 Welcome back lecture 8 today. We are a little abbreviated in numbers today. Actually, I'm kind of surprised that all of you made it We have had some snow for those of you that are not kind of right up with us at the moment some snow and ice and Made classes canceled yesterday and classes were canceled. I know you were sorry to see that canceled until 10 Right and this one starts at 1015 But most of us are here. We will have to make some slight adjustments to the syllabus possibly I don't know if I mentioned this but the tests will take place on the day when they've been scheduled If we do adjust anything it'll be the content that's included by that test But I'm not going to change the test days unless we happen to be out on the test days and then obviously we'd have to change them but We kind of have a little built-in cushion. There are some problem-solving They call them focus on problem-solving sections that it's built into the syllabus if time permits So maybe time won't permit, but we can kind of get back on the schedule that way so the goal today is to start in the chapter 6 and We will see what happens These are different applications of integration so we've learned a lot of techniques of how to integrate We kind of know one application and that is that it is area under a curve And now we're going to learn some other applications of integration John is all excited because his Pittsburgh Steelers won and they're in the Super Bowl And I actually have an outside chance of going to the Super Bowl So I'm kind of waiting on a couple phone calls So that could be exciting and no you're not going to get my ticket if I get one Okay, 6.1 talks about the area between two curves. So that seems like it's pretty simplistic Step forward from the area under a curve But from time to time depending on how the curves are shaped in the plane or if they're parametric in nature It can get a little More complicated than just simply area under a curve So there might be more lines or curves, but basically we'll take them two at a time kind of which one is the Upward upper part of that a region that's bounded and which one is the lower part of the region that's bounded so let's say we have some f of x up here and Some g of x I'm going to start them without their intersection at all So we'll have to decide where to start the process and where to end this Area gathering process. So there's the region that we want to know its area There are two very good ways of remembering this Let's briefly go through both of them and then you can kind of choose which one you hang on to to Remember how to find the area between two curves You could look at this kind of as two pictures The area under the f of x all the way down to the x axis We know how to do that That'll give us what we want there right and then obviously that's too much We want to throw out the region that we don't want that's part of that So if you think about the area under the g of x curve, isn't that exactly the region we want to throw out? So we want to take this This region and basically come back up here and get rid of it So we want to subtract out The area under the g of x and that'll leave us exactly what we want right here in this shaded region So that's one way to remember it. It's the area under the upper curve y equals f of x Subtract from it the area under the lower curve Y equals g of x and we could put those together Because the limits are the same and we're integrating with respect to x So from a to b f of x minus g of x So that'll get us the area between two curves Doesn't really matter where they are, but it does matter that we can kind of establish The region without any overlap and I'll try to get a diagram that does have overlap in a minute but we want to be able to basically define the upper curve the f of x curve and That's defined separately and differently from the lower curve of this region Which is the g of x so it's y value on the upper curve Minus y value on the lower curve now another way of coming up with that same thing is to think of this in terms of skinny little rectangles like we did initially with area and We would say how tall is this rectangle? Well when it was just one curve. It was f of x tall Now we've got two curves, so we need to know this distance from here to here What would you call that distance this curve up here to this curve down here? Okay, isn't that f of x minus g of x isn't that the height of that rectangle and The width is still our old friend delta x, so you could define it in terms of height times width The height is f of x minus g of x. That's how tall it is And the width is delta x and you come up with the same Formula so this actually this developing a Piece of it like we did with the skinny little rectangles. That's actually helpful in the whole process as we go through chapter 6 So but either way that helps you remember this Formula to take the y value on the upper curve Subtract from it the y value on the lower curve. You're either going to be given Starting and stopping point or it's going to be your job to find them So what do I mean by it's going to be your job to find them? so let's say we have a parabola that opens down and A parabola that opens up and they actually have a bounded region So we want to know where to start this process and Where to end this process So let's say the parabola opening down is f of x the parabola opening up Is g of x? How could you find the starting and stopping points for this bounded region? Where they intersect and how could you find where they intersect? Say them equal to each other. So when you want to find this value this value set them equal to each other and Solve and that should give you some Initial x value and then some x value a little bit later down the road So if you're not given them you can set them equal to each other and Solve for them now. Here's what I was talking about you want to make sure that your Regions in this case your skinny little rectangles all have the top on the upper curve in the bottom on the Lower curve and that that never changes as we go from this x value to this x value I think that is the case here right the top of each one is on the parabola that opens down and The bottom of each one is on the parabola that opens up And if that switches then we need to interchange what constitutes the upper y value and the lower y value So if we were setting this up From the points of intersection that we get right here x1 to x2 whatever they are we don't know the function, so we can't say what they are the y value on the upper curve is F of x the y value on the lower curve the lower curve is in fact the g of x and Then we would do any subtraction that we could Before we integrate integrate and evaluate all right. Let's take a look at an example like this in fact, let's look at One very similar to that And I'm not going to call them f of x and g of x we can decide which is the upper and which is the lower couple of the examples we're going to do today are in the book because they're kind of We've either done something with them before like the cycloid when we get to parametric equations Or they're just kind of good introductory problems, I don't think this one is in the book, but it's I don't know my is it Anyway, you've got a variety I think of ones in the book and ones that will not be in the book that you can refer back to Tell me what we have here with these two curves What's the first one? Parabola that opens down right because of the negative coefficient of the x squared. What about the next one? Opens up Probably somewhat similar to what I just sketched on the previous diagram Without a graphing calculator. Let's kind of quickly come up with a Sketch of 4x minus x squared. What can we do to get a rough sketch? All we want to do is see Which curve is above the other one we probably it's probably going to be the parabola that opens down, right? It's going to be the upper curve and the parabola that opens up will be the other one, but let's Get a quick sketch of this That x equals to Okay, how did you know that? Okay, good. I mean we've know some calculus, right? We might as well use it. So the derivative would be what? 4 minus 2x and we want to know the vertex so we want to know where the derivative is 0 and That happens at x equals 2 so we could actually plot that point Or you could remember from a pre-calculus course that the vertex is what negative b over? 2a you remember that No, but we have a better way. We can use calculus to get it So we would put that into the function and see what it kicks out If you're plotting points the nice thing about this equation You can kind of put together some of these use some of these not use some of these But this factors real nicely x factors out right and If we want to know where y is 0 Well, that's at x equals 0 right and x equals 4 Now doesn't that also then add fuel to the fire that the vertex is that too? Right if we're here at 0 0 and 4 0 the vertex better be at 2 something because we've got to have that natural symmetry and to what is the Top function evaluated at x equals 2 4 so there's our parabola that opens down How about the next one does that factor x squared minus 6x plus 8 minus 4 minus 2 I think that's right and when x is What 2 and 4 y is 0 so 2 something and And 4 something actually to 0 right And 4 0 are the points and where's the vertex? It goes through the point 2 0 and 4 0 3 something and what is the Put 3 in there, what do we get? 9 Plus 8 17 minus 18 negative 1 that sound right Now one point I do know because we actually plotted that point on both functions is x equals 4 So I know that's the upper Limit of integration the lower limit might be one, but I'm not necessarily going to trust my graph here that it is One maybe one might be one point one What can we do to make sure we have the exact point of intersection set them equal? So we can add x squared to both sides That's gone Got 2x squared We can subtract 4x From both sides that's gone. So we have 0 over here minus 10x Plus 8 and then factor. I think it factors minus 4 minus 1 So it looks like 4 which we already knew 4 was one of the points we plotted that on both curves and it is 1 Looks like one, but I don't I'm not going to trust my graph, but now I've got a little more Validation other than my picture. So we want the area of this region From 1 to 4 Alright, so we know the limits 1 to 4. We're going to integrate with respect to x What minus what? 4x minus x squared. That's the y value on the upper curve Minus y value on the lower curve Integrated with respect to x Questions or issues about that one. So let's go ahead and do the addition and subtraction we've got Minus x squared minus another x squared Minus two of those we've got 4x minus a negative 6x and Then we've got a what minus 8 you don't have to do the subtraction before you integrate, but it'll save a little bit of work So now we integrate integral of negative 2x squared Integrated with respect to x negative two-thirds x cubed integral of 10x with respect to x 5x squared or 10x squared over 2 and the integral of negative 8 The evaluation is from 1 to 4. Let's at least set up the arithmetic So if we put in 4 we get that if we put in 1 we get that Any question about what to do from that point? Punch the appropriate buttons on your calculator right from that point Questions on that one see if I have that written down the answer. I do not have that written down. Sorry let's suppose That we have a region that is let's say a parabola, but it's type of parabola. That's not a function So let's say we have a Parable that opens to the right. I don't know some other curve. Let's keep it simple. Let's say it's a line Do you see the problem with the same type of mentality is if we split this up into a bunch of skinny little rectangles? We use the same thought process. What's the problem with those two skinny little rectangles? Okay, there the y value up here and the y value down here are on the same curve, right? So it's not like y value on the upper curve minus y value on the lower curve. They're on the same curve Same thing here now eventually when we get past this point our rectangles are good, right? Y value on the upper curve y value on the lower curve and those the rest of the way Those are going to be good, but this part of the region. It's a problem. How can we solve that problem? Okay, let's go rectangles the other way. So how about rectangles that are formed in this fashion? As long as we Describe them properly. We have their height described properly. We describe their width Properly we describe one of them and basically say where they start and where they end then we ought to be in the same exact situation Now as you look at all of these rectangles And we try to determine their height Wouldn't their height be the x value over here? Minus the x value over here. Is that correct? Is that true about all of them? Well those that are pictured and any others that you can visualize The height of this one is x value here minus x value here all of them the same So we want x value of the curve on the right Minus x value of the curve on the left. What's the width or thickness of each one? What would you say that is? Delta x That's a delta y isn't it? Some increment of y Thickness of some delta y when we draw them this way. We've got delta x's right From this x value to the next x value we're crawling along little delta x's along the way now We're going to crawl up so to speak little delta y's along the way So as long as we can describe them in the same fashion All of them which we can x value curve on the right minus x value curve on the left We would start this at some initial y value And we would end this process at some y value a little bit larger than that So we kind of just reorient the whole process instead of going from x zero to x one we go from y zero to y one We're integrating with respect to y that kind of makes sense doesn't it we're going to have a dy in there So we want x value. I'll just put a little subscripted curve on the right minus x value curve on the left We want a delta y or dy From our initial y value to our final y value Same process just change how we're constructing those rectangles And you'll you'll know there's a problem because you'll try to do it the other way Which we can most of the time But the y value up here and the y value down here. We're on the same curve So it's got to be a problem. Let's see if we can reorient the rectangles and have success the other direction Let's see where my example is Now I've always wanted to do my part in here. Just keep that in mind as we go through problems So I'll graph the second one x equals y squared But I'm not sure And then you guys can tell me how to graph the other one So isn't that our old friend y equals x squared just kind of shifted over Opening to the right. How are we going to graph this? We could actually solve it for y. Where does this start? It is a non-function type of parabola just like this one is Linear in terms of x quadratic in terms of y just like this one linear in terms of x quadratic in terms of y x plus 4 what kind of effect does that have? Shifts it how many units in which direction? Left four units is that correct? Don't worry about the square root. Just worry about what's added to or subtracted from x So we're going to shift left four units Now if you put negative four in there, let's just check it if you put negative four in for x You get zero over here So that means y has to be Zero right? So that's got to be the vertex I don't know. Is it open right or left? It opens to the right. Is it a little fatter? Then our old friend right here are a little skinnier. Maybe that's a bad question for a day after a snow holiday Let's where does it hit the axis? Square root of two So if x is zero, this is four Divide both sides by two right? y squared equals two square root of two So there's one there's two square root of two is about what 1.4 So there's our region region bounded between two functions neither one of which is a Two curves neither one of which is a function So can I draw my rectangles this way? Not not going to work So I'm going to have to construct my rectangles left to right. There's one of them. There's one of them Let's describe the height of one of those rectangles the height would be x value of the curve on the right What's the x value of the curve on the right? Which curve is on the right? This one's the one on the right. This one's the one on the left So what's the x value of the curve on the right? Y squared What's the x value of the curve on the left good to y squared? Minus four now the word solve it for x. What is x equal to on this one? This one's already solved for x. So we didn't have to do any work. Just write it down So that should be x value curve on the right check it out This should be x value curve on the left. We want to integrate each one has thickness Delta y or dy and we want to integrate from some y value down here Whatever this y value is right here. I don't know And whatever this y value is up here. I don't know it kind of looks like to but again I'm not going to trust my graph How do I find out the limits of integration? Is that an equal? So 2y squared equals x plus 4 if we solve that for x x equals 2y squared minus 4 The other one is already solved for x So if we solve them both for x and then set those End results equal to each other 2y squared minus 4 equals y squared It's going to be plus or minus 2 in it subtract y squared from each side So 2y squared minus y squared is oh, that's just 2 isn't it the y squareds are gone No, that would be 2 y squared minus one of the y squared is what One of the y squareds remain Sorry, I shouldn't say that stuff out loud, but I've been told that for so many years that Kind of on some level. I want to say it, but I know it's not true So y equals plus or minus 2 So it looks like we're going to go from negative 2 to 2 So this intersection point right here in terms of y is negative 2 and This intersection point right here in terms of y is positive 2. So there's our work Let's take that to another page and at least get it to the point where we've got integrated and evaluated and Not necessarily doing the arithmetic, but we can if it looks pretty simple So we have y squared minus 2 y squared, which would be negative y squared And we have minus a negative 4 which would be plus 4 Does that look right? Let's integrate Negative y squared would be negative y cubed over 3 Integral of 4 with respect to y would be 4 y the negative of 2 cubed over 3 plus 4 twos negative of negative 2 cubed over 3 Plus 4 negative 2 Let's go ahead and do the arithmetic on this one So we have what negative 8 thirds Plus 8 when we cube a negative we get a negative we're going to subtract it Which is positive and then we're going to subtract that again, which is back to negative, right? So negative 8 thirds that's going to be negative 8 which we're going to subtract and get 8 so 16 minus 16 thirds 16 is 48 thirds minus 16 thirds 32 thirds or 10 and 2 thirds square units right of area bounded between these two curves Hopefully that arithmetic went well So after you get a rough sketch of the regions See which orientation as far as the rectangles are concerned is going to give you The correct area as long as you're not bounded on the top and the bottom by the same curve or left and Right by the same curve Okay, let's at least set this one up. There is a reason for looking at a problem like this Which we'll get to when we get a picture We have a cubic polynomial We're not expected to factor this cubic polynomial But we want to get an idea of where it is in the plane I think it's going to be helpful if we do that again I could do my part and graph the second one. Okay, if you guys can graph the first one Let's use derivatives to get a sketch of the first curve. So the derivative is 3x squared Plus 6x minus 9 Why do we want the derivative? What purpose is that going to serve to sketch this curve? Okay, like in the flat places where it kind of reaches a relative max where it has a relative min It's probably going to look something like this Something like that right this cubic polynomial So we can factor out of 3 and then we'll set the derivative equal to 0 To find those places where the curve itself is flat and that appears to be what I think this factors minus 3 Plus one does that work? I think that works So what x values Caused the derivative to be zero not right Oh, there we go, okay, that's plus Thank you, and that's still minus right so that makes it plus three minus minus one. Thank you so x is negative three and One and we need to find those points. So we'll put them back into the Equation Negative three something and one Something what do we get when we put one in there? We get one plus three minus nine minus twelve Where do we get? Negative 21 plus 4 negative 17. Does that look right? negative 3 so we've got negative 3 cubed 3 times negative 3 squared minus 9 times negative 3 Minus 12 so there's negative 27 Plus 27 so those knock each other out is that right positive 27 minus 12 15 How's that look? All right, so let's get a sketch of this one Sketch the line, and then we'll see how this problem is different from the other two that we've looked at so negative three 15 It's flat there in a positive one negative 17 It's flat there So we could do the increasing and decreasing stuff I don't think there's any doubt that in order to get from this point down to this point. You have to leave here fairly flat and Enter here fairly flat. I think if we either use the derivative to see where it's increasing and decreasing I think we'll see that this curve Increases into here increases Out of here, so that's not unlike what we thought it was going to look like So we graph the other one the other one is y equals 4x plus 3 So we've got a y intercept of 3 and a slope of 4 So there's a region that's bounded and there's another region that's bound Why is this problem based on this sketch different from the other two examples that we've looked at? The function that's on top changes it changes at this point right here, whatever that is We could set them equal to each other. That's not the issue. The issue is Deciding that these rectangles Although within this region We're good with these kind of rectangles, right? the upper curve is The cubic polynomial the lower curve is the line Then we change at this point the upper curve Changes to the line the lower curve is the Polynomial so it looks like kind of two separate problems. So the first region From this point Let's call it x zero to this point where they meet x one So from x zero to x one It is the polynomial minus That's the y value on the upper curve y value on the lower curve is 4x plus 3 Everybody content with that? That takes care of our first shaded region when we get past x one or right at x one All the way out to here, whatever that is x two now the upper curve Or the y value on the upper curve is 4x plus 3 Y value on the lower curve So sometimes if the curves Intersect and then the other one takes the upper curve roll and they switch the other one as the lower curve roll We do need to break it up now This is probably something you're not expected to do is to identify x zero x one and x two but if we were going to go about that business what would we do? Let's set them equal to each other now. Here's probably why this is Not a fair problem from this point because you would be setting x cubed plus 3x squared minus 9x minus 12 equal to 4x plus 3 So eventually you'd have a cubic polynomial and it'd be your responsibility to factor it and that's I'm not holding you To that level of responsibility. So that would be x cubed What plus 3x squared minus 13x Minus 15 Now if everything had a factor of x we could factor an x out then we have a quadratic that's fair But factoring this and finding x zero x one and x two That's not the purpose of this problem. Anyway, I just wanted to show that we had a way to graph the cubic using some previous calculus and Then we switch the upper curve and the lower curve. So we just basically Stop and start our integral that solves the problem. All right. I don't know that we can finish this next one But let's see if we can at least get it set up Area under a curve if we just have simple area under curve we would have F of x dx from a to b we've done that numerous times That's kind of even taken a step back from what we've done in this class today if it's helpful to think of Area under the curve f of x dx as y dx That's kind of what we're doing Now let's adapt this to a parametric version Let's say we have y equals Now x equals some function of t We'll get an example in a minute and y equals Some other function of t so we have this parameter Introduced into the problem that we don't have x in terms of y or y in terms of x We've got each one defined in terms of t. Maybe t is time so we've got parametric Well, why we've got y Right here We want to kind of imitate this same process y is really g of t dx Well, we've got x all we have to do is take the derivative of x and if we integrate that with respect to t from our initial t value to our Next t value that's either given to us or that we find ourselves Then we can imitate the same thing from area under a curve to area Under a parametric curve, so I don't think we've had that yet I don't think we've had to contend with parametric equations and finding the area under a curve So I think we can get this set up, but it's probably about all we'll do Earlier in this class We had a cycloid and we had the equation of a cycloid Let me try to quickly refresh your memory what this cycloid was we started here With a circle and there was a dot on that circle and then we rolled that Can or that circle Along the x-axis so as we rolled it along the path of that Dot on the circle Was something like that remember we did that earlier in this class if we we not done that maybe we did that in 141 We did that in 141 I got a okay look from a couple of you, but not from all of you, but If this is not something you did in your 141 class Take a look back at the first reference in this book for a cycloid But if you'll have a I wish I'd brought something now Anything that's circular in shape take the end of a cylinder put a dot on it and roll that cylinder along The desk so along the x-axis The path of that dot as you roll along Follow something like this so eventually you've got this as you roll that along it ends up here That's its highest point and as you continue to roll it along the desktop or the x-axis That out ends up back here. So this is the path of the dot and this curve is called a cycloid And the equations of a cycloid Parametric equations are as follows Whatever the radius is of this circle that's generating the cycloid is r That's the x description in terms of r and theta And the y description is this so we have x in terms of theta and y in terms of theta If we picked a particular value of theta which we could do we're not going to do a lot of this, but So let's say we picked zero If zero is the theta value that means that we haven't Started this process yet. By the way, what zero is where we're going to start What is the angle that we've gone through by the time we get this thing all rolled over there? And it's right back where it started That's two pi So zero, let's see. What would we be? Zero minus the sine of zero that's zero right zero times r is zero One minus the cosine of zero cosine of zeros one one minus one is zero So that's our time zero. So we're starting at zero zero. We knew that that validates that point Let's go at the halfway point here If theta is pi We've got pi minus the sine of pi What's the sine of pi zero? That's pi minus zero, which is pi So r whatever r is so we come over Theta excuse me. We come over pi and we go up r. Is that right? So r is not really the radius. Isn't it the diameter? Is that right? We come over and plot this point. It's over pi and up wait a minute So we've got we've got pi minus zero, which is pi and pi times r. There we go Is that right? That's what we're plotting. So that's our x value and our Y value if we put in pi cosine of pi is Negative one so one minus negative one is two Let's make it simple. Let's make r equal to one So if r is equal to one this point that the x value is Pi right pi times one to simplify it a little bit and the y value if r is one this would be two and if R is one this would be one and this would be one so we would go up r and then another r, right? So we're going to go up to r. So we go over pi times r Let's say that r is one Go over pi and up to and we're at this point and you can continue to do points So let's set this up and then we're out of time So we want to set this up so that we have the y value and then Derivative of the x value and then we'll pick up from this point tomorrow our limits are what 0 to 2 pi What's the y value r? One minus cosine theta and what's the derivative of the x value? derivative of theta with respect to theta is one derivative of sine of theta with respect to theta is cosine theta Something's not right here. Yeah. Yeah, we're okay. I was thinking that we needed So here's our y Here's our DX we will pick up from this point because we're out of time today