I have one, We're going to talk about areas between two curves. And the thing to recall is that what is the definite integral actually mean. If I give you the definite integral from a to b of some function, f(x) dx, remember that this is the signed area. So everything above the x-axis gets a plus, everything below it gets a negative if it's subtracted off. This is the signed area of the graph between the graph and the x-axis. So the graph and x-axis. So in terms of our picture, if I put a curve of f(x) in the first quadrant and I label my left-bound a and my right-bound b, then the integral represents the area underneath this curve. Fine, good. So now the question, of course, is, what if I want the area between two curves? So find the area between two curves. And in that case, let's put f(x) somewhere in quadrant one, and then let's put another function, I don't know, g(x) below it. Same thing. Let's put a left-bound of a and a right-bound of b. So I have this shape. This sort of irregular shape between these two curves. And it turns out that definite integrals will be the way to find this. And the way to see or understand the formula that's to follow is, if they want the part in the middle, you could think of the area underneath f. In the f in the x-axis and then you take away. Let's see if I can take this away here. Let's pick a different color. If I take away the part underneath g. So imagine erasing that part. What you're left with is only the part between the two curves. So you take the whole, take f, and just subtract the part away from g. This gives us our formula. So the formula for the area between two curves is the definite integral from a to b, left-bound and right-bound, of the difference between your top function and your bottom function. Now, notice I call the top and bottom. Again, with the formula, try not to get too attached to the letters they use in the formula, because of course, that could change. So think of it as the top function minus the bottom function. And the way you evaluate these, of course, is to use the fundamental theorem of calculus to evaluate these things. So let's jump right to an example. Let's jump right to an example to see this. So I'll give you a function f(x), which will be x squared plus 4. And let me give you g(x), which will be x minus 2. So I have a parabola and a line. On the side over here, it always helps if you can graph them, just so you can kind of see what's going on. I think these functions are easy enough that, hopefully, we can graph them without running to a calculator. X squared plus 4 is the parabola, but it's pulled up 4 units, so it crosses the Y-axis right at 4. And I have a line, x minus 2. That's a line with positive slope, goes through 2. And it looks nice to me. So label your diagram. Parabola is called f(x), and that's our top function. I have g(x) down here. And let me give you some bounds on this thing. For without the bounds it doesn't make any sense. So I'll say between x equals minus 1 and x equals 3. So plot those points on the map somewhere. There's minus 1, and then 1 to 3. Our picture goes across the 2. So 3, okay. So I have this shape. And you can tell this is not a normal shape. This is not something I have, and it's between two curves. So I have a little bit above the X-axis, a little bit below the X-axis. How do I find this area? Well, let's use our formula. This is the definite integral from minus 1 to 3 of the top function minus the bottom function. So x squared plus 4. Let's put it in parentheses. Now, be careful here. It's the entire bottom function. So you have to subtract using parentheses. If you don't use parentheses, you're going to get a wrong number, you're going to miss the minus sign. And this expression, remember what this captures, it is the height between the two curves. It's the difference of the two functions. Limits of integration that the x values, where the region begins and the region ends. And this integral itself is very basic. And once you've simplified this, which I would recommend doing, so we go from minus 1 to 3. X squared minus x and then 4 minus minus 2. It's just 4 plus 2, that's 6. And this is a nice easy integral that you've seen before that you can evaluate if you want to. So I'll leave that to you. But I guess I care more about, in this section, how do we set up the integral. Whereas last section were cared about evaluating. So I'll leave it to you to work this out with areas. But as long as you understand how to set this up, you can go solve these later. Okay, so let's do another problem. Oftentimes, you'll find sometimes they just ask you to set it up, but. So it looks easy, but of course, there's always a trick, or there's always weird ones, so a couple of things you have to watch out for. So when you're evaluating areas of these regions, you have to make sure the curves don't cross within the corresponding open interval. And then the problem is, if they do, you're going to have to evaluate each region separately, okay? So let's do an example, and I'll show you where the pitfall is, the common mistake to avoid. So find the area, find the area between the x-axis and the graph of y = 2x + 4, we're going to go from -5 to 1, okay? I always recommend you draw a picture, especially if the graph is simple enough to do so. In this case, the picture will help see the danger, or the common mistake that a lot of students make. y = 2x + 4, that has a positive slope, goes right through the x-axis at -2. So we do that, sounds good. We only want from -5, so put that on the map. Picture not drawn to scale, but who cares, to 1. And so the curves here in question are the x-axis, right, the line, and then this other line, so we got two lines. And I want the area between the x-axis and this line. And you can see, it's two triangles. Now obviously, we could always just find area of the triangles. But what's happening in this interval is that the lines cross. And what is, I guess there's not one curve that plays the role of the top function. So you can't, it is a mistake, this is a very common thing people do. I'll put it in red, I'll say, no, don't do this. It's very common to say, well, let's go from -5 to 1, and we'll take the line 2x + 4 dx. Do you see why this is bad, why this will not give the actual area, because that's not what this integral represents. This integral represents the net area. This is going to take, if I call the left side, left triangle A1, and I call the right side A2, this will find not the sum of A2 + A1, this finds the difference. This is A2, and then you substitute the piece that's below. So this is not the area, this is not the area, this is the signed area. So we have a problem, if you went kind of naively about this, you would get the wrong answer. So let's do it correctly, let's do it correctly. We have to find the area of each piece, we have to find the area of each piece. So we're going to find A1 first, because they cross. This is the picture helping you out here. So that goes from where, that's going to go, [INAUDIBLE] leaves a little background. So let's go from -1, sorry, -5 to -2, so label your picture, find the points of intersection. And then it's the top curve minus the bottom curve. Remember, the top curve, the x-axis, is defined by y = 0. They don't tell us that, but hopefully you remember the x-axis is the line defined by y = 0. So this would be the top curve, 0, minus the bottom curve, (2x + 4) dx. Clean this up a little bit, you get the integral from -5 to -2 of -2x- 4 dx, and we can work this one out. If you do this, you get- x squared -4x, evaluated from -5 to -2. Plug in your numbers, so we get, be careful of all the minuses here. So -(-2) squared- 4(-2), again, use as much parentheses, as many as you like. And then so we plug in the top, plug in the bottom, you get -(-5) squared, saying minus a lot, -4(-5). All right, clean that all up, -2 squared, minus, so that's plus 8, -25- 20, all right? So you can work that out, that is 9. The other area, let's do a different color, area 2, the second area, same idea. The left bound is -2, the right bound which was given is at 1. And this is the top curve minus the bottom curve. In this case, the top curve is the line 2x + 4, and then the bottom curve is 0. So you can write that down if you want to, or of course you could leave it out, subtracting 0 doesn't matter. But I want you to see the formula in action, and you have to find A twice. There's no shortcut to getting around doing this twice, when you evaluate this integral. It's an easy integral to do, you get x squared + 4x, evaluated from -2 to 1. And you can check, you get 1 + 4 - (-2) squared is 4, -8. So it's that 5- (-4), 5 + 4, that's also 9. And if you want, you could check that the formula's working by finding the area of each triangle and adding them. But the point is, the point is, I add this up to get the total area. So the total area is then A1 + A2. So notice in the example, had you set this up incorrectly, you would have got an area of 0, 9- 9. And hopefully you'd be like, wait a minute, this thing definitely doesn't have 0 area, something's wrong. But the answer, of course, the real answer is 9 + 9, and so the total area is actually 18. I have seen more than one student come back and say, the area is 0. I say, that doesn't make any sense. So hopefully you get something weird, if you get a negative area, that can't happen either, something's weird. So watch out for this common mistake to avoid. And this is the one where the curves cross within the region, and there's not one single top curve throughout. You have to break it up into two regions and work it that way, okay? Another mistake to avoid, this is famous mistake number 2, is when the area is below the x-axis. And this is when people come back and say, the area is negative. I say, are you sure? They say yes, I say, let's try that again. So let's do an example, just to show you. So this is number 2, famous mistake number 2. So find the area, find the area between the x-axis and the graph of y = x squared- 16, and we're going to go from x to 0 to 2. Again, I think this advice was probably given to you as a child, or something like that. If you can draw a picture, draw a picture, draw a picture, draw a picture. This will help you see see what's happening and avoid these mistakes. x squared- 16, this is the parabola pulled down 16 units. So the intercept is right at -16, and if you plug in 0, you get -16, and then it crosses somewhere. But we don't care where it crosses, we're interested only in the part between 0, so x = 0, and 2. And you could check, if you plug in 2, you get 0 again. So you get 4- 16, sorry, so 4- 16, so you get some value, I don't know, but it's not quite where 4, right? This is x = 4, the intercept, so it's somewhere right in the middle, and it gives you some value down below, that's fine. So the region in question that I'm trying to find is this, not quite trapezoid, but trapezoid with a parabolic base, I want this area. And here's the danger, if you don't think about what you're doing. So the wrong way to do it, the wrong way to do it is if I just set up the integral, and I don't think too hard. Thinking hurts my brain, I don't like it, I do this, right? Now, the picture's going to hopefully tell you why this is wrong. Remember what this does, this finds the signed area, this finds the signed area. And what's going to happen is that this number, it's going to get you the right number, but it's going to have a wrong sign. It's going to turn out to be negative, if you do this, you're going to give me a negative area. And I'm going to look at you funny and say, negative area, that's very interesting, tell me more about that. So you have to fix this, and remember, it's the top curve minus the bottom curve. The bounds are fine, 0 to 2, that's the right bound. But it's the top curve, which is the x-axis, so this is defined by y = 0, and then minus the bottom curve. So if you want to think about the formula, it's 0- (x squared- 16), use parentheses, it's minus the entire curve, dx. Clean this up a little bit, you get 0 to 2, -x squared + 16 dx, and that's not a terrible integral to do. If you do this you get, so let's see, x squared becomes -x cubed over 3, and then 16 becomes + 16x, and we evaluate from 0 to 2. And we plug in, so we get minus eight-thirds + 16 times 2, what's that, 32? And then we have some nice fractions to add, and we get 88 over 3. Positive 88 over 3, might I add, and that's perfectly fine. So we're happy with this number, I call this one the negative area mistake. Watch out for this one, don't tell me that area is negative. And the difference here, what's the difference between what's prior or not? So a definite integral A definite integral can be negative. That's fine, that's kind of what they do, is their point. But areas are always positive or nonzero I should say. Always great than or equal 0. They cannot be negative. So if you get a negative area, go back and you might have made an algebra mistake. But more than likely if you were careful with the algebra, you perhaps didn't consider the fact that the area was below the x-axis. So that might be something take some time and draw the graph to do that. Okay, so the other way that you can see these problems where they get a little tricky is sometimes the limits are not given. And so now they're going to start making this a multi-step problem. Which means there's more opportunities to get the wrong answer or something that, or make mistakes. So just be extra careful. But let me show you how these are presented, okay? So the key is how they're presented. Nothing new, just an extra step. So find the area of the region between y equals x squared and y equals 8 minus x squared. Two very nice parabola is one points up, one points down. And I guess the thing that takeaway here, I don't know if you realize it or not. I'll put it in red because this is the point of this question, that the end points are not provided. The endpoints, the region of integration, not provided. Okay? So how do you find that? That's the piece that's missing. It's very hard to do a definite integral if you don't have the endpoints. So before we do anything, let's draw a picture. These are both parabolas, I know how to draw these things y equals x squared, how pretty is that? Very nice. So we go right through the origin, look at that, is one of my better parabolas have drawn. So here's good old y equals x squared. And then the other one is 8 minus x squared. So remember what this is, this is the parabola turned downwards, pushed up, 8. So let's draw this parabola downwards. Pushed up 8, so crosses right at 8. Okay, so you can now see there is a region between the curves. And this they don't cross between these things. So there's a top curve, there's a bottom curve. The piece we're after though, the piece that is not given, because they assume we can go find it, is what is the lower bound and what is the upper bound? And again, that's the whole point of this problem. Its intentionally not given, and the idea is we're going to find the balance, we gotta find A and B. Well, we know how to do this. We have to do algebra. And here's a question for you from algebra. How do you find where two curves intersect? Well, that's when the two functions are equal. So you set the two functions equal to each other, and you get x squared equals 8 minus x squared, and then we do a good old solve for x, okay? So this is the extra piece. Don't come back to say I don't have enough information. Now you can do this. So let's add x squared to both sides. You get two 2x squared equals 8 or x squared equals 4. And of course, x equals, now be careful, you're about to take a square root, so it's plus or minus 2. So lower bound is going to be -2. Your upper bound is also going to be 2. Now we have all the information we need, and now we can go ahead and find your area. So integral, -2 to 2 of the top curve, minus the bottom curve. Now we don't try not to use f and g. Top curve is the downward parabola, so we have 8 minus x squared, put in parentheses just to see it, minus x squared dx. This is not a terribly difficult integral, so we're okay. We can clean this up there a little bit, -2, 8, -2x squared. Combine the x squareds. And now we can do a couple of things here. We can work this out, or you can also realize that I have symmetric bounds. So your little sort of warning lights should be going off about symmetric bounds, and you have an even function, that's nice. So well, I can simplify this a little bit, it's going to be 2 times 0 to 2 of the function, some like this. The only reason I do that is because if I'm not using a calculator, I'd really rather plug in 0, I have less chance of making a mistake, then plug in -2. So if I'm going to find the idea, this is I find the area on the right side and I just double it. And that gives me anything. So there's some symmetry going on here. Let's do the integration. So don't forget that 2 upfront, so it becomes 8x minus two-thirds cubed, that's all in parentheses. Now we can evaluate that from 0 to 2. And when you do that you get, 2 times 16 2 cubed is 8 times 2, so 16 over 3, plug in the 0s, that all goes away. How nice is that? I have to do some fractional stuff, but that's not the end of the world. And when you do this if you do it right, you get 64, yeah 64 over, 3. Nice positive number, so that's good and this is it. So the key with this one is that you know it's not incredibly difficult. There's just this added extra step where the endpoints are not given. So look for questions like that, that tends to happen pretty quickly as you go through. And then just do some algebra to find, and always be careful. It's very easy to make an algebra mistake or even an arithmetic mistake when evaluating these things. So double-check your answer and just make sure it agrees with your sketch of the region. The closer you can draw it to scale, the better off you're able to be. But, yeah, okay, so we'll do some more questions. And then we'll even see regions that are bounded by three curves or more in the next video. So good job on this one, see you next time.