 Welcome back to our lecture series, Math 1220 Calculus II for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. Welcome back everyone. In our lecture three, we're going to talk about area between curves found in section 6.1 of James Stewart's Calculus textbook. Now, before we jump into the details of area between the curve, it's probably useful to remind ourselves, how do we actually find the area under a curve? When we talk about area under the curve, the problem we're describing is something like the following. We have our x-axis and we have some function f right here, and we have some value a to b right here, and we want to find the area of this region that's above the x-axis but below the function itself. If you need a refresher of this, then check out this video for my Calculus I series that reminds you about the meaning of the definite integral. What we want to do is change the problem up a little bit today. It's basically the same situation. We're going to start off with the x-axis. We're going to have our function f like we did before. But in addition, we're going to have a second function g, where g sits below f, and so we get something like this. We're interested in if we have this value a on the left and this value a or b on the right, these are x values of course, what's the area of the region between the two functions f and g? What is this area? This is the question we want to approach for the next little bit right here. It turns out the way that we can handle this problem is the exact same way that we handle the original problem of area and where we got the definite integral in the first place. The idea is take this interval a to b, and we're going to slice it up into small equal sized regions. We'll draw these little tick marks here. We cut up our interval into small, small, small pieces. We started with x1, x2, x3 continue on. We're going to get someplace xi minus 1 in the middle, xi right here, and they continue on until you get to b. b is in fact the last number here, and if you have n of these x values, that's going to be xn. Actually starting the a value is what we often call x0. We have these intervals, the first interval, the second interval, the third interval, up until the nth interval right here. What we're going to do is we're going to form a rectangle between these values here. For the first one from x0 to x1, we're going to create a rectangle, and it's going to look something like this. Instead of the rectangle going from the x-axis all the way up to the f function, we're only going to draw a rectangle that sits between f and g. Then for the second interval, x1 to x2, we do the same basic idea, right? We draw a rectangle like this for the third interval, x2 to x3. We're going to draw something like this, and then we keep on doing this, pattern, dot, dot, dot, and so the ith interval will take a height right here. We'll take a height right here. We connect the dots, and again, we continue in this fashion. Keep on drawing these rectangles where the top and bottom of the rectangles are determined by the function f and g right there. That helps us, much like with definite integrals, we can add up together the area of the ith rectangle, we'll call that ai right there. We know that the sum of all these ai's, as i goes from one to n, this will be an approximation of the area between the curves. So if we can calculate ai, we can approximate the area between the two curves here. So how do we get ai? Well, it's the area of a rectangle, so its value will be length times width. The width of this thing, just like with the definite integral, is gonna be delta x. Well, delta x has its usual meaning. Delta x is gonna equal b minus a over n, and that's because we selected all of the rectangles to be equidistant, equal in size there. Now, the length of this thing, we have to be a little bit more careful about, I'm gonna give myself a little bit more space right here. The length of this rectangle is gonna depend on a couple things. Where is the top of this thing, and where is the bottom of this thing? It should have some respect of the function itself. And so if we pick some delegate that sits inside of our interval in the textbook, they often call this xi star, we're gonna take the top of the rectangle to be f of xi star. So it's a representative of how high up f is. And on the bottom, we're gonna take our representative to be g of xi star, we're gonna pick the same value right there. And so the length of the function of the rectangle, then be the difference of these values. So we take f of xi star minus g of xi star, kind of got a little crowded right there. And then you need to times that by delta x. And that would give you the area of your rectangle. Now, let's write this a little bit below when I have a little bit more space here. So what we're seeing here is the following, that the area can be computed by taking the sum as I goes from one to n of f of xi star minus g of xi star times delta x. That delta kind of looks like an a delta x, there we go. And so we take this Riemann sum of f of xi star minus g of xi star times delta x. And this will give us an approximation of the area of the curve. Well, just like the original area problem though, if we wanna improve upon this, we can take more rectangles. The more rectangles we take, the better the approximation. If 10 rectangles does good, 100 will do better. 10 trillion will do even better. And in fact, if we wanna get the true area between the curves, we take the limit as n approaches infinity here. And this will then equal the area between the curves. Now this expression right here, the limit of the sum of f of xi star minus g of xi star. If we think of f minus g as a single function, then this right here is the limit as n goes to infinity of the Riemann sum associated to the function f minus g. And so this is in fact equal to an integral, which is this integral right here, the integral from a to b of f of x minus g of x dx. So the area between the curves is just the integral of f minus g. We can see that by going to the original limit definition of the integral, but we can also think of it in the following manner. If we look at the area that's under f and we subtract from it the area that's under g, the difference will give us the area between the curves. Because by integral properties, this thing is the same thing as the integral from a to b of f of x dx minus the integral from a to b of g of x dx. So there's sort of two ways of visualizing this. We can just use this limit properties here and take the difference, or we just integrate the difference in the first place. And I did want to go through the details of this so you can see the comparison with the original limit definition of integrals with the, to find the area under the curve and how we're able to modify it to find the area between two curves. I wanted to show this example because it turns out in calculus too, we're gonna see this situation all over the place. This technique of integration are what's sometimes called accumulation. People often attribute Sir Isaac Newton for inventing the calculus. And don't get me wrong, his contributions to calculus were immense. Newton was essentially the one who gave us the fundamental theorem of calculus. He made the connection between differential calculus aka derivatives and integral calculus. So I mean, we can't undermine how important that is there. But things prior to Newton, I mean, for thousands of years, people understood calculus notions, primitive notions. But I mean, this proto-calculus existed even in the ancient Greeks. Archimedes was doing calculus problems where he was calculating areas and volumes of geometric regions very much like we'll see in forthcoming lectures in this series here. And the idea is you take a problem and slice it into smaller problems like we do here. You slice, slice, slice, slice your problem into smaller linear problems. And then you add up those linear approximations and take the limit. This accumulation technique we're gonna see a lot in this chapter six in the future as well. Now, this diagram works pretty well if the function f stays entirely above the function g. There are some of course situations where maybe your function f is right here, but then your function g might switch places with it. So it starts off below, then it goes above like something like that. In which case you have to be careful because if you're taking f minus g, this region will be considered a positive area, but then the other region will be considered a negative one as you switch sides there. So you have to be careful. If you wanna find the geometric area of this region where all regions are considered positive, you have to make sure to take absolute value of this thing. But if you're interested in a net area, then you can leave it as it is. So let's see a quick example of this type of thing. Let's find the area between the functions f and g, which are given by the following functions. f is the quadratic function negative x squared plus one. g is the linear function, two x plus four. And let's find the area between the functions from x equals negative one to x equals two. And so you can see this region illustrated on the screen right here. I do wanna point out to you that in order to find the area between two curves, much like the area under a curve, it isn't necessary to actually draw the graph, but drawing pictures I see think does help us, it helps us with our intuition, understanding. So if you feel comfortable drawing these things, I would recommend doing. So even if you feel a little uncomfortable, I would try drawing these things as well. No one's gonna be graded based upon how good their graph is. And honestly, these things do not have to be drawn perfectly to scale. The graph itself can be a good intuition. Like if I were to try to draw this picture, I might notice all f of x is negative x squared plus one. The negative x squared means it's a concave down parabola. The plus one means it was shifted up by one. So you get this basic shape right here. For g of x, it's two x plus four. It's a line, it's slope is two, y intercept four, things like that. We can get a pretty good picture of what's going on right here. And so you see this region, you wanna find the area between the curves. Well, from the formula we saw a moment ago, the area between the curves, we're gonna integrate on the boundary, negative one to two, that was given to us. And we have to take the difference of the function. Now, if we want this to be positive area, we should take the bigger function minus the smaller function. So we're gonna take g of x minus f of x right here, dx. Don't forget your differential dx there. Now, g of x was given as two x plus four and f of x was given as negative x squared plus one. So we get this right here. Be aware that you are subtracting f and so you need to make sure you subtract the entire function. It might be helpful to put parentheses around f so that you make sure that you subtract both the negative x squared and the one right here. So I would recommend to distribute that negative sign before you forget to do so and then combine like terms. And so integrating from negative one to two, we're gonna get a negative negative x squared, so that's a positive x squared. We're gonna get a two x and then we're gonna get four minus one, which is a three, like so. And so now to find the area between these two curves, we need to integrate the following quadratic function by the power rule for anti-differentiation. We're gonna increase the power by one and divide by that new power. So x squared becomes x cubed over three. Two x becomes an x squared. You notice I took the liberty of adding two, or moving the exponent up to two and then I divided by two. That cancels out the coefficient story there. And then the plus three will become a three x. For an anti-derivative, there's often a plus C that goes on right here, but for definite integrals, it just cancels out you don't need it. We get a negative one and a two. Now we reach the most challenging part of calculus, which we call arithmetic. We got a plug in the numbers two and negative one, remember by the fundamental theorem. And so if we plug in two, we're gonna get eight thirds plus four, plus six. And then we subtract from that negative one third. Oh, what's happening to my pen there? Negative one third plus one minus three. Try to combine some like terms. And I mean, there's no variables here, but the fractions is what I meant. We have some common denominators. You have an eight thirds plus a one third. That gives us a nine third. That's kind of nice because that'll become a three. We get four plus one, which is a five. I'm sorry, not four plus one. It's four minus one. This is the danger I was mentioning before. You don't wanna forget to distribute that negative sign. Do it either in pencil or with your head. Either one is appropriate. Four minus one would give us a three. And then six plus three gives us a nine. Three over three, like it's a three. Three plus nine is 12. And then add three more. We should get an area of 15, which is the correct answer there. And so area between two curves is really not that different than finding the area under a curve. Be aware that we just made a slight modification to the original area problem. We can very easily adapt it to this area between the curve type of situation. Check out the next video to see some more example of finding area between the curve. There are a few subtleties we have to look out for, but it really doesn't turn out to be much more complicated than what we saw in this example.