 reminder there's a review session for the exam there's one on Sunday I will film the one on Sunday you can watch it on the web if you can't make it however it's probably very interesting to go because really the best way for a review session to work is for you to ask questions and say I really don't understand or I can't be asinable how to review it rather than tell me everything that I never learned because I don't know anything so watching it is probably less useful than attending but I understand people have all the occasions on the web page for the class I put up old exams from math 132 one actual exam two practicing exams and an exam for math 126 which covers some of the same material the drawback of all of these is the 132 exam we're behind so they all cover volumes which will not be on the exam although we will cover it just before the exam so they all cover volumes of revolutions we will not have for all volumes of revolutions on the exam I think one of them will cover the arc length again if we want to get to arc length before the exam so that means that well and the math 126 exam doesn't have any stuff on areas which we'll do today more does it have anything on improper integral so none of the four or five or however many I put up there samples are perfect but it gives you an idea of what the exams are like and if you look at all of them you will notice that they differ quite a lot the appropriate ones how's that so look on the web page buy your recitation most of the people in this lecture are in ESS over one arrow there but one recitation yes but child's recitation which I think is she's not sitting there today is number seven is number seven is and I think old engineering 143 I think so recitation seven six there is no six and then all the other ones if you're in the other lecture well look on one page other questions okay so I apologize at the end of the last class I put a problem of which the answer was right but you can't do that problem so I meant to put up a different problem so let me put up the here's what I here's where I put up the question was does this converge or diverge and really I meant to put up this problem or maybe I meant to put up this problem you should be able to do both of them do you want me to ask them questions no nobody likes the questions no okay so I will just ask this converges diverge why okay so that's wrong but seems sounds good okay yes you have a 50-50 shot made some argument that we can get anybody not anybody but anybody in this room but me so okay this one actually converges the reason it converges is because this is like one over x square so this is so since sine of x over x square is certainly less than one over x square and it's greater than minus one over x square and the integrals from one to infinity of one over x square converges yeah so the sign is between one and minus one all the time so this is never bigger than one on the top and never less than minus one on the top sometimes it's zero sometimes it's a half sometimes it's 0.06 but it's never bigger than one right okay so a lot of times in calculus when you see a sine or a cosine you should think oh maybe I can replace that with one you can't always but sometimes you can so here this is true what I felt okay it's less than or equal to because for certain values of x the sine is actually one but the sine is always between plus one and minus one and x square is always x quick okay so this this integral if it exists at all is always less than this integral well in absolute value because sometimes it's not but this integral we know converges I did that last time right no yes yeah I did because when you integrate x to the minus two you get negative x to the minus one you take the limit of x to the minus one as x goes to infinity you get zero so we're cool so this guy in it in terms of a picture the guy on top so here's one over x square here's minus one over x square and sine x over x square does something like this this area here we know this line it's one you can do this out more carefully if you like okay so your argument was wrong but it sounded really it gave me you know twice the time okay what about this guy I didn't laugh at him I liked his argument it's just wrong nobody's going to tell me diverged well yeah because the other one converges so jeez why would I put you in the okay yeah why is it diverged one over x diverges and one plus sine square of x is always bigger than one so by the same reasoning about similar reasoning one plus sine square of x well you take it you take it over you square it is always positive this is certainly bigger than one for all x's and x is always the same as x so we have that and this integral which also we did this diverges because of what he said this is a log log of infinity is infinity is bigger this is too bad too big and we have something bigger and since this is bigger and that gets as big as you like this gets as big as you like I know this comparison theorem is a little tricky for people it's a good skill to learn be able to look at things and compare them to other things that you know of course you don't necessarily know what you know yet but when you do it enough you'll have a better feel but in general you know one over x to a power there was a homework problem that asked about that and we're going to infinity and the power is bigger than one it converges we're going to be zero and the power is less than one so yeah discuss well it isn't obvious from the picture here you go I mean I can't really draw the picture how it diverges because it looks like it should and in fact if you sit down with a calculator or a computer and you start numerically integrating one over x you're going to get a number because it diverges so slowly that you won't see it but it does diverge because it's like the log and the log grows to infinity but the log grows very slowly the log is an exponent so it grows because it grows slowly I mean the log of a thousand is not that big an over and if we're doing base ten the log of a thousand is three that's not a very big number and from a thousand to a million we're doing base ten base the similar thousand to a million grows from three to six you can't just look at the graph you can look at the graph and draw conclusions but conclusions are warranted so here I can look at the graph of this and compare it to the graph of this I know that one over x is too big and I know that this guy one plus sine squared over x does something I'm going to do up here it touches but it does something up here so this is too big certainly that's even bigger so that really the comparison theorem is where you look at the picture but you just look at the integral and say for sure it diverges you know if I give you a penny every day for a billion years you'll have a lot of money well maybe a trillion here but you know if I always give you a penny every day your amount of money is unbounded but it takes so long that you can't really tell depends on whether we're going to zero or infinity so there's a little more problem like this which asks I forget whether the homework is to zero or to infinity because it's like the last one I don't know if it's on it should have done it because it's zero it's something like find the value for the piece in which this diverges or maybe it's one to infinity is it infinity or zero it doesn't matter okay so this guy for what values of people this diverges so I mean so part of the point of homework is you're supposed to do that and sort of remember how it worked seems nobody remembered how did this work or they didn't do it one or the other yes p less than one why because when you do at the integral if p is less than one then the x goes to the top and so that means that this is zero but all of these one over x to the p's look like that you can't tell just by looking and this on zero one to infinity dx over let's put a q here to differentiate it this guy converges q bigger than one for the same reason q and when q equals one it diverges in both places I mean you did just do this unless you did do your homework and maintain it okay so let me move along to the next topic which is so you are now supposedly experts on improper integrals and techniques of integration let's find out next week whether that's true and then the next topic that we're doing is returning to the idea of finding let's say suppose I have two curves here like y equals x and of course since I've looked at y equals at one minus one no that's not the right one can I do this how about four minus x squared let's just do this one four minus x squared looks like that and what I want here is this area this is not a hard problem because you just have to view it properly this is an integral and in order to find the area between the curves you just subtract the bottom curve from the top curve so the area between is just going to be well I mean let's do it a little formally if we make a little rectangle here it's height because remember when we're integrating we're adding up the areas of little rectangles it's height is going to be from here to here so this is four minus x squared minus x so the height of my rectangles is four minus x squared minus x top curve minus the bottom curve and so the area integral from somewhere to somewhere of four minus x squared minus x the x now we have to figure out where the somewhere and the somewhere are there's a well-defined area here I'm not telling you where they cross but you've got to figure it out we have to find this point in order to do this problem we have to find those two points so how do we find those two points right? we wait for me to do it so how do we find those two points? okay right instead of equal to each other so we need to know so that means that we can rewrite this as let me bring it over there because I like when we need it x squared plus x times four so I need to solve that well that doesn't factor nice but that's okay I know the quadratic formula is negative one plus or minus the square root of one minus four c over so it's two eighty numbers what are numbers? negative one plus or minus the square root of seventeen over two that's where they cross nothing wrong with the square root of seventeen we'll do a non-music calculator just like that and so now we know where they cross this is where x is this one is negative one minus the square root of seventeen over two and this one is negative one plus the square root of seventeen over two okay so that means that the interval we want to do is from negative one minus root of seventeen over two is negative one plus root of seventeen over two of that and that's easy that is four x minus one third x cubed minus minus x squared plus two evaluated from block to block to block to block which is which is four times negative one plus root of seventeen over two minus one third negative one plus root of seventeen over two minus or minus one minus root of seventeen these numbers are ugly is there a third symbol? because it's hard to see around this it's easier for you to see and then you simplify this and you get a number and I'm sorry I chose to be here which I can simplify if you really insist but I don't think you're so big because if you're watching me do okay so that's a pretty easy idea you can have two curves to cross one another to find the area between them you just subtract notice that you can crank this idea up a little bit sometimes let me just give an example that it works I want that one okay and four x squared minus three take the next one square x squared terrible having to shorten that one y squared you want to find the area between those two guys this is y squared so this is a parabola of volume of cyclics right so if we look at the picture here y equals x minus one works like this y squared equals two x plus six this is the same as saying y squared minus six two equals x so if it makes you unhappy so either I can do that or if I prefer I can say y is plus or minus square root of two x plus six it's the same to me okay so when two x plus six is so here this is a parabola well this is the top ten sorry either one they're the same to me so this guy is zero when x is negative three and so is this guy so this here at negative three looks like this I mean it's not level and then this piece so that's this piece and then this piece like that and we're looking for the area here now let's think about so let me not do it this way let me start it this way the way wherever he is there you are you suggested even though we didn't know this before so let me do it this way this isn't a hard way why is this way hard number one square root of suck but number two let's look at this so the area is the top curve minus the bottom curve the answer top curve here is always this guy let me write it up here y equals plus square root of two x minus six and the bottom curve is not always x minus one because it changes right here so in this region I have a different bottom curve that I have in this region so I can set this up let me not even find the points yet as the integral from negative three to some point here let me just call it a top curve is the positive square root bottom curve is the negative square root so it's minus the minus so that's this area and then the other area plus the integral from a to 20 I just didn't even find yet hb of the top curve is still x square root and the bottom curve is x minus one so that's certainly represents the area the intervals aren't too bad I can do those intervals they're both easy substitutions I can find the points a and b without too much trouble but I have to do two intervals I have to put my brain to think about it you want me to do this? no? so you can do this it's fine, you'll get the right answer let's find a and b well let's find a and b anything? so a is where is where the negative square root well negative square root 2x plus 6 equals x minus one that's going to get a and b is where x minus one plus the square root 2x from being unable to read they're all pluses I just can't read they're pluses okay so I can find these points a and b that's the area this is the sucky way to do this problem I mean it's okay but we can also do it another way we can just lay down lay down so sometimes it's good to be lazy of course we lay down on the other side on the other side if we look at this problem if we look at this problem from the side notice that these guys always have the same curves from here to here if we look at just the y values x minus one is always bigger than the parabola you have to lay down going up this is increasing so if you slice it sideways instead of up and down this curve the right curve so the right curve is x minus one so x is y plus one x equals y plus one and the curve on the left is I wrote it down and erased it x is y squared minus six all over two and this area is the integral from y plus one minus one-half y squared minus six d y we already used a and b so let's use we integrate from c with the y values we want to find the y values so y plus one equals one-half y squared minus six so part of this and in fact you should be thinking this way this is part of why it's useful to think about the little rectangles here my rectangles are laying down and equal this way makes my life easier to cut sideways to cut the curve I guess I need to solve this so that's the same thing because y plus two is y squared minus six which is the same thing y squared minus two y minus eight so that means that my integral is four how we're slicing up the object and if it's more convenient to slice it horizontally go ahead in fact if it's more convenient to slice it whichever way it's more convenient to slice it now sometimes maybe the curves cross three times does anybody want me to finish this integral I'm happy to do it if somebody wants me to do it yes you do she wants me to do it so this is do the algebra here this is minus two to four minus one half y squared plus three plus one is four plus y d y which is minus one to six y two plus four y plus one half y squared the value is negative two to four which is minus one six four cubed is 64 four is 16 does that mean it's a minus sign somewhere no and half of sixteen is eight and then we subtract off negative two cubed is eight so this is eight over six negative two times four is eight negative two squared is plus four and then you add let me point out this is not a class in arithmetic so if you have things like one half plus five twelve plus three nineteen on the test you can state it that way care that you get the one half plus five twelve plus three nineteen and whether you can add fractions or not really my concern I can't I mean I can but I don't want to so my concern is that you understand the material well enough to get an answer that does not mean that you cannot that you could just leave something like the sign of pot as the sign of pot you should know that the sign of pot is zero so I need you to do the obvious and I would prefer that you know you not leave eight over four is eight over four you should know that eight over four is two but if you don't want to add five twelve and forty seven nineteen's I can understand why not okay so this is some stuff and I'm gonna leave it to you to do that I guess I have time for another one um I think we're done I have to write something down this is on the midterm and unfortunately I was so worried about all this technology garbage that I left my notes sitting on my desk and I can't remember the problem that I would tell okay so notice that sometimes maybe your curves cross more than once so sometimes your curves cross more than once in which case right it's always important to remember that your area is the top curve minus the bottom curve so in this case let's just call this f call this g this point is zero and g dx plus from zero to a plus the integral from a g minus f this is really what you did if your area is under curves it's just that this is now pulled up it's the same thing so let me do another so say I want to do this one what curve is on top so let's call this one f why do you wrote this? this is a good sign if you just look at zero and zero this is zero and this is negative two that's a hint so y equals x squared minus two and the absolute value looks like a b this is f now both of these are even functions so I can do this in two ways I can either integrate from here to here and then from here to here or I can just realize that the area over here is exactly equal to the area over there so I can just do this half and double either way is okay so what did these cross? so I guess one thing let me point out remind you the absolute value of x is x if x is positive and negative x if x is negative this is another way to read this if what's inside the absolute value is positive, leave it alone it's negative and change the sum so students mess this up all the time don't panic when you see absolute values just remember that they change whenever what's inside changes and so here we need to know length is x equal x squared times two that will tell me the center section points over here so that is x squared times x times two equals zero square root here so that's right so x minus two times x plus one okay good that makes it easier and so that happens when x is two or x is negative one but the negative one one I don't want that's because I only looked when x is equal and also if I looked on the other side I would get negative two so this is the one I want that means that this area is the integral of zero to two of x times x squared plus two dx and then I can double it because it's the same as the integral from negative two times to zero and that's probably a good place to stop before you pack up and leave I'm going to remind you some things we have a holiday now today is Friday the calendar may say Wednesday but today is Friday so if you do a recitation on Wednesday no you don't today if you do a recitation on Friday yes you do paper homework is due after the break there's another web assigned view we're going to have seven problems so this is stuff we're going to have a new session