 Welcome back to Math 241. This class, the one here this semester, has their first test tomorrow. And we started the review process, kind of got bogged down a little bit, but we'll hopefully be able to finish that today in class. The test for this class will cover the review material 5.7 and 5.8, as well as all the new material covered up to and including average value of function in 6.4. Now on the syllabus for us, it says that there's a potential of a focus on problem solving. If time permits, time did not permit that. So we are not doing that. But everything else on the syllabus that is listed prior to test one, we are covering. So for cable TV people, it will be the same for your test. For future semesters on DVD, make sure you check the pacing guide that's supplied to you to see where your first test material ends, because it might be slightly different. You said you had some web assign things. Are they from 5.7 or 5.8? Yes. Okay. I have one. Okay. Nicole, go ahead. It says make a substitution to express the integrand as a rational function and evaluate the integral. Okay. And the problem is the integral of 1 over x minus the square root of x plus 2. Now I know I've done something after class. I think with somebody on this, did we do anything with this in class? Yes? I think we did. But we didn't finish it. We didn't finish it? Well, I'm probably not going to finish it again. But let's get it to a point where you can integrate it. What's probably the most stubborn looking term right there? Square root of x plus 2. Square root of x plus 2. So let's see if we can get anywhere. Now if you let u equal the radical term, and then you also have that term outside of the radical. If this is going to be a u thing, what's this going to be? Isn't that going to be a u squared? Right? If u is a square root of x, then x ought to be u squared more or less. So let's see if we squared both sides. So u squared would be x plus 2. And if we need to know what x is, which we do, because there's an x here, x is equal to what? And we're going to have to get rid of dx and get it in terms of du. So I guess use this one. What's the derivative of the left side? The u squared is 2u du. And derivative of x plus 2 is 1 dx, right? Derivative of x is 1 dx derivative of 2 is 0, so just dx. So there's our substitutions. We'll go ahead and leave that one. The dx is going to be 2u du. So that's technically going to be in the numerator. For x, we can replace u squared minus 2. And for the square root of x plus 2, we can replace that with u. You can bring the two out in front or leave the two in here. It's not going to make much difference. So there's the new problem. Hopefully a little easier looking than the beginning problem. Any suggestions on what to do with this? I've got one, but I think I'll wait and see if you have one. Partial fractions? So off to the side or on a next sheet of paper. We'll take this thing, factor it, if possible. I think it factors, doesn't it? So how would that decomposition look? Constance because of linear factors. No repeated linear factors. So here's the partial fractions decomposition that we want to solve. Get a common denominator. I think we can probably stop here to make sure we get through everything. But a is going to be a number, right? I don't care how ugly the number is. What's this going to be when you integrate it? That's going to be a natural log. b is going to be a number. Again, the ugliness of the number is irrelevant. You bring it out in front. This is also a natural log. So you end up with natural logs that have u in them. And then to finish the problem, everywhere you see a u, you will replace it with what? Square root of x plus 2. Does that work? Is that okay, Nicole? For that one to get it to that point? Yeah. Any others from 5.7 and 5.8 that you want to take a brief look at before we go further in the review? Okay. So in 5.9, I think when we went through this, and remember, we did not spend time with the midpoint rule. So the two rules that we did were the trapezoidal rule, which is using, we don't have a way to do the integration, so we're going to do numerical or approximate integration. So we're going to split up each region and basically call each region the trapezoid and the more we have, the closer we'll get to the actual area under the curve. Similar fashion. With Simpson's rule, it does not use trapezoidal regions, but it uses little parabolic regions that are actually twice as wide, but you can again make them as narrow as you want to make them. But we're forcing the parabola that serves as the lid of this region to not only capture the first point and the last point, but also the point in the middle. So that gives us a higher level of accuracy. So I mentioned when we went through here that on a test question, I'll probably give you the choice of which method to use. They're both pretty good. Simpson's rule is, I think we looked at error estimates and we determined that Simpson's rule is going to do better. But I'm not going to stipulate. You can look at the old test that I have given you and that I'll post for people in the DVD class. Where is it? Number four. Use the trapezoidal rule or Simpson's rule. N equals four. Now it's technically, I guess, N equals four for both. But what N equals four means to the Simpson's rule is that we're technically going to have two parabolic regions. So this is where you're going to start and this is where you're going to end. It kind of looks like four sub-areas, but this is really one parabolic region. Parabola captures that point and that point and we force it to catch the point in the middle. Same thing over here, the first point, the last point and the point in the middle. So technically it's kind of N equals four, but it means two parabolic regions. That particular curve that I had on this test in the fall, E to the one over X, I don't think we're going to be able to find that area any other way, are we? If I know of another way other than some type of approximate integration or numerical integration, you could say let U equal one over X. It's not going to work because you're not going to have DU in the integrand. I think that's a problem that we're without another weapon on this. Now did I on the solution include both methods? Yes. So as you look through this, the solutions that you see here to number four kind of compare and contrast. The H's are found the same way. Those are really delta X's, right? Which is B minus A over N. So the width of your sub-interval divided by N where N is the number of subdivisions. So the differences out in front for trapezoidal rule you're going to have what? H over two, right? And out in front for Simpson's rule you're going to have a H over three, right? What are the differences inside the quantities in parentheses or brackets? How does this one work? The initial value, X0, what do you do with the next one? Two times. Two F of X1, two F of X2. In fact, all those get doubled, right? Except your terminal value F of XN. How does this run inside the brackets for numerical integration? F of X0, what's next? Four. Four F of X1, two F of X2. Alternating fours and twos. The next to last is four, right? Coefficient. And the final coefficient is, again, one for F of XN. So take a look at those. There will be some computation. I guess that is a pretty good reason to have at least a scientific calculator. It's going to be difficult to do E to the four fifths without that. And we've talked about calculators, I think. As long as the calculator does not do calculus, it's fine. So a TI-89 does calculus, so you cannot use that on a test. TI-92 does calculus, so you can't use that. It gives you an unfair advantage, so you can't use those. And any other calculator that's been programmed to do exact integration or differentiation, you can't use that either, Nicole. I don't know if this WebAssign question is directly to this. But I guess I just didn't know how to set it up, because it's asking to use Simpson's rule to estimate arc length. Do you just use Simpson's rule the way you normally would, or do you have to incorporate the arc length? Okay, that's probably a pretty good question to talk about, so it does deal with this. And I have in tests before combined a couple of questions like that. So let's look at that, and then I'll also mention another way that I've asked a Simpson's rule or trapezoidal rule question on a test. Any questions, issues with kind of the basic format of these? If I remember right, didn't we, after class, Daniel somebody, we looked at this problem. So it's an arc length problem, which is 6.3. Is that right? Yes. So to do an arc length, and was it y in terms of x? Was it tangent? And 0 to pi over 4? So we want the length of the tangent curve from 0 to pi over 4. So that's an arc length problem. And we've got it y in terms of x, so we're probably going to use this. Do you have that one committed to memory? I hope. Hopefully by tomorrow you do. So that's kind of our little hypotenuse. We're adding together an infinite number of them from, in this case, 0 to pi over 4. So what I think this problem is, it starts out being an arc length problem. We can't integrate it, so in order to integrate it, we have to switch horses to an approximate integration technique, and that kind of forces us to use Simpson's rule. I think they actually say Simpson's rule. Which I'll give you a choice on that. Derivative squared. What's the derivative of tangent? In that squared is secant to the fourth. Does anybody know a good way to integrate the square root of 1 plus secant to the fourth? I guess Simpson's rule would be a nice little side suggestion, but another exact method. Separated into secant, or 1 plus secant squared. Secant to the third. What was that 1 plus secant squared? Same thing as tangent squared. But where is a 1 plus secant squared there? 1 plus secant squared squared. The reason I'm hesitating on all those is I don't like any of them. They're not legal. All those things that were mentioned are not legal. I could say that this is illegal. I could say that this is the square root of 1 plus the square root of secant to the fourth. I have the pen, and I have the little blue pieces of paper, so I get to say whatever I want to, but that's not right. You can't take the square root of a sum as the sum of the square roots. If that were legal, look at how the rules that we have fall apart. Is that the square root of 9 plus the square root of 16? Well, the square root of 9 is 3, the square root of 16 is 4. This is the square root of 25. Oh, we have a new system today. This started. The square root of 25 is 7. It doesn't work that way. So you can't take the square root of a sum as the sum of the square root. So that's not right. So does everything under the radical have secant squared in it? So we can't factor out a secant squared. Is this by itself a perfect square? Can you square something and get 1 plus secant to the fourth? What is that that you would square? 1 minus secant squared. Okay. Wouldn't there be a middle term? Right? Which this one doesn't have a middle term? Okay, I knew that was coming next. Wouldn't that have a middle term? This one doesn't have a middle term. This is not a perfect square. So if we can't factor something out like a secant squared, the square root of secant squared would be secant, we're kind of without a weapon on this problem, except to do approximate integration. So that's not going to work, that's not going to work. It's okay to think of those things to kind of mentally say, is that, can I do that? No. Am I going to be able to do that? Then you kind of quickly rule it out and then you say, gosh, I've tried everything I know how to do. I don't think we have a weapon. You could look at the table. We don't, I mean, that's not an option on this problem. So what we have to do is get an answer or an approximate answer to this problem and we don't really have another way to do it. So that's when this arc length problem actually becomes a Simpson's rule problem. This is now what we want the answer to. It happens to be the length of this arc. Okay. Secant squared. Well, what is the identity? It'd be tangent squared plus one equals secant squared, right? So it'd be secant squared minus tangent squared. Is that what you said? So we're going to plug that in there for one. These are good. These are valid things we could try. Now some of them aren't valid. That's not valid. This was valid, but we ruled it out. We tried this. It's worth a try. We ruled it out. So we want to plug that in there for one. Does that help the cause? It doesn't really do much. I think we're kind of... The weaponry that we're now trying is, you know, like we've got this Philistine giant out here and we've got a little rubber band. Okay. You know, of course it's five smooth stones. It's the real story, but we don't... You know, we're kind of running out. But it tells us what to do anyway. It tells us to use Simpson's rule. The question is, are we going to be able to check it with some other technique? And we're probably not going to be able to check it. So we know it's going to give us the length of the arc of the tangent curve from zero to pi over four. But to get that, we can't integrate it. So we're going to switch to an approximate integration technique. So we want to do this problem. It happens to be an arc length problem, but use Simpson's rule to get us an answer. So from zero to pi over four. Now if it bothers you that we've switched this question to an area under a curve question, that's in essence what we have to evaluate to find the length of the tangent curve. It ends up being an area under a curve. So the area under this curve happens to be the length of the tangent curve from zero to pi over four. Did it suggest an n? Ten. I knew that was coming. So n equals ten, I guess technically means that there'd be five parabolic regions, right? So our h or delta x is b minus a over n, which is pi over four minus zero, which is pi over four, divided by ten, which is pi over 40, right? Over three. Right. So if we were setting this up, we'll get it started. So our h over three, so our first value is f of zero. We know for Simpson's rule the next coefficient is four. So we're going to start at zero and we're going to slide over this h, which is pi over 40. So if we start at zero and go over pi over 40, we are at pi over 40. And then we're going to slide over another pi over 40. So we're at two pi over 40 or pi over 20, and so on, right? Each time we're going over a pi over 40, well, let's take one of these off to the side. What is the f of pi over 40? Here's our f. So to evaluate the f of pi over 40, that's what we need. Is that correct? And then when we do this, we do all the arithmetic, we add it all up, we multiply by pi over 120. That then is the answer to an area problem, but the answer to that area problem is the length of the curve. Is that enough on that one? I mean, that's too lengthy. N equals 10 is too lengthy for a test question. Is secant? Secant is one over cosine, right? Yes. So when we type it in our calculator, we do type in the square root of one plus one over cosine to the fourth. Yes. Okay. Anything else for we leave that one? Anything else from that material, Simpson's Rule or Trapezoidal Rule? Another way I could ask that question is to have you do a, let's say a Simpson's Rule or Trapezoidal Rule problem that also happens to be in the table of integrals, you do the approximate integration, you'll get an answer, use the same problem, look it up in the table of integrals, the subset that I provide for you, check and see if they're, I like that kind of problem because you can see if the answers are pretty close, you have a pretty good feeling that you did them both right or you did them both wrong, I guess would be the other end of that, but normally that means you probably did them right and then you can move on with confidence to the next problem, but I've asked it in that fashion before. All right, improper integrals. Improper, the simple type or the basic type is because one of the values, let's say we're going from one to infinity, that's not proper, that we would let this thing run all the way to infinity and think and expect to get an answer, but some of them have an answer, we say they converge, some of them don't have an answer, we say they don't converge or they diverge, so if we had let's say 8 over x cubed and we want to see if we could get an answer, the first thing we do is leave the, kind of the bad part analysis to the end, so we're going to go from one to a and then the very last step will let a approach infinity. I don't know, maybe I should have used the one on the test that I have one, number six, so you've got another pretty good example, I think that's going to converge also, isn't it? Yeah, you get an answer, you get a finite answer of one fourth, so we would say that's a convergent integral. You can look through that, 8x to the negative third, so the integral is 8x to the, so we have in the numerator, what, negative four and in the denominator, x squared, so if we put in a and put in negative, and put in one, so if that's what we have, the second term we're done with, minus the negative is actually plus, so that's really what, plus four, so we get four there, what happens to this term, as a gets infinitely large, approaches zero, so we get zero minus a negative four, since we got an answer other than does not exist or no limit, then we would say the integral converges, so that's if one of the limits of integration is infinite, the other is that the function itself gets infinite at either one of the values or somewhere in between the two limits of integration, so let's kind of combine those, let's say we're going to go from zero to three and we have five over x minus two dx, so everything looks kind of tame, except that what's our bad value here? Two. Two. So we would probably want to split this up and go from zero to two, since two is the bad value, pick up at two and go to three, now do we need to do both of them? No. We can start with one of them, and if this doesn't have a limit which it's not going to, somebody run through that thought process real quickly, you're going to get natural log, so we're going to be in trouble because we're going to have the natural log of zero, a natural log curve looks like this, so as you get close to zero, you're going to head toward zero, what's happening to the natural log? Negative infinity. Right, so this one's going to diverge, there's no reason to evaluate this one, it's not going to help it or save it, so this one's going to diverge, so therefore this is a divergent integral. Any questions, any web assigned questions that are related to improper integrals? None? Any experts? A bunch of experts on this, a couple of you are until, I'm kind of looking forward to how you will fare on this test, I think some of you are going to do really well. Chapter six, this should be a little bit more fresh on our minds, area between two curves, split it up into our skinny little rectangles, each rectangle is y value on the upper curve and this y value on the lower curve, that's how tall it is, delta x or delta y wide. So we may have to set them equal to each other to see where exactly it is that they intersect, that's algebra, shouldn't be an issue. One of the pieces would be a skinny little rectangle, so this would be g of x up here, this is f of x down here, g of x is the upper curve and this diagram, f of x is the lower curve and we would go from whatever their initial point of intersection is to their point of intersection that's later. This I think, I encourage that type of mentality to remember this because I think it'll serve you well through everything we get in chapter six, what's one of the pieces look like? It's a skinny little rectangle, what's the area of that skinny little rectangle? There's how tall it is, there's how wide it is, so it's a description of one of the representative pieces, we can add an infinite number of these together with the use of the definite integral. So that I think will serve you well through all of chapter six. What else did we have in here? Parametric areas of two areas under a curve that is described by two parametric equations. Normally it's this, if we have y equals f of x, how does that translate into parametric equations? This is really, I guess, y dx. So let's say we have y as a function of t and x as some other function of t. So we want to use our parametric equations. What do I plug in here for y, which is the old f of x, now it's y and what is y equal to? f of t. And what about derivative of x? Well, here's x, so derivative of x becomes what? g prime. We're going to integrate with respect to t from some t value to some other t value. So if you know area under a curve, you just make the kind of translation or transition from that to our parametric equations and you ought to be in business. Anything from areas, that's a web assigned question, Nicole. Find the area of the region x plus y squared equals two and x plus y equals zero. Area of that bounded region? Yes. So it doesn't give me any limits? No. Set this up pretty rapidly, I hope. So this is really y equals negative x, right? And this is equal to what? x equals negative y squared plus two. You can plot points however you want to go about it, but y equals negative x is this line. And what is that? Okay, non-function parabola, right? And it opens to the left because of the negative coefficient. And let's just plot a point. If x is zero, what is y? Sorry, that's not what I want. If y is zero, what's x? Two. So two zero. Now, is that the vertex of this? No. So I guess maybe another form would be x minus two. So we have the x plus h format and negative y squared. So if you had that x minus two equals negative y minus zero or plus zero squared. So the vertex ought to be hk, is that correct? Two zero. So we could plot points, but I think this is what we want, isn't it? That region. That's good that we're getting this because we can't use the same method in kind of the generic diagram that I used because if you try to do rectangles this way and this is not a function, we're in trouble. So let's construct our rectangles this way. And I think we're okay. We're always working with the curve on the left and the curve on the right. So isn't it x value curve on the right minus x value curve on the left? Is that correct? Yes. For the height of that skinny little rectangle, that should be the case with all of them. So what's an x value of the curve on the right? Right here? Yeah. That's x value curve on the right. From that, we're going to subtract x value curve on the left, negative y. So that ought to be the height of each of the skinny little rectangles. How wide or in this case, tall is this? What is that? Delta y? So there's the height, there's the width. Where do we start forming these? Down here would be our first one, right? And we keep getting all these skinny little rectangles all the way up to here. So we're integrating with respect to y. So we need a y value here and a y value here. How do you get those? Set them equal to each other. I think it was negative 2 and 2, but I can't really... You know, the accuracy that I have diagrammed this with, I think it is negative 2 to 2. It's highly, highly accurate. Anybody remember? Negative 2 to 2. Set these two equal to each other. So it would be negative y squared plus 2 equals... That's what x is here, x here is what? Negative y. So the two solutions to that ought to be your limits. Is that okay? And they are 2 and negative 1. Now, sometimes it's beneficial to actually do the subtraction of these things before you integrate. I mean, I'd at least make this plus y, but you're not going to gain anything by... You can't combine any of those terms. That worked. Volumes. So with volumes, we need to pay strict attention on the problem to what is the axis of revolution. So if we're going around the x-axis, kind of get a visual of that region together with its symmetric image having gone around the x-axis. So you could have a region, let's say it looks like this, and if you go around the x-axis, you'll want to get an image to see what each slice is going to look like. These are washers. So for a washer, and then this depends on how we're kind of forming those washers, dx or dy. We have a thickness of delta x. So this would be dx. Here'd be our initial x value right here, all the way over to here. So I think pictures help. Just again, where's the axis of revolution? If it is this method, and we are using where we're chopping it up, perpendicular to the axis of revolution, to form solid disks or washers, then I think it helps to get a look at a picture of it. When you're deciding on the inner radius, go from the axis of revolution up to the first curve. If you go from the x-axis up, that's typically what? For this point, don't we go over its x value and up its y value? So from the x-axis up to that point is the y value of this point. And for the outer radius, from the x-axis up to this point. And that would be the y value on this curve. Let's say for test purposes, I think it's okay to look at in class and even a web assign problem where the axis of revolution is something other than the x-axis or the y-axis. But for the test, the axis of revolution will be either the x-axis or the y-axis. Now, did I have one of those on here? Number eight on the old test. And it's around the x-axis. Now, this is possible. I don't know if I mentioned this. Set up only, do not integrate nor evaluate. That's possible. On a problem or two, the reason that's possible is that that saves time. And although being able to integrate and evaluate is part of what we're doing, we don't have to necessarily do that on every problem. So it could be washer. It could be solid disc. If the discs are solid, then they're just little short, squatty cylinders on edge. And that's the volume of a cylinder. So the h is really the thickness. So that's dy or dx. And the other one was then cylindrical shells, right? Take an element of area of skinny little rectangle, this time parallel. And this is a potential test question. If we can't slice it up or chop it up perpendicular to the axis of revolution and get a solid disc or a washer that we can actually evaluate, then we don't have a choice except to go to this, 2 pi r h times the thickness. That's how much stuff there is in each shell of a cylinder. Thickness would be dx or dy. So I think we did an example that looked something like this. That was the curve. We were going around the x-axis. So we took a little piece of area at this time parallel to the axis of revolution. We wrapped that around and we got this cylindrical shell. Pretty cluttered diagram. That's how much stuff there is in a cylindrical shell. So what you need to figure out is what is the radius? The radius is what? From the x-axis up is the y-value. The height would be from the y-axis over to here. That is an x-value. So we can decide what the radius and height are. This particular thickness, this is an increment of y. Is that correct? We've got little delta y's that we're wrapping around the x-axis. So the fact that we have a delta y or a dy says that if the radius was, when we say it was y, that's good, right? We can leave that y because we're integrating with respect to y. If the height of each one is x, then aren't we going to have to rename x in terms of that function what is x in terms of y? So you can't integrate x things with respect to y. So you can decide what they are and write them down, but that doesn't mean they're going to necessarily hang around when we're actually solving the problem. Keep things around that are okay. The y's are okay. Because we have delta y's or dy's, the x's have to go back to what x's actually are according to the function. Web assigned question there. Katie. I have one, but it's not really like a test question type thing. Okay. It's the frustrum of a pyramid one, the volume of it. Yeah, those are weird. I got the one before it. Okay, here's... Now, who did I went through that? You and I went through that? The cone one. It's a cone and we don't want all the cone, right? I got the cone one, but I can't get the... I did the pyramid one the same way. Okay. So I think we have a figure that looks something like this. And this is capital R. This is little r. That one? No, that's the one that I figured out. Okay, and this was h. All right. So the other one is... This is a pyramid. Okay. And it has a... So that's probably not a test question, but let me just say that whatever it is, what you can do, like here you can look at one of the slices in this fashion. You should be able to, or at least attempt to describe one of the slices of that. Okay, so slice that pyramid up the same way you would slice this up. So you're not going to be able to do the axis of revolution kind of thing with that one. It's just a different kind of problem. But that's not a test type question. But we can set it up when class is over if you want to try to set it up. Now you could generate a picture that's going to generate this figure with a solid of revolution. So you could actually come here with a line that goes through this point, which is right here. So that would be R. And then another line that comes out here to capital R. So really this is a line that serves as the edge of this, right? When you revolve it around. Get the symmetric image over here. Get the equation of this line that serves as the outer edge. And then look at your slices which are solid discs and say well the radius is the, what, X value? Is that correct? Of every point on this line. Now you say, how do I figure the equation of that line? Well, I think this was H. So that's that point over R and up H. Is that point? I think that's correct with the WebAssign diagram. And this is a point that's over capital R and up zero. So you can get the equation of that line by using those two points and find the slope and generate point slope to find the equation. Alright, we are out of time but we did talk briefly today about arc length and the most recent topic which should be the freshest in your mind, I hope. Average value of a function which then led us to find the X value such that when we put that X value, we called it C, into the function, it generated the height or the average height or the average Y value. Alright, see you tomorrow. We'll start a couple minutes before class normally starts if you so choose.