WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about area under a curve--also known as **integrals**.
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The other major idea in calculus is the notion of the integral, a way to find the area underneath some portion of a curve.
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Like the derivative, at first glance, this might not seem terribly useful.
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Of course, it is somewhat interesting to be able to know the area under a curve.
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But what can we do with it that has any real use?--as it turns out, a huge amount.
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It allows us to consider what a function has done, in total, over the span of some interval.
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For example, if we have a function that gives the velocity of an object, how fast an object is moving around, the integral will tell us the object's location.
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The velocity of an object, as a function--we can take the integral of that, and we can get location out of it.
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Why is that? Well, velocity tells us motion--how fast we are moving around.
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Well, if we look at how fast something has moved around in total, that ends up telling us where the thing lands at the end.
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Motion, looked at over the long term--if we put all of the motion together--if we look at all it has done in total,
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all of the motion in total, we end up seeing where we end up; we end up seeing location out of it.
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So, that is just one example of how useful integrals can be.
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Trust me on this: this is really useful stuff--let's check it out.
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Consider if we had some function f(x) that has the graph right here.
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We could ask, "What is the area between the curve and the x-axis on the interval going from a to b?"
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So, we can fill that in; and the shaded-in portion that we have right here represents that area.
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But how could we go about finding how much area that actually is?
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Let's look for some way to approximate it; and notice that that is what we did with derivatives, as well.
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What we did is thought, "Is there some way to approximate close to what the slope is?" and then,
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"Is there a way to improve on that approximation?" and then,
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"Is there a way to basically see what that improvement will eventually go to in the infinite version?"
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As we got closer and closer to being right up against it with the derivative, as h went to 0, we were able to see
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what it became perfectly--what the slope was in that instant.
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And we will see sort of a similar idea as we work with integrals.
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All right, we are getting a little ahead of ourselves; so let's start looking at our approximation.
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The easiest way to find area is with a rectangle; rectangles are a great way to find area, because it is so easy to figure out what the area of a rectangle is.
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It is simply width times height: area is just width times height.
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Let's approximate the area under our curve with rectangles.
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We begin by breaking our interval into n equal sub-intervals.
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In this case, we have broken it into n = 4; here are 1, 2, 3, 4 equal-width sub-intervals.
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So, we go from a to b, and we break it into n equal-width chunks.
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In each one of these sub-intervals, we will base the height of each rectangle off of some x<font size="-6">i</font> in each sub-interval.
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That is to say, we will choose some horizontal location in the sub-interval; and then we will see what height that horizontal location goes to.
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That horizontal location, as a point--what height is that point at?
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And then, that is going to set the height of that rectangle for that sub-interval.
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So, for the ith sub-interval, we will use x<font size="-6">i</font> to determine that height.
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The ith rectangle has a height of f evaluated at x<font size="-6">i</font>;
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so x<font size="-6">i</font> is the horizontal reference location that we use to determine the height for the ith rectangle.
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For example, in our third sub-interval, we might look at x₃, the horizontal reference location for our third sub-interval.
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We check, and we see what height that ends up going to; and so, that would be f evaluated at x₃.
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And then, that determines the height for our third rectangle.
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We use x<font size="-6">i</font> to determine the horizontal reference location for our ith sub-interval,
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which then determines the height for our ith rectangle.
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We use this x<font size="-6">i</font> to determine height by getting f(x<font size="-6">i</font>) to tell us what that rectangle's height is.
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Now, there is a variety of ways to choose our x<font size="-6">i</font>.
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In any given sub-interval, there is a continuum: there are infinitely many different locations we could pick to be our x that we are going to choose.
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We could choose this location, but we could also choose this location.
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Or we could choose the location between them, or the location in between them, or over here,
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or at the very far edge, or the other very far edge, or somewhere else.
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There are all sorts of different ways that we can choose this.
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And notice: the way we decide to choose our x<font size="-6">i</font>, for each one of these sub-intervals:
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because the x<font size="-6">i</font>, that horizontal reference location, determines the height for that rectangle,
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it is going to affect the height of the ith rectangle,
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because height is just f evaluated at whatever horizontal reference location we chose, our x<font size="-6">i</font>.
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By using a different method to choose our x<font size="-6">i</font>, we will get a different approximation for the area,
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because we are going to change the height of that rectangle; and if we change the heights of all of our rectangles,
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we will end up having a totally new version for our area approximation.
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So, we are going to now look at some of the most common methods to choose x<font size="-6">i</font> for creating the heights of our rectangles.
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Our first common method that we see is the leftmost point method,
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where we end up evaluating the height of each rectangle by just evaluating where the leftmost point gets mapped to.
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How high is the left side of each sub-interval?
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We can see an example of this right here: what we do for our first sub-interval--
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we look: where would the leftmost point of that sub-interval get mapped to?
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It gets mapped to a height of here; and so, that determines the height of our entire rectangle.
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For our second sub-interval, we look at the leftmost location here; it gets mapped to this height, and that determines the height of that rectangle.
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For our third sub-interval, we look at this leftmost location; that determines that height.
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Our fourth and final: we look at the leftmost location, and that determines that height.
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We are determining the height of each rectangle based on the leftmost location in each of the rectangles.
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Basically the same idea, but flipped: we can look at the rightmost point.
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We can say, "For each sub-interval, what is the height that the rightmost horizontal location gets mapped to?"
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For our first sub-interval, we look at the rightmost location for that one;
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that ends up getting mapped to here, so that determines the height of our first rectangle.
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For our second sub-interval, we look at the rightmost location in that one; that is here, so that determines the height of our second rectangle.
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For our third sub-interval, we look at the rightmost location of that sub-interval; that determines the height of our third rectangle.
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And for our fourth and final one, we look at the rightmost location, and that determines the height of that fourth and final rectangle.
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We are just looking at the rightmost horizontal location to determine the height for each one of our rectangles.
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We could also do this based on the horizontal midpoint.
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So far, we have looked at the two extremes: the far left side of the sub-interval and the far right side of the sub-interval.
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But we could also ask what happens to the one in the middle.
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For our first sub-interval, we look at the one that is in the middle of that; we go up, and the midpoint gets mapped to this height.
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So, that determines the height of that first rectangle.
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For our second sub-interval, we look at the midpoint of that sub-interval.
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We go up, and that tells us the height for our rectangle here.
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For our third sub-interval, we look at the midpoint; we see what height that horizontal midpoint gets mapped to.
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And that determines the height of that entire third rectangle.
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And the same thing for our fourth one: we look at the midpoint.
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What value does that midpoint get mapped to? That determines the height of that entire rectangle.
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So, so far, we have looked at the far left side, the far right side, and the middle.
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But there is also another way of looking at this; we can also look at the maximum height in each one of these.
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We look at the sub-interval, and we see which one of these points gets mapped to the highest possible location.
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We look over the entire sub-interval, and we see which one is highest in this portion.
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So, in this case, the highest location that we get for this sub-interval is here; so that determines the height of the entire rectangle.
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For our second sub-interval, the highest possible location we reach in this sub-interval is this location right here.
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So, that determines the height of that entire second rectangle.
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For our third sub-interval, we see that the highest possible height we reach in that sub-interval is here.
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So, that determines the height of the entire third rectangle.
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And for our fourth and final one, the highest possible point we get is here; so that determines the height of the entire rectangle.
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The maximum height method is also sometimes known as the upper sum, because, notice:
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by choosing the maximum height in each of our sub-intervals, we will always end up getting an approximation
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that is going to be at least the value of the area under the curve, and more than that area, usually.
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We can see all of the places that we ended up going above it.
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And so, we end up having something that is above what we are actually going to end up having as the area underneath the curve.
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And so, we call it an **upper sum**, because we are getting something that is above the value of the actual area underneath the curve.
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We can also talk about the minimum height for each one of our sub-intervals.
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We can look and see, for our first sub-interval, what is the lowest possible height that we have in here.
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We see that it is here, and so we end up getting this as the height for it.
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We can also do this for each one of our sub-intervals.
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We can have our second sub-interval; the lowest possible height in that second sub-interval is right here,
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so we end up evaluating that as the height of our second rectangle.
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For our third sub-interval, the lowest possible height is right here; so that gives us the height of our third rectangle.
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And for our fourth and final sub-interval, the lowest possible height over that sub-interval is right there.
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And so, that gives us the height of our fourth rectangle.
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This one is sometimes called the **lower sum**, because it is always going to end up giving us a value
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that is below the actual area underneath the curve, because, since we are choosing based on minimum heights,
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each of our rectangles is actually under-cutting the area.
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We can see area that we are missing each time.
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For each of our rectangles, we are not fully filling out that area, because we are always below it.
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So, we end up having less than the area that is actually underneath the curve.
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And so, it is sometimes also called the lower sum.
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How can we find what this approximation actually ends up giving us?
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How can we find this area? Well, first we want to figure out what the width of each one of these rectangles is.
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That is the easier thing to figure out: we split the interval into n sub-intervals.
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That is how we did this, right from the beginning; we split it into n equal-width sub-intervals.
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So, what is the total length of the interval we have?
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Well, that is going to be b - a, where we end minus where we started.
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If we have split it into n equal-width sub-intervals, the width of each one of them must be the total length, divided by how many intervals we have.
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So, our width is equal to b - a, divided by n; that is the width of one of our rectangles, b - a over n.
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To figure out the height for each rectangle...well, the ith rectangle's height
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depends on the specific x<font size="-6">i</font> we chose for it, with the different methods that we were just talking about.
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There are various different ways that we can choose that x<font size="-6">i</font>.
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But whatever x<font size="-6">i</font> we end up choosing will tell us the height; so just by definition,
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our height is equal to f evaluated at the various x<font size="-6">i</font> that we chose.
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So, with these two things in mind, we now have our width, and we have our height.
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The area for the ith rectangle, any given rectangle, is going to be that rectangle's width,
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which is b - a divided by n, times that rectangle's height, which was determined by the x<font size="-6">i</font> we chose, so our height right here.
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So, the area of our ith rectangle is equal to the width, b - a over n, times the height of that ith rectangle,
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which is going to be f evaluated at x<font size="-6">i</font>--whatever horizontal reference location we chose.
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If we want to figure out what the total approximation is, we need to add up each one of our rectangles.
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We aren't concerned with just one of our rectangles; we want to add up all of the rectangles in our approximation.
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So, we sum them all up; and we can use sigma notation for that.
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So, we have i = 1, our first rectangle, to n, the nth rectangle, which is our last rectangle,
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since we are using a total of n sub-intervals; and then we just end up adding up the area from each one of these.
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So, it is going to be b - a over n, the width of each one of these, times the height of each one of these, f(x<font size="-6">i</font>).
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Notice that whatever method we choose to determine the height of the rectangles, the more rectangles we use, the better our approximation becomes.
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Over here, we have n = 4; but over here, we have n = 8.
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And notice: in this second one, we end up having a closer approximation.
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We have less missing chunks; the more rectangles we use for our approximation,
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the closer our approximation becomes to actually giving us the area underneath the curve.
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Thus, our approximation becomes ever more accurate as the number of rectangles, n, goes off to infinity.
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So, as we increase our n-value higher and higher and higher and higher, our approximation becomes better and better and better and better.
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As our n slides off to infinity, we will be able to get, effectively, what the perfect value is underneath that curve.
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With this realization in mind, remember: the approximation we just figured out for the area with n rectangles
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was the sum of i = 1 up until n of our width for each rectangle, (b - a)/n,
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times the height of each rectangle, f evaluated at its specific x<font size="-6">i</font>.
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And this approximation becomes more accurate as our n goes off to infinity,
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as the number of rectangles we have becomes more and more and more and more.
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We get a finer and finer sense of the area underneath the curve.
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With that idea in mind, we take the limit at infinity to find the area underneath the curve.
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The area under our function f(x) from a to b, in that interval from a to b, is the limit as n goes to infinity,
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as our number of rectangles slides off to infinity, of our approximation formula, the sum from i = 1 to n of the width, (b - a)/n, times the height, f(x<font size="-6">i</font>).
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It is the limit as our number of rectangles slides off to infinity of our normal approximation formula.
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If this limit exists (and it might not for some weird functions, but for most of the functions we are used to,
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it will end up existing), what that limit is: it is called the **integral** from a to b.
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It is denoted with the integral from a to b...that is the integral sign there,
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that new sign that you probably haven't seen before: integral from a to b of f(x)dx.
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The process of finding integrals is called **integration**.
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The integral from a to b tells us the area underneath the curve, f(x), from a to b (that interval a to b)--really cool stuff here.
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Here is the really amazing part: the integral from a to b of f(x)dx is based on the anti-derivative of f(x)--
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that is, the derivative process done in reverse on f(x).
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So, we talked about the idea of taking the derivative of some function, of being able to see some function,
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and then turn it into another function that talks about the rate of change of that.
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For that, we had f(x) become, through the derivative, f'(x).
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But we can also talk about if we did the reverse of this process.
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Instead of taking a derivative, we did the anti-derivative, where we worked our way up the chain.
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And we symbolized that with the capital F(x); F(x) is the one that you can take the derivative of, and it becomes f(x).
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So, F(x), the anti-derivative, is the thing where, if you take its derivative, you just get f(x),
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the function that we are starting with, what is inside of our integral.
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We can take the derivative process and reverse it, and we are able to start talking about the area underneath the curve.
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This is really cool stuff; armed with this notation of F(x), it turns out that the integral of a to b of f(x)dx--
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that is, the area underneath the curve from a to b of some f(x) function--
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is equal to the anti-derivative of f, evaluated at b, minus the anti-derivative of f, evaluated at a.
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Wow, that is amazing; it is just so incredibly elegant that there is this way to talk about the area underneath the curve
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with this thing that we just talked about that had rate of change.
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And it seems really shocking that these things are connected at all.
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But it is absolutely amazing that it allows us to really easily find area for something that would be otherwise very difficult to work through,
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that infinite limit that we were just talking about.
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So, you might be wondering why this happens; why is there this connection between the area underneath the curve,
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and this anti-derivative, where we are talking about what the height is...the anti-derivative of the height of it
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gives us the area underneath the curve...it seems really surprising at first.
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In short, what we can do is think of the area underneath the curve as being a special function, a(x).
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Now, notice: we have this area underneath the curve, the shaded portion, as being a(x).
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Notice that the rate of change for the area is based on the height of the function at any given location.
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For example, if we consider this x location right here, how fast our area function is changing is based on how tall our function is in that moment.
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And how tall is a function? Well, that is just what you get when you evaluate the function, f(x).
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Notice: if we had gone to some other horizontal location, where we had a different height,
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we would end up having a very different speed that our area was growing at.
00:17:24.100 --> 00:17:28.400
If the height of the function is small, our area grows at a slow rate.
00:17:28.400 --> 00:17:32.500
If the height of the function is high, our area grows at a fast rate.
00:17:32.500 --> 00:17:36.200
So, the height of the function changes how fast our area grows.
00:17:36.200 --> 00:17:40.400
How fast does something grow? Well, that is what we are talking about: rate of change.
00:17:40.400 --> 00:17:49.200
That means that since height is connected to area through rate of change, since height gives us the rate of change of area,
00:17:49.200 --> 00:17:52.800
well, rate of change was a derivative that we were just talking about.
00:17:52.800 --> 00:17:58.100
So, height is the derivative of area, because height is the rate of change of our area.
00:17:58.100 --> 00:18:02.800
That means that the reverse works as well; we can look at the symmetric version of this.
00:18:02.800 --> 00:18:08.100
Area is based on the anti-derivative of height; since height is the derivative of area,
00:18:08.100 --> 00:18:10.800
that means that area must be the anti-derivative of height.
00:18:10.800 --> 00:18:15.600
Since we go in one direction, if we go in the opposite direction, doing the opposite thing, we get the other version.
00:18:15.600 --> 00:18:23.800
Since a'(x) = f(x)--that is to say, the derivative of x equals little f(x), if we do the anti-derivative of f,
00:18:23.800 --> 00:18:29.000
we end up getting the anti-derivative of a', which is just our area function that we started with.
00:18:29.000 --> 00:18:33.700
If this a little bit confusing to you now, don't worry; it very well might be, and it would be perfectly reasonable.
00:18:33.700 --> 00:18:38.900
Just think about it when you end up getting exposed to this new idea when you actually take a calculus class.
00:18:38.900 --> 00:18:43.500
This ends up being a fair bit of a way into calculus class, but I think it is a really cool idea.
00:18:43.500 --> 00:18:50.100
And now that you understand what is going on intuitively, or at least have the seed ready to blossom at a future date,
00:18:50.100 --> 00:18:54.800
when you see it later, you will think, "Oh, now I understand what that guy was talking about--
00:18:54.800 --> 00:18:57.800
it is now starting to make sense!" and there is really cool stuff here.
00:18:57.800 --> 00:19:04.100
I really, really love this stuff; and I hope that you are getting some sense of just how amazingly beautiful all of this stuff in math is.
00:19:04.100 --> 00:19:07.300
All right, let's start looking at some examples.
00:19:07.300 --> 00:19:15.200
Find an approximation to the area under f(x) = x² from a = 0 to b = 3 by using three equal-width rectangles--
00:19:15.200 --> 00:19:19.300
that is, n = 3--on the left-hand point of each interval.
00:19:19.300 --> 00:19:23.100
The first thing: let's just get a quick sketch here, so that we can see what is going on.
00:19:23.100 --> 00:19:27.500
f(x) = x²: we will just, really quickly, sketch what we are looking at here.
00:19:27.500 --> 00:19:33.200
And so, let's say here is a = 0; here is b = 3.
00:19:33.200 --> 00:19:38.800
So, if we are going to use three equal-width rectangles, n = 3, we are breaking it up into three chunks.
00:19:38.800 --> 00:19:43.400
And we are going to evaluate each one of these rectangles on the left-hand point of each interval.
00:19:43.400 --> 00:19:56.500
Three equal chunks here...if it is the left-hand point, then that first one there, this one here, and this one here give us the area for each one of these.
00:19:56.500 --> 00:20:03.400
Notice that that first rectangle won't have any area at all, because we are evaluating the left-hand point of each interval.
00:20:03.400 --> 00:20:08.000
First, let's check and see what the width is: the width is going to be...
00:20:08.000 --> 00:20:11.700
since it is equal width for each one of these...how long is our interval?
00:20:11.700 --> 00:20:16.400
That is b - a, divided by...how many sub-intervals do we break it into? n.
00:20:16.400 --> 00:20:23.600
So, we have 3 for our ending location, minus 0 for our starting location, divided by 3 (is the total number of sub-intervals).
00:20:23.600 --> 00:20:27.900
3/3 gets us 1, so we have a width of 1 for each one of these.
00:20:27.900 --> 00:20:35.000
So, at this point, we can go in and see where we have our first location; the sub-interval will go from 0 to 1;
00:20:35.000 --> 00:20:39.900
the next one will go from 1 to 2; the next one, the last one, goes from 2 to 3.
00:20:39.900 --> 00:20:44.400
All right, with all of this in mind, we can now see about evaluating each one of these rectangles.
00:20:44.400 --> 00:20:53.000
Our first rectangle will be i = 1; our second rectangle, i = 2; and our last, final rectangle, i = 3.
00:20:53.000 --> 00:20:57.500
Where will we evaluate our first rectangle? Well, that is the left-hand point.
00:20:57.500 --> 00:21:05.700
If our first rectangle...remember: it is going from 0 to 1; 0 to 1 is the sub-interval it is evaluating.
00:21:05.700 --> 00:21:11.000
The left-hand point is 0; so it is going to have a height of f at 0.
00:21:11.000 --> 00:21:15.600
What is the width? The width is 1, so 1 times f(0).
00:21:15.600 --> 00:21:26.900
In general, notice that that is also just the same thing as saying the width, (b - a)/n, times our function evaluated at...however we determined our x<font size="-6">i</font>.
00:21:26.900 --> 00:21:29.900
In this case, we are determining based on left-hand points.
00:21:29.900 --> 00:21:35.800
So, 1 times f at 0...our f(x) equals x², so we have 1(0)²,
00:21:35.800 --> 00:21:41.000
which comes out to be just 0 for the area of our first rectangle, this right here.
00:21:41.000 --> 00:21:44.800
And we can see that it is going to have to be completely flat, not really a rectangle,
00:21:44.800 --> 00:21:49.500
just a chunk of line, because it doesn't have any height, because we are evaluating at the left-hand side.
00:21:49.500 --> 00:21:56.400
For our second rectangle, once again, there is a width of 1, times the height (will be evaluated at the left-hand side here); it will be 1.
00:21:56.400 --> 00:22:04.200
So, f evaluated at 1...1 times 1² gets us just 1; so there is an area of 1 for our second rectangle.
00:22:04.200 --> 00:22:12.100
And our third and final rectangle: 1 times f evaluated at 2, because the left-hand from 2 to 3 is going to be 2;
00:22:12.100 --> 00:22:18.000
1 times f(2) is 1 times 2², when we evaluate that function; and that comes out to 4.
00:22:18.000 --> 00:22:23.100
So, the total amount of area that we get for our approximation, the total approximation we get,
00:22:23.100 --> 00:22:29.500
is going to be equal to each of these added up together--the first rectangle, 0,
00:22:29.500 --> 00:22:36.300
plus the second rectangle, 1, plus the third, final rectangle (4 there)--each of our areas.
00:22:36.300 --> 00:22:48.300
0 + 1 + 4 gets us a total area of 5 for our entire approximation of a = 0 to b = 3 with three sub-intervals and the left-hand point for each one.
00:22:48.300 --> 00:22:51.200
Our second example is very similar to our previous example.
00:22:51.200 --> 00:22:57.800
We are finding an approximation to the area under f(x) = x² from a = 0 to b = 3
00:22:57.800 --> 00:23:02.900
by using four equal-width rectangles, n = 4, on the maximum point of each interval.
00:23:02.900 --> 00:23:09.800
The only difference here is that we are now using four rectangles, and we are doing it based on the maximum point,
00:23:09.800 --> 00:23:12.200
the highest location for each one of these intervals.
00:23:12.200 --> 00:23:19.800
We draw in our curve...x²...we have something like this; we are going, once again, from a = 0 to b = 3.
00:23:19.800 --> 00:23:24.600
So, there is our interval; and we are going to be looking for the maximum point of each interval.
00:23:24.600 --> 00:23:29.100
If it is four equal-width rectangles (let's draw that in really quickly: 1, 2, 3, 4),
00:23:29.100 --> 00:23:34.500
notice: the maximum point of each interval--where is that going to be here?--well, that is going to be here.
00:23:34.500 --> 00:23:36.600
Where is that going to be for this one?--that will be here.
00:23:36.600 --> 00:23:39.900
Where will that be for this one?--that will be here; and where for this one?--that will be here.
00:23:39.900 --> 00:23:47.700
The maximum point for this one is basically the same thing as saying the right-hand side.
00:23:47.700 --> 00:23:50.200
Now, I want to point out that this is not always true.
00:23:50.200 --> 00:23:55.900
As we saw when we were working through this lesson, right-hand point and maximum can give us totally different rectangle pictures.
00:23:55.900 --> 00:24:02.100
However, for something like f(x) = x², where it is just constantly growing, constantly growing, constantly growing,
00:24:02.100 --> 00:24:05.800
since it is always growing as it goes off to the right, that means that for any sub-interval,
00:24:05.800 --> 00:24:09.600
the right-most point will be at the highest height; so the right-most point
00:24:09.600 --> 00:24:14.000
is the same thing as maximum for the specific case of the function x².
00:24:14.000 --> 00:24:17.900
With a different function, you might end up having different things; so it is something that you have to think about.
00:24:17.900 --> 00:24:21.700
But in this specific case, it will be the same thing as just evaluating at the right-hand side.
00:24:21.700 --> 00:24:24.100
What is the width of each one of these rectangles?
00:24:24.100 --> 00:24:29.800
Well, the width is (b - a)/n; in that case, we have a width of 3 total, divided by 4 for each one.
00:24:29.800 --> 00:24:34.200
And let's put that in decimal for ease, because we are going to end up using a calculator to crunch these numbers,
00:24:34.200 --> 00:24:36.800
because we will have a lot of decimals showing up otherwise.
00:24:36.800 --> 00:24:41.700
Our very first horizontal location is 0, because we start at 0.
00:24:41.700 --> 00:24:46.200
The next one will be 0.75; we have a width of 0.75.
00:24:46.200 --> 00:24:51.300
The next one will be at 1.5, because we are another 0.75 step ahead of that.
00:24:51.300 --> 00:24:58.300
The next one will be at 2.25, another 0.75 step ahead of that; and finally, we finish at 3, which makes sense.
00:24:58.300 --> 00:25:06.200
We have to start at 0 and end at 3; and we work by step after step of our width, 0.75, to make it up each time.
00:25:06.200 --> 00:25:14.600
Our first rectangle...figure out the area for that one; our second rectangle; our third rectangle; and our fourth rectangle.
00:25:14.600 --> 00:25:23.000
OK, so our first rectangle: remember, the right-hand point is the same thing as the maximum point in the specific case of the function x².
00:25:23.000 --> 00:25:30.400
So, what is the width here? The width is 3/4; remember, in general, the area of any rectangle is the width times the height.
00:25:30.400 --> 00:25:37.600
So, in this case, it will be (b - a)/n, because that is always going to be the width if we have n equal sub-intervals going from a to b.
00:25:37.600 --> 00:25:40.700
b - a is the total length; divide by n for the width.
00:25:40.700 --> 00:25:44.700
Multiply that times the height of each one of them, our f evaluated at x<font size="-6">i</font>.
00:25:44.700 --> 00:25:48.300
And x<font size="-6">i</font> will depend on how we are choosing the point to look at.
00:25:48.300 --> 00:25:51.600
In this case, we are looking at the maximum point, which ends up being the right-hand point;
00:25:51.600 --> 00:25:55.400
so we will always end up looking at the right side of each one of these sub-intervals.
00:25:55.400 --> 00:26:00.700
OK, back to our first one: 3/4, our width, times the height--where does the height end up getting evaluated?
00:26:00.700 --> 00:26:10.400
Well, if we are going from 0 to 0.75 (that is our first sub-interval), we are going to end up looking at the most right-hand part, which is 0.75.
00:26:10.400 --> 00:26:23.100
So, f...plug in 0.75; 3/4 times f(0.75) is the same thing as 0.75 times 0.75².
00:26:23.100 --> 00:26:28.800
Since f is just x², just a squaring function, it is 0.75 times 0.75².
00:26:28.800 --> 00:26:33.800
We work that out with a calculator, and we get 0.422; great.
00:26:33.800 --> 00:26:41.300
The second rectangle: once again, we evaluate 3/4, the width, times f evaluated now at 1.5,
00:26:41.300 --> 00:26:47.100
because the right side of this one, the right side of our second rectangle, will be 1.5.
00:26:47.100 --> 00:26:52.500
Here was our first rectangle; now we are on our second rectangle; its right side is 1.5.
00:26:52.500 --> 00:27:02.800
So, 3/4 times 1.5²: work that out with a calculator; we get 1.688.
00:27:02.800 --> 00:27:09.000
The third rectangle is this one right here: the right side of that rectangle, the maximum height, is 2.25.
00:27:09.000 --> 00:27:13.800
It is still the same width; the width will never end up changing, because we set the width to be equal for all of them.
00:27:13.800 --> 00:27:21.500
So, f...plug in 2.25 from the right side (in this specific case);3/4...our function is squared, so 2.25²;
00:27:21.500 --> 00:27:29.700
that comes out to be 2.797; and our fourth and final rectangle...width times the height that it evaluates at...
00:27:29.700 --> 00:27:34.300
that is going to be 3, because the far right-hand side of our final interval is 3.
00:27:34.300 --> 00:27:40.700
3/4 times f(3)...remember, the right-hand side, in this specific case, happens to be the maximum.
00:27:40.700 --> 00:27:45.700
If you worked with a different function, you would have to think about what was going to be the maximum location there.
00:27:45.700 --> 00:27:51.300
3/4 times 3² comes out to be 6.75.
00:27:51.300 --> 00:27:56.300
If we want to know what the total area approximation is, we add up all of these numbers.
00:27:56.300 --> 00:28:06.600
So, it is going to be area equals 0.422 + 1.688 + 3.797 + 6.75.
00:28:06.600 --> 00:28:13.400
We add up all four of the rectangles, and that tells us the total approximation we got by using this specific method.
00:28:13.400 --> 00:28:21.200
And that comes out to be 12.657; so that is the total approximation we end up getting here.
00:28:21.200 --> 00:28:24.900
Notice: at this point, we now have the lower sum and the upper sum.
00:28:24.900 --> 00:28:31.600
In our first example, we chose minimum, because we chose the left-hand side; and we ended up getting 5 out of that.
00:28:31.600 --> 00:28:35.300
At this point, we now have just chosen the maximum, so we got an upper sum,
00:28:35.300 --> 00:28:40.700
something that is the most area it could possibly be; and we ended up getting 12.657.
00:28:40.700 --> 00:28:45.000
So, whatever the actual area underneath that curve is, we know that it has to be between 5
00:28:45.000 --> 00:28:51.200
(because that was the lower sum in our first example) and 12.657 (because we just got an upper sum in this, our second example).
00:28:51.200 --> 00:28:59.200
So, whatever the actual value of the area underneath that curve between 0 and 3 is, it has to be somewhere between 5 and 12.657.
00:28:59.200 --> 00:29:05.100
Now, in our third example, we are going to actually figure out what it is precisely by taking an infinite limit.
00:29:05.100 --> 00:29:11.100
For our third example, find the precise area under f(x) = x² from a = 0 to b = 3,
00:29:11.100 --> 00:29:14.700
using the limit as the number of rectangles, n, goes off to infinity.
00:29:14.700 --> 00:29:25.900
To do this, we will also end up needing this specific identity, this sum of i = 1 to n of i² = n(n + 1)(2n + 1), all divided by 6.
00:29:25.900 --> 00:29:29.300
But we won't actually end up using that until we get about halfway through this thing.
00:29:29.300 --> 00:29:32.000
So, first, how do we set this thing up?
00:29:32.000 --> 00:29:36.700
Once again, here is just a quick picture to help us clarify things.
00:29:36.700 --> 00:29:44.800
We have x²; we are working our way from a = 0 to b = 3.
00:29:44.800 --> 00:29:48.800
All right, and we are going to end up cutting it into some number of sub-intervals.
00:29:48.800 --> 00:29:50.500
We are going to end up cutting it into n sub-intervals.
00:29:50.500 --> 00:29:53.400
And then, from there, we will let n go off to infinity.
00:29:53.400 --> 00:30:00.700
But we have to start by figuring out what the area would be if we had just some actual value for n, and then we let n run off to infinity.
00:30:00.700 --> 00:30:02.900
First, what is our width going to end up being?
00:30:02.900 --> 00:30:09.300
Our width will end up being (b - a)/n, because that is always the case: (b - a)/n.
00:30:09.300 --> 00:30:16.500
In this case, a = 0; b = 3; so our width is going to be 3, divided by the number of rectangles we choose to use, 3 divided by n.
00:30:16.500 --> 00:30:18.500
That is the width of each one of these.
00:30:18.500 --> 00:30:23.500
Now, we need to decide how we are going to determine where we choose our x<font size="-6">i</font>.
00:30:23.500 --> 00:30:30.800
It is going to end up actually making our notation just a little bit easier if we choose right-most; so we are just going to arbitrarily choose right-most.
00:30:30.800 --> 00:30:38.300
I want to point out to you, though, that it doesn't actually matter which one we choose: right-most, left-most, midpoint, upper sum, lower sum...
00:30:38.300 --> 00:30:46.000
If it does eventually converge to a single value, if our limit does exist as n goes off to infinity, then we will have found the area underneath it.
00:30:46.000 --> 00:30:53.700
And no matter how we choose to set up our cuts and our heights, as the number of rectangles goes farther and farther off to infinity,
00:30:53.700 --> 00:30:55.700
our area has to get closer and closer to the actual thing,
00:30:55.700 --> 00:31:00.800
no matter how we set up the heights, if it can get to an actual value under the area.
00:31:00.800 --> 00:31:02.800
So, it doesn't matter which one we choose, specifically.
00:31:02.800 --> 00:31:07.400
Right-most is nice, because it is easy to do it notation-wise; so if you end up having to do a similar problem,
00:31:07.400 --> 00:31:10.500
you can basically just copy the method I am doing here to set up.
00:31:10.500 --> 00:31:12.900
And it will end up working for you in notation, as well.
00:31:12.900 --> 00:31:18.600
All right, let's get back to this: we will set up our right-most x<font size="-6">i</font> for each one.
00:31:18.600 --> 00:31:25.700
So, what will x<font size="-6">i</font> end up being? Well, our first x<font size="-6">i</font> will be...
00:31:25.700 --> 00:31:28.200
a is the far left side, so what would be the next one?
00:31:28.200 --> 00:31:35.900
Well, that would be a plus...what is our width? Our width is 3/n...3/n for the first x<font size="-6">i</font>.
00:31:35.900 --> 00:31:39.700
So, in general, to get out to the ith x<font size="-6">i</font>, we start at a.
00:31:39.700 --> 00:31:45.000
And then, if we are at the ith x<font size="-6">i</font>, we will be...3/n is our width each time we go forward a step.
00:31:45.000 --> 00:31:48.200
How many steps did we take forward? i steps.
00:31:48.200 --> 00:31:52.600
If we start at a, and then we want to get to the x<font size="-6">i</font>, which is the far right side of any one,
00:31:52.600 --> 00:32:03.900
well, the first one would show up at +1(3/n); the second one would show up at +2(3/n); and the third one would show up at +3(3/n).
00:32:03.900 --> 00:32:08.000
It is the number of steps, times our width each time, 3/n.
00:32:08.000 --> 00:32:14.800
So, the number of steps that we have taken, if we are at the right-most of each side, is just going to be i, whatever sub-interval we are at.
00:32:14.800 --> 00:32:18.500
If we are in our first sub-interval, we have taken one width-step to get to the right side.
00:32:18.500 --> 00:32:23.900
If we are in our tenth sub-interval, we have taken ten width-steps to get to the right side of that tenth sub-interval.
00:32:23.900 --> 00:32:29.900
In general, x<font size="-6">i</font> is equal to a, our starting location, plus 3/n, times i.
00:32:29.900 --> 00:32:34.700
In the specific case of this problem, we can plug in what our a equals, 0.
00:32:34.700 --> 00:32:43.900
We have x<font size="-6">i</font> = 0 + 3/n(i); so 3 over n times i gives us our x<font size="-6">i</font> for this specific problem.
00:32:43.900 --> 00:32:49.000
But if you were doing just any problem in general, you would want to use that first part, a plus...
00:32:49.000 --> 00:32:55.300
and also, you wouldn't use 3/n; you would end up using whatever your specific width ended up being, which would be (b - a)/n.
00:32:55.300 --> 00:32:59.900
It is not necessarily 3/n; that is this specific problem that we are working right here.
00:32:59.900 --> 00:33:05.000
All right, if we have x<font size="-6">i</font> for each one, what will end up being the height for each one of our rectangles?
00:33:05.000 --> 00:33:21.000
The height of the ith rectangle is going to be f evaluated at x<font size="-6">i</font>, which to say f evaluated at 3/n times i.
00:33:21.000 --> 00:33:29.300
So, in general, what we do next is set up our limit: the limit as n goes off to infinity of our sum.
00:33:29.300 --> 00:33:35.600
The approximation for n rectangles will be to start at our first rectangle, i = 1, and go out to our last rectangle, n.
00:33:35.600 --> 00:33:43.000
And it is going to be the width of each one of these rectangles, (b - a)/n, times the height of each one of these rectangles, f(x<font size="-6">i</font>).
00:33:43.000 --> 00:33:48.100
Now, this formula here will always end up working for any problem that you have set up.
00:33:48.100 --> 00:33:49.900
So, that is a useful thing to work with.
00:33:49.900 --> 00:33:54.100
Now, we are going to start using what we have in our specific problem--we will start plugging things in.
00:33:54.100 --> 00:34:00.100
Our limit is still the same; our summation is still the same; we are going from the first to the last.
00:34:00.100 --> 00:34:06.200
Our (b - a)/n...we figured out that that was a width of 3, divided by the number of rectangles we chose, 3 divided by n,
00:34:06.200 --> 00:34:10.500
times f evaluated at x<font size="-6">i</font>...well, what is f evaluated at x<font size="-6">i</font>?
00:34:10.500 --> 00:34:20.700
Well, f(x) = x²; so if our x<font size="-6">i</font> is 3/n times i (that was what we figured out that our x<font size="-6">i</font> has to be),
00:34:20.700 --> 00:34:30.300
then f evaluated at 3/n, times i, is 3/n times i squared.
00:34:30.300 --> 00:34:37.200
And we have that right here; we now plug that in: 3/n times i, the whole thing squared.
00:34:37.200 --> 00:34:46.500
At this point, we can now simplify that just a little bit, and we have the limit as n goes off to infinity of the sum, i = 1 to n,
00:34:46.500 --> 00:34:53.500
our first rectangle to our last rectangle; the square distributes--we have 3²/n² times i².
00:34:53.500 --> 00:35:03.200
We also have that 3/n there; so we simplify that to 27, over n³, times i².
00:35:03.200 --> 00:35:06.800
All right, at this point, we are now ready to move on to the second half of this.
00:35:06.800 --> 00:35:12.800
We can now start working to figure out what this infinite series ends up coming out to be.
00:35:12.800 --> 00:35:21.100
All right, continuing on with our example: we have that the area is equal to the limit as n goes off to infinity from i = 1 to n
00:35:21.100 --> 00:35:26.400
of 27/n³ times i², whatever that ends up happening to be.
00:35:26.400 --> 00:35:31.300
And at a later point, we will end up putting this identity into the problem so that we can actually solve this thing out.
00:35:31.300 --> 00:35:40.000
All right, the first thing to notice is that the 27/n³ part doesn't actually do anything inside of the sum.
00:35:40.000 --> 00:35:43.900
The n, as far as the sum is concerned, is actually a fixed value.
00:35:43.900 --> 00:35:49.400
The n here is just a constant; remember when we were working with sigma notation--the number on top was just some number.
00:35:49.400 --> 00:35:52.100
It didn't change around during the course of doing things.
00:35:52.100 --> 00:35:58.100
So, since the n isn't changing inside of the sigma (it will change over here, because n will go off to infinity,
00:35:58.100 --> 00:36:02.200
but as far as the sigma is concerned, it doesn't change--the limit has to do something--
00:36:02.200 --> 00:36:07.700
so since the sigma doesn't have anything happening), we can actually pull it outside of the sigma.
00:36:07.700 --> 00:36:13.200
The first step here is to see that what we really have is: we can write this as limit as n goes off to infinity.
00:36:13.200 --> 00:36:18.800
We pull out the 27/n³, because it is not affected if it does it on the inside or the outside.
00:36:18.800 --> 00:36:24.600
It is just a scalar constant, as far as our series here is concerned.
00:36:24.600 --> 00:36:34.900
So, i = 1 up to n of i²...at this point, look: we have the sum as i = 1 goes off to n of i².
00:36:34.900 --> 00:36:39.300
So, we can now swap it out for this portion right here.
00:36:39.300 --> 00:36:47.200
So, we swap it out for that portion right here; we have the limit as n goes off to infinity of 27/n³;
00:36:47.200 --> 00:36:56.700
and we swap out n times n + 1 times 2n + 1, all over 6.
00:36:56.700 --> 00:37:02.400
Great; let's work to simplify things out a bit: the limit as n goes off to infinity...
00:37:02.400 --> 00:37:07.700
we will keep our 27/n³ off to the side for just a moment.
00:37:07.700 --> 00:37:10.900
Well, let's actually put it into the thing, but we will deal with this part first.
00:37:10.900 --> 00:37:14.100
n times n + 1 times 2n + 1...let's expand this a little bit.
00:37:14.100 --> 00:37:28.700
We do a little bit of expanding; we have n² + n, times 2n + 1, all over n³ times 6; great.
00:37:28.700 --> 00:37:32.800
Limit as n goes off to infinity...we can finish expanding our factors there.
00:37:32.800 --> 00:37:44.200
We have 27, times n² times 2n (becomes 2n³); n² + 1...n² times +1; n times 2n gets 2n²...
00:37:44.200 --> 00:37:54.200
So, we have a total of +3n²; and n times 1 gets us + n, all over 6n³.
00:37:54.200 --> 00:38:05.500
Limit as n goes off to infinity...we have 27 times 2n³; that comes out to be 54n³;
00:38:05.500 --> 00:38:20.000
plus 3 times n²...that comes out to be 81n²; plus 27n, all divided by 6n³.
00:38:20.000 --> 00:38:26.100
All right, at this point, let's move this all up here, so that we can keep working on it.
00:38:26.100 --> 00:38:30.600
Notice that, if we want, we can break this into two separate fractions.
00:38:30.600 --> 00:38:51.100
We have the limit as n goes to infinity of 54n³, over 6n³, plus 81n², plus 27n, all over 6n³.
00:38:51.100 --> 00:38:55.900
Now, this portion right here, the n³ and the n³, will end up canceling out.
00:38:55.900 --> 00:39:03.000
We have 54 divided by 6; but the limit as n goes off to infinity for this portion...
00:39:03.000 --> 00:39:10.000
well, we have n³ on the bottom; that is a 3 degree on the bottom; but on the top, we only have n² and n.
00:39:10.000 --> 00:39:15.400
So, that is a 2 degree and a 1 degree; those can't compete with an n³ on the bottom.
00:39:15.400 --> 00:39:19.100
A degree of 3 on the bottom is going to end up crushing that in the long term.
00:39:19.100 --> 00:39:26.400
As n rides off to infinity, our denominator is going to get so much bigger than our numerator that this whole part here just crushes down to 0.
00:39:26.400 --> 00:39:32.100
That means that we are left, as our n goes off to infinity, with simply having 54 divided by 6.
00:39:32.100 --> 00:39:41.200
And 54 divided by 6 gets us 9; so the total area underneath that curve is actually precisely equal to 9--pretty cool.
00:39:41.200 --> 00:39:46.700
Notice that it does take a little bit of challenging effort to work through this limit as n goes to infinity.
00:39:46.700 --> 00:39:49.500
We can work through it slowly, but it is not easy.
00:39:49.500 --> 00:39:52.300
We had to pull up this kind of arcane formula.
00:39:52.300 --> 00:39:56.400
It is not a very difficult summation formula, but it is not one that we probably know immediately.
00:39:56.400 --> 00:40:02.900
So, this stuff isn't super easy to work with (limit as n goes off to infinity), if we are trying to do this infinite cut method.
00:40:02.900 --> 00:40:08.700
And that is why that fundamental theorem of calculus, that integral with anti-derivative stuff, is so useful.
00:40:08.700 --> 00:40:12.800
Let's see just how powerful that is now in our final example of the course.
00:40:12.800 --> 00:40:21.100
Using the amazing fact that the integral of a to b of f(x)dx is equal to the anti-derivative of f evaluated at b,
00:40:21.100 --> 00:40:27.800
minus the anti-derivative of f evaluated at a (and remember: the integral is just the area underneath the curve, from a to b,
00:40:27.800 --> 00:40:34.800
for some curve defined by f(x)), find the area under f(x) = x² from a = 0 to b = 3.
00:40:34.800 --> 00:40:41.400
This part right here is the area that we are looking for.
00:40:41.400 --> 00:40:51.800
So, what we want to do is figure out what F(b) - F(a) is; F(x) is the anti-derivative of f(x).
00:40:51.800 --> 00:40:56.800
So, the very first step that we need to figure out here is what the anti-derivative is to f(x).
00:40:56.800 --> 00:41:05.300
f(x) = x²; when we worked through the derivative examples, we talked about the power rule,
00:41:05.300 --> 00:41:07.600
where you take the exponent, and you move it down.
00:41:07.600 --> 00:41:20.900
So, for example, if we have f(x) = x⁵, the derivative of x is equal to...move that exponent, the 5, down...
00:41:20.900 --> 00:41:29.100
we have 5 times x, and then we subtract 1 from the exponent; so - 1 from it at that point...it will be 5x⁴.
00:41:29.100 --> 00:41:32.700
There is this nice, easy rule for taking derivatives with the power rule.
00:41:32.700 --> 00:41:38.600
If we have x², and we want to reverse the process, well, since it drops down by 1
00:41:38.600 --> 00:41:45.900
each time we end up taking a derivative, that means that the anti-derivative must push our exponent up by 1.
00:41:45.900 --> 00:41:52.700
Since it will drop down one when we take the derivative, when we go to the derivative of F(x), remember:
00:41:52.700 --> 00:42:00.500
since we can take the derivative of F(x) to just get f(x), we have to have this relationship
00:42:00.500 --> 00:42:04.900
of our exponent dropping down when we take the derivative of F(x).
00:42:04.900 --> 00:42:10.900
The derivative of F(x), that 3, must drop down to a square, since 3 - 1 will go down to 2.
00:42:10.900 --> 00:42:14.300
So, we know that the exponent that must start there must be 3.
00:42:14.300 --> 00:42:19.500
However, if we were to work this out, we would end up getting x³; if we took the derivative of this,
00:42:19.500 --> 00:42:25.400
we would bring the 3 down, and we would get 3 times x^3 - 1, or 3x².
00:42:25.400 --> 00:42:29.600
But there is this problem here: it is not 3x²; it is just plain x².
00:42:29.600 --> 00:42:35.100
So, how can we get rid of this 3 coefficient at the front? We just divide the whole thing by 3.
00:42:35.100 --> 00:42:39.700
So, we divide the whole thing by 3, and now we have the anti-derivative to it.
00:42:39.700 --> 00:42:50.100
Let's check and make sure that that works: let's check: if F(x) is equal to x³/3, then what is the derivative of F(x)?
00:42:50.100 --> 00:42:56.500
Well, it should turn out to be f(x); we have this cubed exponent; we bring that down and out to the front.
00:42:56.500 --> 00:43:04.300
That is going to be equal to 3 times what we started with; 3 - 1 becomes a 2; it is still divided by 3;
00:43:04.300 --> 00:43:10.900
3 and 3 on the bottom cancel out, and we have x², which is what we started with; that checks out--great.
00:43:10.900 --> 00:43:13.700
So now, we have figured out what our anti-derivative is.
00:43:13.700 --> 00:43:18.300
At this point, we can now actually use this portion of the formula:
00:43:18.300 --> 00:43:28.000
F evaluated at...what is our b?...3, minus F evaluated at...what is our a?...0:
00:43:28.000 --> 00:43:36.000
so, F(3) - F(0)...the anti-derivative of our function evaluated at the far end, minus the anti-derivative at our starting location.
00:43:36.000 --> 00:43:39.500
The ending location, minus the starting location--what is our thing?
00:43:39.500 --> 00:43:49.900
It is x³/3; that is what our F(x) is; so we have 3³/3 - 0³/3.
00:43:49.900 --> 00:43:56.900
0³/3 just disappears; so we have 3³/3; 3 cancels this into a 2;
00:43:56.900 --> 00:44:05.300
the 3 on the bottom cancels our top exponent down by 1; we have just 3²; 3² is 9--nice.
00:44:05.300 --> 00:44:10.200
And that is the exact same thing that we got by working through that long infinite series method,
00:44:10.200 --> 00:44:13.500
where we had the limit as n goes to infinity of this approximation.
00:44:13.500 --> 00:44:16.600
And notice how much faster this was than that previous method.
00:44:16.600 --> 00:44:20.800
And this was with me carefully explaining a bunch of ideas that we have just seen for the very first time.
00:44:20.800 --> 00:44:24.400
If you were used to doing this...when you get used to this method, you can fly through it.
00:44:24.400 --> 00:44:27.200
You can do it so much faster than trying to work through the limits.
00:44:27.200 --> 00:44:34.200
That precise, formal limit method is something that you learn at the very beginning, just so we can introduce this really, really amazing fact.
00:44:34.200 --> 00:44:39.400
This is the fundamental theorem of calculus; this idea is so important that it gets the name "fundamental theorem of calculus."
00:44:39.400 --> 00:44:40.900
It is really cool stuff here.
00:44:40.900 --> 00:44:44.600
All right, that finishes this course; it has been a pleasure teaching you.
00:44:44.600 --> 00:44:48.600
I hope that you have not just learned how these things work and how to use them,
00:44:48.600 --> 00:44:53.200
but that you have also started to gain some idea of how math works on a deeper, more intuitive level.
00:44:53.200 --> 00:44:56.300
Things are starting to make more sense to you than they were at the beginning of the course.
00:44:56.300 --> 00:45:02.000
Beyond just the specific material that you have worked with, you are starting to get a better sense of just how math works in a personal way.
00:45:02.000 --> 00:45:05.400
And even if you don't decide to continue on with math, I hope you have started to develop
00:45:05.400 --> 00:45:09.400
a little bit more of an appreciation for just how cool it is--how many uses it ends up having.
00:45:09.400 --> 00:45:13.900
This stuff is really great, and it forms the basis for so much of the society that we have.
00:45:13.900 --> 00:45:21.200
The basis of technology, in many ways, is mathematics; plus, I think that this stuff is just beautiful and cool for itself, just to study it.
00:45:21.200 --> 00:45:23.600
All right, we will see you at Educator.com later.
00:45:23.600 --> 00:45:26.000
And I wish you luck in whatever you end up doing--goodbye!