here. And then if we want to draw the rectangle for the second interval here, interval number two, we know that the width goes from 2 to 3, but the height is going to be determined by the left edge. So if we look at the left edge, which is an X equals two, and we come up to the point where that meets the curve, that's this point right here. So what we can say then is that the rectangle is going to look like this. It's going to be that height and that width. And now this is the rectangle that we're going to use to approximate this area right here, the area underneath the curve. And so if we kept drawing our rectangles, what we'd have here is this rectangle. At 3 to 4 and then this rectangle from 4 to 5 over the interval, 4 to 5. And so we would use these four rectangles to approximate the area underneath the curve. Now, since we're using left end points, we're interested in the left endpoints. What we want to do is look at the left end points for each of these rectangles. So the left end point for this first rectangle is X equals one. We can call that x sub one. The left end point for the second rectangle is here at x equals two. We call that x sub to the left end point for the third rectangle is x sub three and the left end point for the fourth rectangle is x sub four. Now we don't put any value here an x equals five because remember we're only interested in left end points and x equals five only represents the right end point of the last rectangle. So we just want these left end points. And if we look then at these left end points, what we can say is that this point right here on the curve would just be f of x sub one. In other words, if we wanted to find the value of that point, we would plug x sub one into our original function and it would give us this value right here. So at that point is F of x sub one, this point here is going to be F of x sub two. This point will be F of x sub three and this point will be F of x sub four. Now you can see that for our area formula and this is going to be the area underneath the curve approximated using these rectangles. All we need is to take Delta X, which we already found. We said Delta X is equal to one and multiply it by the sum of all these values here. F of x one, f of x two, f of x three and F of x four. So if we were going to plug some things into our formula, what we would say is that area is going to be equal to Delta X we know is one multiplied by F of x sub one will x sub one we know is one. So we'll just say f of one x sub two is two. So we'll say F of two. Then we have x sub three which is at three. So we say plus f of three and x sub four is at four. So we're going to say plus F of four and that's our last left end point. So this is going to be then the equation that approximates area underneath the curve. So now what we need to do is find the values of F of one, F of two, F of three and F of four. And the way that we do that, remember that F of one is just whatever we get when we plug one into our original function here. So if we plug one in four x, we're going to get F of one is equal to plugging in one for x, we get one over one or one to find F of two will say F of two is going to be equal to one over two to find F of three plugging three into our original function for X, we get one over three and then F of four is going to be one over four. Now we just plug those back into our area equation and we get area is equal to one multiplied by F of one, which is one plus F of two which is one half plus F of three which is one third plus F of four, which is one fourth. Now keep in mind that multiplying by one doesn't have any effect. So we can go ahead and just cancel that out. And now we're just looking at one plus one half plus one third plus one fourth. We need to go ahead and find a common denominator. Our common denominator is going to be 12. So this one is going to become 12 over 12, one half is going to become six over 12, one third is going to become four over 12 and one fourth is going to become three over 12. And so then when we add all those together, 12 plus six gives us 18 plus four gives us 22 plus three gives us 25. So we end up with 25 over 12 as an approximation for the area underneath this curve over the interval, 1 to 5 using for rectangles and left end points to approximate the area.