Riemann sums, left endpoints

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English [Auto] In this video we're talking about how to use a Riemann sum and left end points to find the area underneath the 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 Riemann 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 1 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 5 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 under the curve. So whenever you're dealing with a Riemann sum problem the first thing you're going to want to do is find a delta x and the formula that we always use to find delta x of Delta axis the width of each one of our rectangles. The formula is going to be B minus a divided by and 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. N is this value we've been given here. The number of rectangles or the number of sub intervals and is equal to 4. So if we want to find delta x we can just say B minus say or in our case 5 minus 1 divided by N which we know is 4. And then when we simplify. Five minus one is four. So we end up with four over 4 or just one. So delta x is going to be equal to 1. That means that the width of each of our rectangles is going to be 1 unit. So before we get into the formula that we'll use for the Riemann sum. Let's take a look at the curve f of x equals 1 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 sequels 1 divided by X and we just zoomed in here we've got the origin and you can see that we have the interval here. X equals 1 to X equals 5. 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 1 or that the width of each rectangle was equal to 1. So our first rectangle is always going to start at the left edge of our interval which here is X equals 1. If the width of every rectangle is 1 then the width of the first rectangle has to go from 1 to 2 and end right here because the width here between 1 and 2 would be 1. So this interval from 1 to 2 is going to define the first rectangle and then the second rectangle would be to find 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 n 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 Riemann sum to approximate area which means that the height of each rectangle we are you know the width of each rectangles 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 two to three but the height is going to be determined by the left edge. So if we look at the left edge which is X equals 2 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 three to four then this rectangle from four to five over the interval four to five. 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 end points what we want to do is look at the left end points for each of these rectangles. So the left and point for this first rectangle is X equals 1. We can call that x 1 the left end point for the second rectangle is here. X equals 2. We call that X to the left and point for the third rectangle is x 3 and the left end point for the fourth rectangle is x. For now we don't put any value here in X equals 5 because remember we're only interested in left end points and X equals 5 only represents the right endpoint of the last rectangle. So we just want these left end points and if we look at these left end points what we can say is that this point right here on the curve would just be half of x 1. In other words if we wanted to find the value of that point we would plug X someone into our original function and it would give us this value right here so that point is f of x 1 this point here is going to be f of x up to this point will be f x 3. And this point will be f of x 4. Now you can see that for our area of 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 was equal to 1 and multiply it by the sum of all these values here f of x 1 of of X to f of x 3 and f of x 4. So if we were going to plug somethings into our formula what you would say is that area is going to be equal to delta x we know is one multiplied by f of x of 1. Someone we know is one. So we'll just say F of 1 x up to is 2. So we'll say F of 2. Then we have X sub 3 which is at 3. So we say plus F of 3 and X up 4 is 4 so we're going to say plus F of 4 and that's our last left endpoint. 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 of two F of 3 and 4. And the way that we do that remember that f of 1 is just whatever we get when we plug one into our original function here. So if we plug one in for x we're gonna get f of one is equal to plugging in one for x we get 1 over 1 or 1 to find f of 2 we'll say F of 2. It's going to be equal to 1 over 2 to find f of 3 pugging 3 into our original function for X. We get 1 over 3 and then f 4 is going to be 1 over 4. Now we just plug those back into our area equation and we get areas equal to 1 multiplied by half of 1 which is 1 plus F of 2 which is 1 1/2 plus F of 3 which is one third plus F or 4 which is one fourth. Now keep in mind that multiplying by 1 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 for 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 one to five using four rectangles on left end points to approximate the area.